PLZZZZ HELPPPPPPPPPppppp​

PLZZZZ HELPPPPPPPPPppppp

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Answer 1
What grade are you in?!!??

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Jane, looking for Tarzan, is running at top speed (6.0 m/s ) and grabs a vine hanging vertically from a tall tree in the jungle.
How high can she swing upward?
Express your answer to two significant figures and include the appropriate units.
Does the length of the vine affect your answer?

Answers

(a) The maximum height reached by Jane is 1.8 m.

(b) The length of the vine will affect the time of her motion, which will impact on speed and maximum height attained.

Maximum height Jane can swing

apply the principle of conservation of energy;

P.E = K.E

mgh = ¹/₂mv²

h = v²/2g

where;

v is speed of janeg is acceleration due to gravity

h = (6²)/(2 x 9.8)

h = 1.84 m

Time of motion of Jane

Assuming Jane to be in simple harmonic motion, the time of motion is calculated as;

T =  2π√(L/g)

where;

L is the length of the vine

Thus, the length of the vine will affect the time of her motion, which will impact on speed and maximum height attained.

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In physics, when a baseball player catches a ball, which one of newtons laws is it an example of?

A. 1st law

B. 2nd law

C. 3rd law

Answers

Its C newtons 3rd law, because both the ball and the player are exerting a force. The ball is exerting a force on the player and the player is exerting a force to bring the ball to a rest or a state where it isn’t moving.

A cart of mass 0.5 kg sits at rest on a table on which it can roll without friction. It is attached to an unstretched spring. You give the mass a push with a constant force over a distance of 5 cm in the direction that compresses the spring, after which the mass starts undergoing simple harmonic motion with a frequency of 0.5 complete oscillations per second and an amplitude of 15 cm.
A) What is the spring constant of the spring?
B) How fast was the cart moving at the instant when you finished pushing it?
C) What force did you exert on the cart?

Answers

(A) The spring constant of the spring is 4.94 N.

(B) The speed of the cart after pushing it is 0.47 m/s.

(C) The force applied to the cart is 0.75 N.

Spring constant

ω = √k/m

where;

ω is angular frequencyk is spring constantm is mass

0.5 rev/s = 0.5(2π) rad/s = π rad/s = 3.142 rad/s

ω² = k/m

k = mω²

k = 0.5 x (3.142)²

k = 4.94 N/m

Energy stored in the spring

E = ¹/₂kA²

where;

A is amplitude

E = ¹/₂(4.94)(0.15)²

E = 0.056 J

Speed of the cart

E = ¹/₂mv²

2E = mv²

v² = 2E/m

v² = (2 x 0.056)/(0.5)

v² = 0.224

v = √0.224

v = 0.47 m/s

Force exerted on the cart

E = ¹/₂FA

2E = FA

F = 2E/A

F = (2 x 0.056)/(0.15)

F = 0.75 N

Thus, the spring constant of the spring is 4.94 N. The speed of the cart after pushing it is 0.47 m/s. The force applied to the cart is 0.75 N.

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For light of wavelength 589 nm, calculate the critical angles for the following substances when surrounded by air.
polystyrene

flint glass

Answers

(1) The critical angles for the air is 90⁰,

(2) The critical angles for the polystyrene is 39.1 ⁰ and

(3) The critical angles for the flint glass is 37.2 ⁰.

Critical angles for the different medium

θc = sin⁻¹( 1/η)

where;

η is the refractive index

Refractive index of air = 1

Refractive index of polystyrene = 1.5865

Refractive index of flint glass = 1.655

Critical angles for air

θc = sin⁻¹( 1/1)

θc = 90⁰

Critical angles for polystyrene

θc = sin⁻¹( 1/1.5865)

θc =  39.1 ⁰

Critical angles for flint glass

θc = sin⁻¹( 1/1.655)

θc = 37.2 ⁰

Thus, the critical angles for the air is 90⁰, the critical angles for the polystyrene is 39.1 ⁰ and the critical angles for the flint glass is 37.2 ⁰.

