A ball is travelling 32° above the horizontal at a speed of 24 m/s. What is the horizontal component of its speed

A. 12.7 m/s
B. 13.0 m/s
C. 29.2 m/s
D. 20.4 m/s

Answers

Answer 1

Answer:

Since the ball is travelling 32 degrees above the horizontal, the value of Θ is 32

In the figure, v vector is the vertical component whereas h vector is the horizontal component

Using trigonometry:

CosΘ = h /24

Cos 32 = h/  24

0.85 = h / 24

h = 24*0.85

h = 20.4 m/s

A Ball Is Travelling 32 Above The Horizontal At A Speed Of 24 M/s. What Is The Horizontal Component Of

Related Questions

What is the final velocity of a drag racer that has constant acceleration and finishes a
1/4 mile race in 15 seconds?

Answers

Answer:

120 mph

Explanation:

Given:

Δx = 0.25 mi

v₀ = 0 mi/s

t = 15 s

Find: v

Δx = ½ (v + v₀) t

0.25 mi = ½ (v + 0 mi/s) (15 s)

v = 0.0333 mi/s

v = 120 mi/h

A skydiver has a mass of 80 kg. When she is at a height of 900 m, her GPE is
705,600 J. She is falling at a velocity of 32 m/s downward. What is her ME at this
point?

Answers

Answer:

11.25 J

Explanation:

GPE=wxh=80kg over 900m=11.25 J

what is lighting write its causes​

Answers

Answer: lightning is the occurrence of a natural electrical discharge of very short duration and high voltage between a cloud and the ground or within a cloud, accompanied by a bright flash and typically also thunder.

the causes of lightning:

Lightning happens when the negative charges (electrons) in the bottom of the cloud are attracted to the positive charges (protons) in the ground.

In the early stages of development, air acts as an insulator between the positive and negative charges in the cloud and between the cloud and the ground. When the opposite charges build up enough, this insulating capacity of the air breaks down and there is a rapid discharge of electricity that we know as lightning.

A train has a mass of 1.50 x 107 kg. If the engine can
exert a net force of 7.50 x 105 N on the train, how much
time is required for the train to reach a speed of 80.0
km/h, if the train begins from rest?

Answers

vi=0, vf=80km/hr=22m/s, m=1.5x107kg, F = 7.5x105N.

a=F/m, t=(vf-vi)/a= (vf-vi)m/F = (22m/s)(1.5x107kg)/7.5x105N = 440 kg m/sN =  440 s or 7mins

5. A capacitor is discharging through a resistor. The initial voltage across the capacitor is 3.2 V at t = 0. The voltage across the capacitor at time t = 20 ms is 0.8 V. The time it takes for the voltage across the capacitor to drop from 0.8 V to 0.2 V is ​

Answers

Hi there!

Recall the equation for the voltage of a discharging capacitor:

[tex]V_C(t) = V_0e^{-\frac{t}{\tau}}[/tex]

V₀ = Initial voltage of the capacitor (V)
t = Time (s)
τ or RC = Time Constant (s)

With the given information, we can plug in the value for V₀:
[tex]V_C(t) = 3.2e^{-\frac{t}{\tau}}[/tex]

We are given that at t = 20 ms (0.02 s), the voltage of the capacitor is 0.8V. We can use this to solve for the time constant (τ).

[tex]0.8 = 3.2e^{-\frac{0.02}{\tau}}\\\\0.25 = e^{-\frac{0.02}{\tau}}[/tex]

Take the natural log of both sides and solve.

