Jumbo is back! Jumbo is the 4.8-Mg elephant. This time he's standing at the outer edge of a 11-Mg turntable of radius 8.5 m, rotating with angular velocity 0.15 s?1 on friction less bearings. Jumbo then walks to the center of the turntable. Treat Jumbo as a point mass and the turntable as a solid disk.
Part A: Find the magnitude of the angular velocity of the turntable once Jumbo reaches the center. ?t = ? rad/s
Part B: Find the work Jumbo does in walking to the center. W = ?

The highest point of the wheel is the position of the wheel when its 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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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.

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

                   [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?

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

                               [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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Answer:

Explanation:

the value is -8 cm

Answer:

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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]

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.

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.

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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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

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

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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(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.

U = qΔV

where;

U = q(Ed)

U = (1.6 x 10⁻¹⁹)(1500 x 1.5 x 10⁻²)

U = 3.6 x 10⁻¹⁸ J

ΔV = E(0.5d)

ΔV = 0.5Ed

ΔV = 0.5 (1500)(1.5 x 10⁻²)

ΔV = 11.25 V

U = ¹/₂mv²

U = mv²

v² = 2U/m

where;

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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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

Change in potential energy = 1.02 * 10⁶ J

Minimum work required of the hiker = 1.89 * 10⁶ J

The potential energy of a body is calculated as follows:

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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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

(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.

ω = √k/m

where;

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

E = ¹/₂kA²

where;

A is amplitude

E = ¹/₂(4.94)(0.15)²

E = 0.056 J

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

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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Answer:

D. Impulse

Explanation: Hope this helps

The tension in the upper rope is 50.53 N.

The tension in the upper rope is calculated as follows;

T(up) = T(dn) + mg

where;

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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(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.

g = GM/R²

where;

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

g = 0.13 m/s²

v² = u² - 2gh

where;

at maximum height, v = 0

u² = 2gh

u = √2gh

u = √(2 x 0.13 x 1,440)

u = 19.35 m/s

v = √GM/r

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

apply the principle of conservation of energy;

P.E = K.E

mgh = ¹/₂mv²

h = v²/2g

where;

h = (6²)/(2 x 9.8)

h = 1.84 m

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

T =  2π√(L/g)

where;

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

The two conditions are;

The given parameters are;

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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(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.

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

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.

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.

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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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.

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

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:

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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The solution for the three questions  is mathematically given as

Parts A and C are both zeros.

Part B B = 0.001855Tesla

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

The angle of the red light is mathematically given as

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

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

A = πr²

where;

A = π(4.3 x 10⁴)²

A = 5.81 x 10⁹ cm²

Let the error in measurement = 1 x 10⁴ cm

A =  π(1 x 10⁴)² = 3.14 x 10⁸ cm²

% = (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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Answer:

Create space

Watch your breath

Watch your thoughts

Be gentle with yourself

Affirm what you want

Explanation:

(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.

θc = sin⁻¹( 1/η)

where;

Refractive index of air = 1

Refractive index of polystyrene = 1.5865

Refractive index of flint glass = 1.655

θc = sin⁻¹( 1/1)

θc = 90⁰

θc = sin⁻¹( 1/1.5865)

θc =  39.1 ⁰

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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C. The opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.

The law of increasing opportunity cost is an economic principle that describes how opportunity costs increase as resources are applied.

As the student gives up his sleep or night rest in the place of his exam preparation, we say that the opportunity cost is the sleep or rest.

Thus, the opportunity cost of a student who is staying up all night to study for an exam that he has to take in the early morning is sleep or rest.

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