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An ideal toroidal solenoid (Figure 1) has inner radius r1 = 15.1 cm and outer radius r2 = 18.3 cm. The solenoid has 180 turns and carries a current of 8.40 A.
a) What is the magnitude of the magnetic field at 12.6 cm from the center of the torus?
b) What is the magnitude of the magnetic field at 16.3 cm from the center of the torus?
c) What is the magnitude of the magnetic field at 20.7 cm from the center of the torus?

Answers

The solution for the three questions  is mathematically given as

Parts A and C are both zeros.

Part B B = 0.001855Tesla

What is the magnitude of the magnetic field at 12.6 cm from the center of the torus?

Parts A and C are both zeros.

For component A, the magnetic field is zero since 12.6 cm is still inside the toroidal solenoid. Part C has no magnetic field since it is 20.7 cm outside of the toroidal solenoid.

Generally, the equation for  magnetic field is  mathematically given as

B = (mu_0*N*I)/(2*pi*r)

Therefore

B = ((4*pi*10^{-7})*180*8.40)/(2*pi*0.163)

B = 0.001855Tesla

In conclusion, the magnitude of the magnetic field at 16.3 cm from the center of the torus is

B = 0.001855Tesla

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Answer:

Part A & C: 0 T

Part B: 1.855*10^-3 T

Explanation:

The formula that models the magnetic field of a toroid is B=μ0*N*I/2π*r .

Note: Toroids keep their magnetic field rotating within the coils that carry current.

Part A & C: Thus the B field magnitude 12.6 cm & 20.7 cm away from the center is 0 T.

Part B: [tex]B=\frac{(4\pi *10x^{-7} )(180)(8.4 A)}{(2\pi )(16.3*10x^{-2} m)}[/tex]

B=1.855*10^-3 T

An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.

a) Find the magnitude of the magnetic field this electron produces at the point A .
b) Find the magnitude of the magnetic field this electron produces at the point B .
c) Find the magnitude of the magnetic field this electron produces at the point C .
d) Find the magnitude of the magnetic field this electron produces at the point D

Answers

Hi there!

We can use Biot-Savart's Law for a moving particle:
[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }[/tex]

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }[/tex]

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

[tex]B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }[/tex]

[tex]B = \boxed{7.07 *10^{-10} T}[/tex]

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

c)

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

[tex]B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}[/tex]

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

[tex]\boxed{B = 0 T}[/tex]

The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×10^3 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10^-11 Nm2/kg2.)
2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×10^5 m around the planet Glob. Calculate the speed of the satellite.

Answers

(a) The initial speed of the rock as it left the astronaut's hand is 19.35 m/s.

(b) The speed of the satellite is 50.24 m/s.

Acceleration due to gravity of the planet

g = GM/R²

where;

M is mass of the planetR is radius of the planet

g = (6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(63200)²

g = 0.13 m/s²

Initial speed of the rock

v² = u² - 2gh

where;

v is final velocityu is initial velocity

at maximum height, v = 0

u² = 2gh

u = √2gh

u = √(2 x 0.13 x 1,440)

u = 19.35 m/s

Speed of the satellite

v = √GM/r

M is mass of the planet Globr is the total distance from the center of the planet Glob

r = radius of planet Glob + radius of the satellite

r = 63200 m + 145,000 m = 208,200 m

v = √[(6.67 x 10⁻¹¹ x 7.88 x 10¹⁸)/(208,200)]

v = 50.24 m/s

Thus, the initial speed of the rock as it left the astronaut's hand is 19.35 m/s.

The speed of the satellite is 50.24 m/s.

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in the absence of friction, the output power of a winding engine is 100kw but thus is reduced by friction to 90kw . how much oil initially at 120° is required per second to to keep the Temperature of the bearing down to 70°C ? specific heat capacity of oil is 2100 j/kg°C.

please I need it now Bosses ‍♀️​

Answers

Answer:

Explanation:

The angular speed of the rotor is 200 rad/s.

The torque needed to be transmitted by the engine is 180 Nm.

The power of the rotor required to transmit energy to apply a torque τ to rotate a motor with angular speed ω,

P=τω

=180×200W

=36kW

What is the speed of light while traveling through (a) a vacuum, (b) air at 30°C, and (c) air at 0°C?