[tex]ln(0.25) = -\frac{0.02}{\tau}\\\\-1.3863 = -\frac{0.02}{\tau}\\\\\tau = \frac{0.02}{1.3863} = 0.0144 s[/tex]

Now, we can use this time constant to solve for the time taken for the voltage to drop from 0.8 V to 0.2 V. Solve for the time taken for the capacitor's voltage to drop to 0.2 V:

[tex]0.2= 3.2e^{-\frac{t}{0.0144}}\\\\0.0625 = e^{-\frac{t}{0.0144}}\\\\ln(0.0625) = -\frac{t}{0.0144}\\\\t = (-2.773)(-0.0144) = 0.04 s[/tex]

Now, subtract the times:
[tex]0.04 - 0.02 = 0.02 = \boxed{20 ms}[/tex]

PHYSICS 1403
Lab Homework - Friction on a Ramp
A laborer wants to move crates containing bottles of olive oil from a truck to the ground by sliding them
along a ramp. The ramp is 6 m long and is at an angle of 25º. There is friction on the ramp for the first
crate. The laborer doesn't know that there is a small leak in one of the bottles. The leak leaves a layer of
oil on the ramp. The oil creates a frictionless surface for the second crate Wayne sends down the ramp. At the bottom of the ramp, the speed of the second crate (without friction) is 2.5 the speed of the first crate (with friction). Find the coefficient of kinetic friction. Hint: this is a multistep problem that is
be solved using only energy equations. Do not use kinematics or you will not receive full
credit, even if your answer is correct. Use conservation of energy and start with the frictionless case.

Answers

Hi there!

Hi there!

We can begin by simplifying the work-energy theorem for Crate 2.

Since there is no friction, there is no energy dissipated. Thus, the initial energy is equal to the final energy.

Initially, we only have gravitational potential energy (U = mgh), and when the box has fully slid down, it only has kinetic energy (KE = 1/2mv²), therefore:
[tex]E_i = E_f\\\\mgh = \frac{1}{2}mv^2[/tex]

We can cancel out the mass and solve for velocity.

[tex]gh = \frac{1}{2}v^2\\\\v^2 = 2gh \\\\v = \sqrt{2gh}[/tex]

We must use right triangle trigonometry to solve for the HEIGHT given the ramp's length (hypotenuse).

We can use sine:
[tex]sin\theta = \frac{\text{h}}{L} \\\\Lsin\theta = h = 6 * sin(25) = 2.5357 m[/tex]

Now, solve for velocity.

[tex]v = \sqrt{2(9.8)(2.5357)} = 7.05 \frac{m}{s}[/tex]

Since this is 2.5 times the speed of the first crate, we know that the final velocity of crate 1 is:


[tex]v_1 = \frac{v}{2.5} = 2.82 \frac{m}{s}[/tex]

Crate 1:
In this instance, we have friction. Recall the following.

[tex]F_f = \mu N[/tex]

On an incline, the normal force is equivalent to the cosine of the force of gravity, so:
[tex]N = mgcos\theta[/tex]

Now, create an equation for the force due to friction.

[tex]F_f = \mu mgcos\theta[/tex]

The work done by any force is:
[tex]W = F \cdot d\\\\W_f = \mu mgdcos\theta[/tex]

In this instance, d = the ramp's length, or 6 m.

Now, we can use the work-energy theorem.

Ei = Ef

However, there is energy dissipated; we can call this Wf (Work due to friction). Therefore:
Ei - Wf = Ef

Now, we can rearrange to solve for Wf:
Ei - Ef = Wf

Like above, there is initially only GPE (U = mgh) and finally only KE (K = 1/2mv²), so:
[tex]mgh - \frac{1}{2}mv^2 = \mu mgdcos\theta[/tex]

Solve for the coefficient of friction. Begin by canceling out the mass and multiplying all terms by 2:
[tex]2mgh - mv^2 = 2\mu mgdcos\theta\\\\2gh - v^2 = 2\mu gdcos\theta\\\\\mu = \frac{2gh - v^2}{2gdcos\theta}\\\\\mu = \frac{2(9.8)( 2.5357)- (2.82)^2}{2(9.8)(6)cos(25)}[/tex]

Evaluate:
[tex]\boxed{\mu = 0.39}[/tex]

A woman 1.2 m tall lies along the axis of a space vehicle traveling at 0.87c. What is her height as measured by a stationary observer?