The speed of light through a transparent substance is 2.00 × 108 m/s. What is the substance?

The wavelength of light from a monochromatic source is measured to be 6.80 × 10−7 m. (a) What is the frequency of this light? (b) What color would you observe?

What is the energy of a photon of red light with a frequency of 4.3 × 1014 Hz?

Answers

(1) The speed of light through vacuum is 3 x 10⁸ m/s.

(2) The speed of light in air at 30⁰C is 350 m/s.

(3) The speed of light in air at 0⁰C is 330 m/s.

(4) The transparent substance is glass.

(5) The color of light observed is red.

(6) The energy of a photon of red light is 2.85 x 10⁻¹⁹ J.

Speed of light

The speed of light through vacuum is 3 x 10⁸ m/s.

The speed of light in air at 30⁰C is 350 m/s.

The speed of light in air at 0⁰C is 330 m/s

Refractive index of the substance

n = speed of light in vacuum/speed of light in the substance

n = ( 3 x 10⁸ m/s) / ( 2 x 10⁸ m/s)

n = 1.5

The refractive index of glass is 1.5, thus, the transparent substance is glass.

Frequency of light

f = v/λ

f = (3 x 10⁸) / (6.8 x 10⁻⁷)

f = 4.41 x 10¹⁴ Hz

The color of light observed is red based on the frequency and wavelength.

Energy of the photon

E = hf

E = (6.626 x 10⁻³⁴)(4.3 x 10¹⁴)

E = 2.85 x 10⁻¹⁹ J

Thus, the speed of light through vacuum is 3 x 10⁸ m/s.

The speed of light in air at 30⁰C is 350 m/s.

The speed of light in air at 0⁰C is 330 m/s.

The transparent substance is glass.

The color of light observed is red.

The energy of a photon of red light is 2.85 x 10⁻¹⁹ J.

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An object with a mass of m = 3.85 kg is suspended at rest between the ceiling and the floor by two thin vertical ropes.
The magnitude of the tension in the lower rope is 12.8 N. Calculate the magnitude of the tension in the upper rope.

Answers

The tension in the upper rope is 50.53 N.

Tension in the upper rope

The tension in the upper rope is calculated as follows;

T(up) = T(dn) + mg

where;

T(dn) is the tension in the lower ropemg is the weight of the object

T(up) = 12.8 N + (3.85 x 9.8) N

T(up) = 50.53 N

Thus, the tension in the upper rope is 50.53 N.

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What is the approximate uncertainty in the area of a circle of radius 4.3×104cm ?

Answers

The approximate uncertainty in the area of a circle is 5.4%.

Area of the circle

A = πr²

where;

r is radius of the circle

A = π(4.3 x 10⁴)²

A = 5.81 x 10⁹ cm²

Let the error in measurement = 1 x 10⁴ cm

Error in Area measurement

A =  π(1 x 10⁴)² = 3.14 x 10⁸ cm²

Uncertainty in measurement

% = (error/actual area) x 100%

% = ( 3.14 x 10⁸) / (5.81 x 10⁹) x 100%

% = 5.4 %

Thus, the approximate uncertainty in the area of a circle is 5.4%.

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The image shows a wheel that’s wound up and released. The wheel moves up and down as shown. Identify the position of the wheel when its potential energy is greatest.

Answers

The highest point of the wheel is the position of the wheel when its potential energy is greatest.

At what position of the wheel potential energy is greatest?

The position of the wheel when its potential energy is greatest when it is at the highest point because potential energy depends on the height of an object. If the object is at more height then it has more potential energy and vice versa.

So we can conclude that the highest point of the wheel is the position of the wheel when its potential energy is greatest.

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When a scientific calculator shows the quantity below, what does it mean?
1.5E8

Answers

1.5 with 8 exponent of 10.
1.5 x 100000000
150000000

an 8 kilogram ball is fired horizontally form a 1 times 10^3 kilogram cannon initially at rest after having been fired the momentum of the ball is 2.4 time 10^3 kilogram meters per second east calculate the magnitude of the cannons velocity after the ball is fired

Answers

The magnitude of the cannons velocity after the ball is fired will be 2.4m/s.