Answers

Answer:

L = L0 (1 - v^2/c^2)     where L0 is proper length and L the measured length

L = 1.2 (1 - .87^2)^1/2 m = .59 m

what is a everyday activities examples of newtons 1 law of motion

Answers

Answer:

bouncing  a baskletball

Explanation:

Answer:someone kicking a soccer ball.

Explanation:

Because the ball isn’t in motion until acted on by another object (the foot)

A fish scale, consisting of a spring with spring constant k=200N/m, is hung vertically from the ceiling. A 2.6 kg fish is attached to the end of the unstretched spring and then released. The fish moves downward until the spring is fully stretched, then starts to move back up as the spring begins to contract.
What is the maximum distance through which the fish falls?

Answers

Answer:

Explanation:

The fish is initially at rest and it is also at rest when the spring is fully stretched at the maximum distance.

Change in gravity potential energy = change in spring potential energy

mgh = 1/2kh^2

Assume gravity constant g is 10m/s^2

2.6*10*h = 1/2*200*h^2

100h^2 - 26h = 0

2h(50h - 13) = 0

h = 0 or h = 13/50 = 0.65m

h = 0 is before the spring is stretched

So the maximum distance is 0.65m.

The maximum distance through which the fish falls is 0.25m.

What is law of conservation of energy?

Energy cannot be created or destroyed, according to the law of conservation of energy. However, it is capable of change from one form to another. An isolated system's total energy is constant regardless of the types of energy present.

Given parameters:

Spring constant: k = 200 N/m.

Mass of the fish: m =2.6 kg.

Let the maximum distance through which the fish falls, i.e. the maximum stretched distance of the spring = x.

From law of conservation of energy:

Change in gravity potential energy  = change in potential energy

mgx = 1/2kx^2

2.6×9.8×h = 1/2*200*h^2

h(100h - 2.6×9.8) = 0

h = 0 or h = 2.6×9.8/100 = 0.25m

As h = 0 is the equilibrium position; the maximum distance through which the fish falls is 0.25m.

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WHO WANTS BRAINLY? PLEASE ANSWER THIS PHYSICS QUESTION!
force of 187 N acts on a 7.3 kg bowling ball for 0.40 s.
1. What is the change in momentum of the ball?
2. What is its change in velocity?

Answers

Change in momentum of the ball is 74.8 kg m/s

change in velocity of the ball is 0.25 m/s

What is change in momentum?

Change in momentum is the product of mass and change in velocity this is also equal to impusle which ia a product of force and time.

Change in momentum = m ( v - u ) = Impulse = f * t

m = mass

change in velocity = v - u

v = final velocity

u = initial velocity

f = force

t = time

Given:

f  = 187 N

m = 7.3 kg

t = 0.40 s

!.

Change in momentum = impulse = f * t

Change in momentum = 187 * 0.40

Change in momentum = 74.8 kg m/s

2.

Change in momentum = m * ( v - u )

( v - u ) = change in momentum / m

( v - u ) = 187 / 7.3

( v - u ) = 10.25 m/s

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An athlete needs to get across a river. She must reach from point A on one bank to point B, which is directly across from point A on the opposite bank. If she wishes to minimize the total time T it would take to do so, at what angle upstreams (measured from the line AB) should she swim? Let her swimming/rowing speed relative to the water be 2.0 mi/h, and her running speed along the bank be 5 mi/h. The river flows downstream at 1.5 mi/h. (The rule of the competition stipulates that the athletes must swim first from point A, then run to reach point B.)

Answers

Answer:

θ = 90°

Explanation:

Speed at which she can swim in stationary water; v_m = 2 mi/h

Speed at which the river is flowing downstream; v_r = 1.5 mi/h

This can be illustrated in a form of triangle as attached. Where θ is the angle to the x-axis.

With respect to the bank of the river, the total speed of the girl is;

V_R = v_r + v_m

Now the horizontal component is;

V_R = v_r + v_m(cos θ)

While the vertical component is;

V_R = v_m(sin θ)

Now, the total time taken to get across the river is;

T = width of river/vertical component of total speed

Let's denote width of river as d.

Thus;

T = d/(v_m(sin θ))

Now, we are told she wishes o minimize the total time(T) .

Since the width(d) of the river is constant, and v_m is fixed, then it means the only variable we have is the angle θ.

Now, for the total time to be minimized, the denominator has to be at maximum.

Since angle θ is the only one that is variable, we have to find the value of angle θ that will make sin θ to be at a maximum value.

Now, for sin θ to be maximum, it means it has to be equal to 1.

Thus, θ = sin^(-1) 1

θ = 90°

Total internal reflection will occur when:

light goes from high to low density above the critical angle
light goes from low to high density above the critical angle
light goes from high to low density below the critical angle
light goes from low to high density below the critical angle

Answers

Answer:

Try B or C if I'm wrong sorry

Explanation:

The Earth rotates once every day. What is the velocity of a person on its surface, if the radius of Earth is 6 million meters

Answers

Answer:

v = ω R

number of seconds in 24 hrs = 24 * 3600 = 86400 s / da = T (period)

f = 1 / T

ω  = 2 Π f = 2 Π / T = 2 Π / 86400

v = 2 Π / 8.64E4 * 6.0E6 = 436 m / s

Check:

S = 2 Π * 6.0E6 = 3.77E7 m/da

S / T = 3.77E7 m/da / 8.64E4 s/da = 436 m / s

1) A pile driver with mass 1x10⁴ kg strikes a pile with velocity 10.0 m/s. What is the kinetic energy of the driver as it strikers the pile?​

Answers

Explanation:

please mark me as brainlest

\A thin hoop with a radius of 10 cm and a mass of 3.0 kg is rotating about its center with an angular speed of 3.5 rad/s. What is its kinetic energy

Answers

The rotational or kinetic energy of the thin hoop with the given radius, mass and angular speed is 0.092J

Given the data in the question;

Mass of hoop; [tex]m = 3.0kg[/tex]Radius; [tex]r = 10cm = 0.1m[/tex]Angular speed; [tex]w = 3.5rad/s[/tex]

Rotational or kinetic energy; [tex]E_{rotational} = \ ?[/tex]

Rotational energy or angular kinetic energy

Rotational energy or angular kinetic energy is simply kinetic energy due to the rotation of a rigid body.

It is expressed as;

[tex]E_{rotational} = \frac{1}{2}Iw^2[/tex]

Where [tex]I[/tex] is the moment of inertia around the axis of rotation and [tex]w[/tex] is the angular speed or velocity.

For the moment of inertia around the axis of rotation.

[tex]I = \frac{1}{2}mr^2[/tex]

Hence

[tex]E_{rotational} = \frac{1}{2}(\frac{1}{2}mr^2)w^2 \\\\E_{rotational} = (\frac{1}{4}mr^2)w^2[/tex]

Now, we substitute our given values into the above equation to find the rotational or kinetic energy.

[tex]E_{rotational} = (\frac{1}{4}*3.0kg * (0.1m)^2) * (3.5rad/s)^2 \\\\E_{rotational} = 0.0075kgm^2 * 12.25rad/s^2\\\\E_{rotational} = 0.092kg.m^2/s^2\\\\E_{rotational} = 0.092J[/tex]

Therefore, the rotational or kinetic energy of the thin hoop with the given radius, mass and angular speed is 0.092J

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Please Help Me
When you apply brakes on a bicycle, the bicycle’s ________________________

energy is not destroyed; instead, the bicycle’s _____________________energy is

transformed into thermal energy. The __________________amount of energy remains the same.

Answers

Answer:

1. Acceleration begins to slow down quickly

2. Kenetic

3. Mass

Answer:

Acceleration slows down

Kinetic

Mass

Explanation:


2
In 1987, the fastest auto race in the United States was the Busch Clash in Daytona, Florida. That year, the
winner's average speed was about 318 km/h. Suppose the kinetic energy of the winning car was 3.80 MJ. What
was the mass of the car and its driver?

Answers

Since the winner's average speed was about 318 km/h and the kinetic energy of the winning car was 3.80 MJ. So, the mass of the car and its driver is 974 kg.