To find the answer, we have to know more about the law of conservation of linear momentum.

How to find the magnitude of the cannons velocity after the ball is fired?The law of conservation of linear momentum states that, the initial momentum of a system will be equal to the final momentum.Here, then initial momentum of the system will be equal to zero, since the initial velocity of the cannon is equal to zero.Given that,

                      [tex]M_B=8kg\\M_C=1*10^3 kg\\P_f=2.4*10^3 kgm/s.\\[/tex]

When the ball is fired from the canon, then the cannon will have a recoil velocity in the opposite direction of motion of the ball.Since it has a final momentum towards east, the recoil momentum will be in the west.Thus, the velocity of the cannon after when the ball is fired will be,

                   [tex]P_f=M_CV_C\\V_C=\frac{P_f}{M_C}=\frac{2.4*10^3}{1*10^3} =2.4m/s \\west[/tex]

Thus, we can conclude that, the magnitude of the cannons velocity after the ball is fired is 2.4m/s.

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After the ball is shot, the cannon will move at a speed of 2.4 m/s.

We must learn more about the rule of conservation of linear momentum in order to locate the solution.

How can I determine the cannon's post-ball velocity magnitude?

According to the law of conservation of linear momentum, a system's starting and ultimate velocities are equal.Since the cannon's initial velocity is 0 in this case, the system's initial momentum will be equal to zero as well.We have,

                          [tex]m=8kg\\M=1000kg\\P_f=2.4*10^3kgm/s[/tex]

When a cannon fires a ball, the cannon will recoil quickly and in the opposite direction of the ball's motion.Given that it is now moving eastward, the recoil momentum is towards the west.As a result, when the ball is fired, the cannon's velocity will be,

                               [tex]P_f=MV\\V=\frac{P_f}{M} =2.4m/s[/tex]

Thus, we can infer that the cannon's maximum speed once the ball is fired is 2.4 m/s.

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. What is the potential energy of a 0.40 kg ball at a height of 9.2 m?​

Answers

Explanation:

gravitational potential energy = mgh (must be in S.I. unit)

m= 0.4 kg ; g= 10m/s (gravitational acceleration occurs); h=9.2 m

hence mgh=0.4×10×9.2= 36.8J

unit for energy is joules and since the variables are in S.I. unit, we can use Joules as the final unit for measurement

Answer:

36.06 J if you define g = 9.8 m/s² or 36.8 J if you define g = 10 m/s²

Explanation:

Potential Energy = Mass x Gravitational Acceleration x Height

It can be expressed as [tex]\displaystyle{PE = mgh}[/tex] where PE is potential energy, m is mass, g is gravitational acceleration and h is height.

In this case, we know mass is 0.40 kg and height of 9.2 m as well as gravitational acceleration is defined to be 9.8 m/s² (You can also define g = 10 m/s²)

Therefore, substitute given information in formula:

[tex]\displaystyle{PE=0.40 \ \times \ 9.8 \ \times 9.2 }\\\\\displaystyle{PE=36.06 \ J}[/tex]

With g = 10 m/s², you'll get:

[tex]\displaystyle{PE = 0.40 \ \times 10 \ \times 9.2}\\\\\displaystyle{PE = 36.8 \ J}[/tex]

Note that J is for joule unit.

Therefore, the answer is 36.06 J if you define g = 9.8 m/s² or 36.8 J if you define g = 10 m/s² - both work.

what is the wavelength of a tubas A2(110hz) tone in air at standard conditions/

Answers

The wavelength of A2 with frequency (110hz) is 3.01 m.

What is the wavelength of a sound wave?

Wavelength of a sound wave is the distance between successive similar points in the wave such as rarefactions or compressions.

A tuba is a musical instrument that produces sound waves.

Wavelength is related to frequency and velocity by the formula below:

Wavelength = velocity/frequency

The wavelength of A2 with frequency (110hz) is 3.01 m.

In conclusion, the wavelength of a wave is inversely proportional to frequency.

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A 21.0 kg uniform beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
If the angle between the beam and the cable is θ = 66.0° what is the tension in the cable?

Answers

Since upward forces must be equal to the downward forces, the tension in the cable is 225.3 N

What are the two conditions for equilibrium ?

The two conditions are;

Sum of the upward forces must be equal to the sum of the downward forces.The sum of the clockwise moment must be equal to the sum of the anticlockwise moment.

The given parameters are;

Mass m = 21 kgangle θ = 66°

Tsinθ = mg

Tsin 66 = 21 x 9.8

T = 205.8 / 0.914

T = 225.3 N

Therefore, the tension in the cable is 225.3 N

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Radiation from the sun hits Earth unequally and is absorbed by different materials in varying amounts. This is called
A. the Coriolis Effect
B. a thermocline
OC. El Nino Southern Oscillation
OD. differential heating

Answers

Radiation from the sun hits earth unequally and is absorbed by different materials in varying amounts. This is called differential heating.

What is differential heating?

Differential heating is the property that causes different surfaces to heat up and cool down at different rates. The earth's surface receives different magnitudes of solar radiation and also the earth's surfaces absorb thermal energy in different magnitudes.

The color, shape, texture, surface and presence of constructions can influence the heating or cooling of the earth.

The earth in the equator line is heated more by solar action than that of the poles, since it receives more amount of radiation per unit area.

In general, dry surfaces heat up and cool down faster than wet ones.

Therefore, we can confirm that when radiation from the sun hits earth unequally and is absorbed by different materials in varying amounts is called differential heating.

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A force of 100 newtons is applied to a box at an angle of 36° with the horizontal. If the mass of the box is 25 kilograms, what is the horizontal
acceleration of the box?
OA
1.52 meters/second²
OB. 3.24 meters/second²
OC. 5.48 meters/second²
O D.
6.87 meters/second²
OE
7.15 meters/second²

Answers

Answer:

See below

Explanation:

I will assume the force is in a DOWNWARD direction ( I believe it makes no answer difference)

Horizontal component is then 100 cos 36° =80.9 N

F = ma

80.9 = 25 kg  *a

a = 3.24 m/s^2

Answer:

See image

Explanation:

Plato

Motorcycle safety helmet extend the time of collision hence decreasing,
a:chance of collision
b:force acting
c: velocity
c:Impulse

Answers

Answer:

D. Impulse

Explanation: Hope this helps

White light is spread out into its spectral components by a diffraction grating. If the grating has 1975 lines per centimeter, at what angle does red light of wavelength 640 nm appear in first-order spectrum? (Assume that the light is incident normally on the grating.)

Answers

The angle of the red light is mathematically given as

[tex]\theta = 7.24 \textdegree[/tex]

What angle does red light of wavelength 640 nm appear in the first-order spectrum?

Generally, the equation for the grating element is  mathematically given as

d= 1 / N

Therefore

d= 1/1965

d= 5.089 * 10^{-6} m

Generally, the equation for the  differential formula is  mathematically given as

[tex]d sin \theta = m\lambda[/tex]

Therefore

[tex]sin \theta = \lambda / d[/tex]

[tex]sin \theta= (640 * 10 ^ {-9} m)/(5.089 * 10 ^ {-6} m)[/tex]

[tex]\theta = 7.24 \textdegree[/tex]

In conclusion, The angle of the red light

[tex]\theta = 7.24 \textdegree[/tex]

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. A bargain hunter purchases a "gold" crown at a flea market. After she gets home, she hangs it from a scale and finds its weight to be 7.84 N. She then weighs the crown while it is immersed in water, as shown below in Figure, and now the scale reads 6.86 N. Is the crown made of pure gold? Find the density of the crown and compare it to the density of gold. wag PODECID LE B water​

Answers

Answer:

Average density of the crown: approximately [tex]8\; {\rm g \cdot mL^{-1}}[/tex].

Hence, if this crown contains no empty space, this crown is not made of pure gold.

Explanation:

Let [tex]m(\text{crown})[/tex] and [tex]V(\text{crown})[/tex] denote the mass and volume of this crown. Let [tex]g[/tex] denote the gravitational field strength.