Kinetic energy

The kinetic energy of the car is given by K = 1/2mv² where

m = mass of car and driver and v = average speed of car = 318 km/h = 318 × 1000 m/3600 s = 88.33 m/sMass of the car and driver

Making m subject of the formula, we have

m = 2K/v²

Since K = 3.80 MJ = 3.80 × 10⁶ J, substituting the values of the variables into the equation, we have

m = 2K/v²

m = 2 × 3.80 × 10⁶ J/(88.33 m/s)²

m = 7.6 × 10⁶ J ÷ (7802.78 m²/s²)

m = 7600000 J ÷ 7802.78 m²/s²

m = 974.01 kg

m ≅ 974 kg

So, the mass of the car and its driver is 974 kg

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Which two elements have the same number of valence electrons?
A. C and o
B. Na and Mg
C. Cl and F
D. Ga and Ge

Answers

The answer is Cl and f.
They belong to group halogens and have 7 valence electrons in their valence shell

Two Earth satellites, A and B, each of mass m = 940 kg , are launched into circular orbits around the Earth's center. Satellite A orbits at an altitude of 4500 km , and satellite B orbits at an altitude of 11100 km .
c) How much work would it require to change the orbit of satellite A to match that of satellite B?

Answers

Answer:

The required work done is [tex]6.5\times10^{9}\ J[/tex]

Explanation:

Given that,

Mass of each satellites = 940 kg

Altitude of A = 4500 km

Altitude of B = 11100 km

We need to calculate the potential energy

Using formula of potential

[tex]U_{A}=-\dfrac{Gm_{A}m_{E}}{r_{A}}[/tex]

Put the value into the formula

[tex]U_{A}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+4.50\times10^{6}}[/tex]

[tex]U_{A}=-3.44\times10^{10}\ J[/tex]

We need to calculate the potential energy

Using formula of potential

[tex]U_{B}=-\dfrac{Gm_{B}m_{E}}{r_{A}}[/tex]

Put the value into the formula

[tex]U_{B}=-\dfrac{6.67\times10^{-11}\times940\times5.98\times10^{24}}{6.38\times10^{6}+11.10\times10^{6}}[/tex]

[tex]U_{B}=-2.14\times10^{10}\ J[/tex]

We need to calculate the value of [tex]k_{A}[/tex]

Using formula of [tex]k_{A}[/tex]

[tex]k_{A}=-\dfrac{1}{2}U_{A}[/tex]

Put the value into the formula

[tex]k_{A}=\dfrac{1}{2}\times3.44\times10^{10}[/tex]

[tex]k_{A}=1.72\times10^{10}\ J[/tex]

We need to calculate the value of [tex]k_{B}[/tex]

Using formula of [tex]k_{B}[/tex]

[tex]k_{B}=-\dfrac{1}{2}U_{B}[/tex]

Put the value into the formula

[tex]k_{B}=\dfrac{1}{2}\times2.14\times10^{10}[/tex]

[tex]k_{B}=1.07\times10^{10}\ J[/tex]

We need to calculate the work done

Using formula of work done

[tex]W=\Delta K+\Delta U[/tex]

[tex]W=(k_{B}-k_{A})+(U_{B}-U_{A})[/tex]

[tex]W=(-\dfrac{U_{B}}{2}+\dfrac{U_{A}}{2})+(U_{B}-U_{A})[/tex]

[tex]W=\dfrac{1}{2}(U_{B}-U_{A})[/tex]

Put the value into the formula

[tex]W=\dfrac{1}{2}\times(-2.14\times10^{10}+3.44\times10^{10})[/tex]

[tex]W=6.5\times10^{9}\ J[/tex]

Hence, The required work done is [tex]6.5\times10^{9}\ J[/tex]

If the period of a wave is 20 s, then what is its frequency?

Answers

The frequency would be 0.05 hertz :)

Please help I only have 20 min left!!!!