Since this crown is fully immersed in water, the volume of water displaced [tex]V(\text{water, displaced})[/tex] is equal to the volume of this crown:

[tex]V(\text{water, displaced}) = V(\text{crown})[/tex].

The mass of water displaced would be:

[tex]\begin{aligned}m(\text{water, displaced}) &= \rho(\text{water}) \, V(\text{water, displaced}) \\ &= \rho(\text{water}) \, V(\text{crown})\end{aligned}[/tex].

The weight of water displaced would be [tex]m(\text{water, displaced})\, g = \rho(\text{water}) \, V(\text{crown})\, g\end{aligned}[/tex].

The buoyancy force on this crown is equal to the weight of water that this crown displaced:

[tex]F(\text{buoyancy}) = \rho(\text{water}) \, V(\text{crown})\, g[/tex].

The magnitude of this buoyancy force is [tex]7.84\; {\rm N} - 6.86\; {\rm N} = 0.98\; {\rm N}[/tex]. Rearrange the equation for buoyancy to find [tex]V(\text{crown})[/tex]:

[tex]\begin{aligned} V(\text{crown}) &= \frac{F(\text{buoyancy}) }{\rho(\text{water}) \, g}\end{aligned}[/tex].

Since the weight of this crown is [tex]\text{weight}(\text{crown}) = m(\text{crown})\, g[/tex], the mass of this crown would be [tex]m(\text{crown})= \text{weight}(\text{crown}) / g[/tex].

The average density of this crown would be:

[tex]\begin{aligned}\rho(\text{crown}) &= \frac{m(\text{crown})}{V(\text{crown})} \\ &= \frac{\text{weight}(\text{crown}) / g}{F(\text{buoyancy}) / (\rho(\text{water})\, g)} \\ &= \frac{\text{weight}(\text{crown})}{F(\text{buoyancy})}\, \rho(\text{water}) \\ &= \frac{7.84\; {\rm N}}{0.98\; {\rm N}}\times 1.000 \; {\rm g\cdot mL^{-1}} \\ &= 8.0\; {\rm g \cdot mL^{-1}}\end{aligned}[/tex].

The density of pure gold is significantly higher than [tex]8.0\; {\rm g\cdot mL^{-1}}[/tex]. Hence, if this crown contains no empty space (i.e., no air bubble within the crown), the crown would not be made of pure gold.

using a weston cadmium cell of 1.0283 v and a standard resistance of .1ohm a potentiometer was adjusted so that 1.0183 m was equivalent to the emf of the cell: when a certain direct current was flowing through the standard resistance, the voltage across it corresponds to 150 cm. what was the value of current​

Answers

Answer:

Explanation:

the value is -8 cm

A 72.5-kg hiker starts at an elevation of 1230 m and climbs to the top of a peak 2660 m high.
What is the hiker's change in potential energy?
Express your answer to three significant figures and include the appropriate units.
What is the minimum work required of the hiker?
Express your answer to three significant figures and include the appropriate units.

Answers

Change in potential energy = 1.02 * 10⁶ J

Minimum work required of the hiker = 1.89 * 10⁶ J

What is the change in potential energy of the hiker?

The potential energy of a body is calculated as follows:

Potential energy = mgh

Change in potential energy = Final PE - Initial PE

Change in potential energy = mg(H - h)

Change in potential energy = 72.5 * 9.81 * (2660 - 1230)

Change in potential energy = 1.02 * 10⁶ J

The minimum work required of the hiker is the potential energy at the highest point.

Minimum work = mgH

Minimum work required of the hiker = 1.89 * 10⁶ J

In conclusion, potential energy is energy due to state or position of a body.

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How do I connect with my higher self?​

Answers

Answer:

Create space

Watch your breath

Watch your thoughts

Be gentle with yourself

Affirm what you want

Explanation:

Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude of the tension in the string labeled C is 56.3 N. Calculate the magnitude of the tension in the string marked A.

Answers

The magnitude of the tension in the string marked A is 39.5 N.