Power: The work performed as a function of time for a process is given by W = at3, where a = 2.4 J/s3. What is the instantaneous power output at t = 3.7 s?
Group of answer choices

99 W

139 W

208 W

69 W

Answers

The instantaneous power output at t = 3.7 s right answer is 99 W.

Power is defined by the amount of energy transferred over a period of time. Instantaneous power, on the other hand, refers to the power consumed at a point in time. Instantaneous power is an important metric in electronics. Instantaneous power is the power measured at a specific point in time.

The basic difference between average effort and instantaneous effort is that average effort is the ratio of total work time to total time. Although the instantaneous power is the limit of the average power. The statement of work does not state that the same amount of work he will complete in a second or an hour. Instantaneous power can be positive or negative. Positive instantaneous power means that energy is flowing from the source to the load, and negative instantaneous power means that energy is flowing from the load to the source.

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The system is released from rest with the spring initially stretched 3 in. Calculate the velocity of the cylinder after it has dropped 0.5 in.

Answers

The velocity of the cylinder after it has dropped 0.5 in is 0.497 m/s.

Conservation of energy

The velocity of the cylinder can be determined by applying the principle of conservation of energy as shown below;

Ei = Ef

¹/₂k(xi)² + mgh = ¹/₂mv² + ¹/₂k(xt)²

where;

k is spring constant = 6 lb/in = 1050 N/mm is mass of the system = 100 lb = 45.35 kgxi is initial extension = 3 in = 0.076 mxt is the total extension = 2xi + 0.5 in = 6.5 in = 0.165 mh is height = 0.5 in = 0.0127 m

¹/₂k(xi)² = -mgh + ¹/₂mv² + ¹/₂k(xt)²

¹/₂(1050)(0.076)² = -(45.35)(9.8)(0.0127) + ¹/₂(45.35)v² + ¹/₂(1050)(0.165)²

3.032 = -5.64 + 22.68v² + 14.29

3.032 = 8.65 + 22.68v²

-5.62 = 22.68v²

|v| = 0.497 m/s

Thus, the velocity of the cylinder after it has dropped 0.5 in is 0.497 m/s.

The complete question is below:

The spring has a stiffness of 6 lb/in and the mass of the load is 100 lb.

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Solar energy can be directly converted to elec-trical energy by which of the following de-vices?​

Answers

Answer: (c) solar cell

Explanation:

Answer:
A solar cell takes solar energy directly from the sunlight and converts it into electrical energy

A horizontal spring is attached to the wall on one end and to a mass on the other end. The mass can slide freely on a frictionless surface below. Suppose you pull the mass so that the spring is stretched out (initial state) and then you release it, so that the mass starts moving towards the spring is unstretched position (final state). The impulse imparted on the spring-mass system by the force that the wall exerts on the spring is zero, since the wall does not move during this process.

Required:
What total percentage of the period does the mass lie in these regions?

Answers

Answer:

a) x=0  %T=0,   b) x= A %T=100%,   c) x=-A %T=50%

Explanation:

This is a simple harmonic movement exercise, which is explained by the expression

          x = A cos (wt + Ф)

where angular velocity is related to frequency and period

         w = 2π f = 2π / T

we can write the equation of the oscillation

         x = A cos θ

When seeing the two equations they are equivalent, so what happens with the angle will also happen with time

We are asked for the percentage of the period at three points: at the maximum elongation and at the point of x = 0, in general the distance is measured from the point of the spring without stretching

The period is defined as the time that the system takes to give a complete oscillation, that is, from x = 0 to x = A and return

a) for the unstretched spring point x = 0

In general, both distance and time are measured from this point, so the percentage of time is zero.