What is the tension in A?

The tension in A is determined thus:

The angle at A, θ = tan⁻¹(3/8) = 20.56

When extrapolated below negative x, the angle at B, α = tan⁻¹(5/4) = 51.34

When extrapolated below negative x, the angle at C, β = tan⁻¹(1/6) = 9.46

Taking the horizontal components of tension;

56.3cos(9.46) = A * cos(20.56) + B * cos(51.34)

0.6247B= 55.53 - 0.936A

B = (55.53 - 0.936A)/0.6247 ----(1)

Taking the vertical components of tension;

56.3 * sin(9.46) + A * sin(20.6) = B * sin(51.3)

9.25 + 0.35A = 0.78B  ---- (2)

substitute the value (1)  in (2)

9.25 + 0.35A = 0.78{(55.53 - 0.936A)/0.6247}

(9.25 + 0.35A) * 0.6247 = 43.31 - 0.73A

0.22A + 0.73A = 43.31 - 5.78

0.93A = 37.53

A = 39.5 N

In conclusion, the tension in A is determined by solving for the vertical and horizontal components of tension.

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The head of a hammer with a mass of 1.5 kg is allowed to fall onto a nail from a height of 0.70 m .
What is the maximum amount of work it could do on the nail?
Express your answer to two significant figures and include the appropriate units.
Why do people not just "let it fall" but add their own force to the hammer as it falls?
a)To make the trajectory of the fall more precise.
b)To increase the power of the hammer.

Answers

Maximum work done by the hammer on the is 10 J.

Force is added to the hammer in order to increase the power of the hammer; option B.

What is work done?

Work done is defined as the product of force and the distance travelled by force.

Energy is used to do work.

The potential energy of the hammer is converted to work done.

Potential energy = mgh

The maximum amount of work it could do on the nail is given below:

Maximum work = 1.5 * 9.81 * 0.7

Maximum work = 10 J

Force is added to the hammer in order to increase the power of the hammer.

In conclusion, the work done by the hammer is obtained from the potential energy of the hammer.

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A proton is moved from the negative to the positive plate of a parallel-plate arrangement. The plates are 1.50 cm apart, and the electric field is uniform with a magnitude of 1 500 N/C.
What is the proton’s potential energy change?
What is the potential difference between the plates?
What is the potential difference between the negative plate and a point midway between the plates?
If the proton is released from rest at the positive plate, what speed will it have just before it hits the negative plate?

Answers

(a) The proton’s potential energy change is 3.6 x 10⁻¹⁸ J.

(b) The potential difference between the negative plate and a point midway between the plates is 11.25 V.

(c) The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.

Potential energy of the proton

U = qΔV

where;

q is charge of the protonΔV is potential difference

U = q(Ed)

U = (1.6 x 10⁻¹⁹)(1500 x 1.5 x 10⁻²)

U = 3.6 x 10⁻¹⁸ J

Potential difference between the negative plate and a point midway

ΔV = E(0.5d)

ΔV = 0.5Ed

ΔV = 0.5 (1500)(1.5 x 10⁻²)

ΔV = 11.25 V

Speed of the proton

U = ¹/₂mv²

U = mv²

v² = 2U/m

where;

m is mass of proton = 1.67 x 10⁻²⁷ kg

v² = (2 x 3.6 x 10⁻¹⁸) / ( 1.67 x 10⁻²⁷)

v² = 4.311 x 10⁹

v = √(4.311 x 10⁹)

v = 6.57 x 10⁴ m/s

Thus, the proton’s potential energy change is 3.6 x 10⁻¹⁸ J.

The potential difference between the negative plate and a point midway between the plates is 11.25 V.

The speed of the proton just before it hits the negative plate is 6.57 x 10⁴ m/s.

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A uniform beam of length L = 2.2 m and mass M = 49 kg has its lower end fixed to pivot at a point P on the floor, making an angle θ = 18° as shown in the diagram. A horizontal cable is attached at its upper end B to a point A on a wall. A box of the same mass M as the beam is suspended from a rope that is attached to the beam one-fourth L from its upper end.