         % T = 0

b) for x = A

 let's find the angle

      cos tea = x / A = 1

therefore the angles tea = 2π rad

when the movement reaches the point of 2π radians it begins to repeat so the period is complete

            % T = 100%

c) the point of maximum compression x = -A

let's look for the angles

      cos tea = x / A = -1

therefore the angles tea = π rad

at this point the movement is halfway so it should take half the time

                % T = 50%

5. Use the algebraic technique for adding vectors to find the total displacement of a person who walks the following three paths (displacements-on a flat field. First, she walks 25.0 m in a direction 49.00 north of east. Then she walks 23.0 m heading 15.00 north of east. Finally, she turns and walks 32.0 m in a direction 68.0° south of east. ​

Answers

Answer:

jjjjjjjjjjjjjjjj

Explanation:

A student is moving a magnet through a coil of wire. Which of these indicates the strength of the electric current in the coil?

Answers

Answer:

Can you help me with another question

Explanation:

The electric current in the coil  depends on changes in the magnetic field of the magnet.

Electricity is generated when there is relative motion between a conductor and a magnetic field. As the magnet is being moved through the coil of wire, the magnetic flux around the conductor changes and current is induced on it.

Hence, the strength of the electric current in the coil is determined by  changes in the magnetic field of the magnet.

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g A satellite is orbiting planet Earth with a linear speed of 3,914 m/s. The orbital radius of the satellite in km is:

Answers

The orbital radius of the satellite orbiting planet Earth at the given linear speed is 26,000 km.

Orbital radius of the planet

The orbital radius of the planet is the distance of the planet from the center of the Earth. The orbital radius of the planet is calculated as follows;

[tex]v = \sqrt{\frac{GM}{r} }[/tex]

where;

M is mass of EarthG is gravitational constantr is the orbital radius

r = GM/v²

r = (6.67 x 10⁻¹¹ x 5.97 x 10²⁴)/(3,914²)

r = 2.6 x 10⁷ m

r = 26,000 km

Thus, the orbital radius of the planet is 26,000 km.

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The equation of state of a certain gas is p(V −b) = RT, where b is a constant. What order of magnitude do you expect b to be? Show that the internal energy of this gas is a function of temperature only

Answers

Answer:

LT 23 don't ask how I got it

Electric field lines moves away from positive to wards negative?

Answers

The vector force on the unit positive charge placed at any location in the field defines the strength of the electric field at that point. The charge used to determine field intensity (field strength) is known as the test charge. Now, a field line is defined as a line to which the previously mentioned field strength vectors are tangents at the relevant places. When we study positive charge field lines, the field strength vectors point away from the positive charge. If there is a negative charge anywhere in the vicinity, the field lines that began from the positive charge will all terminate at the negative charge if the value of the negative charge is the same as the value of the positive charge. Remember that the number of field lines originating from positive charge is proportional to the charge's value, and similarly, the number of field lines terminating at negative charge is proportionate to the charge's value. As a result, if all charges are equivalent, all lines originating from the positive charge terminate at the negative charge. If the value of the positive charge is greater than the value of the negative charge, the number of lines ending at the negative charge will be proportionally less than the number of lines beginning at the positive charge. The remaining lines that do not end at the negative charge will go to infinity. If the positive charge is less, all lines from it terminate at a negative charge, and any other reasonable number of ines terminate at a negative charge from infinity. We should also keep in mind that the number of lines that run perpendicular to the field direction across a surface of unit area is proportional to the field strength at that location. As a result, lines are dense in the strong field zone and sparse in the low intensity region.

A student placed a ladder up against a wall as shown below. The normal force applied by the wall in the ladder will be directed:
Up and to the right
To the left
Down and to the left
Straight up

Answers

The normal force is always perpendicular to the surface. So it would be straight to the left of the wall

The normal force applied by the wall in the ladder will be directed to the left. Hence, option (B) is correct.

What is normal force?

The component of a contact force in mechanics known as the normal force is perpendicular to the surface that an item encounters.

In this context, normal refers to perpendicular in a geometric sense rather than "usual" or "expected" as it does in everyday parlance. Gravity acts on a person standing stationary on a platform; gravity would otherwise draw the person down into the Earth's core absent a countervailing force from the resistance of the platform's molecules, known as the "normal force."

As normal force acts perpendicular to the surface, the normal force applied by the wall in the ladder will be directed to the left.

Learn more about normal force here:

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