What is the y-component Py of the force, in newtons, exerted by the pivot on the beam?

Write an expression for the tension T in the horizontal cable AB.

What is the x-component Px of the force, in newtons, exerted by the pivot on the beam?

I cannot figure part C

Answers

Hi there!

Part A.

To solve this part, all we need to do is a summation of vertical forces.

We have the following acting on the beam :
- Force of gravity (Fg, down)
- Force of tension from the rope holding the box (T, down)
- Force exerted by pivot (Py, up)

These sum to zero because the beam is not accelerating vertically.

[tex]\Sigma F = -F_g - T + P_y = 0[/tex]

[tex]P_y = F_g + T[/tex]

The tension force is equal to the box's weight because the forces on the box are balanced. Let's use values and solve.

[tex]P_y = 49(9.8) + 49(9.8) = \boxed{960.4N}[/tex]

Part B.

We must begin by doing a summation of torques. Placing the pivot point at the pivot, we have the following present:
- Force of gravity acting at the center of mass of the rod (CC, at L/2)

- The tension of the horizontal cable acting at the end of the rod (CCW, at L)

- The force of tension in the rope holding the box (CC, at 3L/4)


Since the rod is not rotating, these torques sum to zero.

The equation for torque is:
[tex]\tau = r \times F[/tex]

This is a cross-product, and you must find the lever arm (perpendicular distance between pivot and line of action of force). We will need to use trigonometry for this.

Now, let's find the torque from all three of these forces.

- Force of gravity at center:

The perpendicular distance between the force of gravity and the pivot point is the cosine with respect to the angle made with the floor.

[tex]\tau = Mg\frac{L}{2}cos(\theta) = 49(9.8)*\frac{2.2}{2} cos(18) = 502.367 Nm[/tex]

- Tension of horizontal cable:

The lever arm is the sine with respect to the angle. We will still have to solve for the value of 'T'.

[tex]\tau = TLsin\theta = T(2.2)sin(18) = 0.68T[/tex]

- Tension of rope holding box:
The tension is equal to the weight of the box since the box isn't accelerating. Thus, the torque would be:
[tex]\tau = Mg(\frac{3L}{4}) = 49(9.8)*\frac{3(2.2)}{4}cos(18) = 753.55 Nm[/tex]

Summing with clockwise torques + and counterclockwise -:
[tex]\Sigma \tau = 502.367 + 753.55 - 0.68 T = 0 \\\\1255.917 = 0.68T\\T = \boxed{1847.38 N}[/tex]

Part C.

This part is a lot easier than it seems. All we need to do is a summation of horizontal forces.

We only have two:
- The horizontal tension in the cable to the left (1847.38 N)

- The horizontal force exerted by the pivot on the beam to the right

These two balance out because there is no acceleration of the beam horizontally, so:
[tex]\Sigma F = P_x - T = 0 \\\\P_x = T\\\\P_x = \boxed{1847.38 N}[/tex]

**to the right

Answer:

a) 960.4 N

b) T= 5/4 Mg CotanΘ

c) 1847. 38

Explanation:

a) Py= 2Mg

=2(49 x 9.8)

= 960.4

b) T= (Mg x 1/2 x cos Θ + Mg x 3/4 x cos Θ) / sin Θ

T= 5/4 X Mg cotanΘ

c) T= (5/4) x (49 x 9.8) cotan (18)

T= 1847.37954

= 1847.38

A 15.0-kg child descends a slide 2.40 m high and reaches the bottom with a speed of 1.10 m/s .
How much thermal energy due to friction was generated in this process?
Express your answer to three significant figures and include the appropriate units.

Answers

The thermal energy that is generated due to friction is 344J.

What is the thermal energy?

Now we know that the total mechanical energy in the system is constant. The loss in energy is given by the loss in energy.

Thus, the kinetic energy is given as;

KE = 0.5 * mv^2 =0.5 * 15.0-kg * (1.10 m/s)^2 = 9.1 J

PE = mgh = 15.0-kg * 9.8 m/s^2 *  2.40 m = 352.8 J

The thermal energy is; 352.8 J - 9.1 J = 344J

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