At what speed was object A moving ?

Answer:

B

Explanation:

i took the test

Explanation:

Since gravitational forces are inversely proportional to the square of their distances, tripling the distance means the reduction of the force by a factor of 9. That means the force experienced by the two objects will be 1/9 smaller than before.

Answer:

b. is the correct answer ....

Answer:

2

Explanation:

Answer:

Less

Explanation:

Weight is a force measurement. The object's mass is 5kg not its weight. To find its weight you have to take the mass of an object and multiply it by the acceleration of gravity. The acceleration of gravity is greater on earth than on the moon so therefore the object will weigh less on the moon.

Answer:

The rotation rate is 15.3 rad/s.

Explanation:

maximum voltage, V = 120 V

Magnetic field, B = 1.5 T

length, L = 10.2 cm

width, W = 12.8 cm

Number of loops, N = 400

Let the rate of rotation is w.

The maximum voltage is given by

V = N B A w

120 = 400 x 1.5 x 0.102 x 0.128 x w

w = 15.3 rad/s

Answer:

.....where's the model-

Answer:

the work done by the motor is 1,044 J.

Explanation:

Given;

the output power of the motor, P = 0.7 hp

duration of the work, t = 2 s

The relationship between horse-power and watt is given as;

1 hp = 745.7 W

0.7 hp = ?

0.7 hp = 522 W = 522 J/s

The work done by the motor is calculated as;

W = Power x time

W = 522 J/s  x  2 s

W = 1,044 J

Therefore, the work done by the motor is 1,044 J.

Answer:

The distance traveled in 1 year is: [tex]3.143*10^{16}ft[/tex]

Explanation:

Given

[tex]s = 982,080,000 ft/s[/tex] --- speed

[tex]t = 32,000,000 s[/tex] --- time

Required

The distance traveled

This is calculated as:

[tex]Speed = \frac{Distance}{Time}[/tex]

So, we have:

[tex]Distance = Speed * Time[/tex]

This gives:

[tex]Distance = 982,080,000 ft/s * 32,000,000 s[/tex]

[tex]Distance = 982,080,000 * 32,000,000ft[/tex]

[tex]Distance = 3.143*10^{16}ft[/tex] -- approximated

Answer:

Explanation:

From the missing image attached below, it is obvious that there no external force. This implies that they cannot change their position by merely just pulling the ropes. As a result, there will be no movement and no net force will exist.

So, if there is no external force;

The center of mass of the two people is:

[tex]X_{cm}= \dfrac{m_1x_1+m_2x_2}{m_1+m_2} \\ \\ X_{cm}= \dfrac{m(-3m)+m(+3m)} {m+m}\\ \\ X_{cm}= \dfrac{0}{2m} \\ \\ X_{cm} =0[/tex]

Thus, In the system, no movement occurs and all forces remain the same.

Answer:

Power = 24.41Watts

Explanation:

Find the diagram attached

Power output  = Force * distance/Time

Given

Force = 120lb

Distance = 9inches

Time = 5sec

Since

1lb = 4.4482216153 N

120lb  = 120 * 4.4482216153

120lb = 533.787N

9in to meters

9in = 0.0254*9

9in = 0.2286N

Power = 533.787*0.2286/5

Power = 24.41Watts

Answer:

Wavelength = [tex]\lambda_p = 3.986 * 10^{-6}[/tex] m

Explanation:

As we know

Fringe width (w) = [tex]\frac{D*\lambda}{d}[/tex]

where

[tex]\lambda[/tex] is the wavelength

D is distance between source and screen

d is the distance between two slits

[tex]\frac{D}{d} = \frac{y}{\lambda}[/tex]

[tex]\frac{D}{d} = \frac{y_r}{\lambda_r} = \frac{y_p}{\lambda_p}\\\frac{y_r}{\lambda_r} = \frac{y_p}{\lambda_p}\\\lambda_p = \frac{y_p* \lambda_r}{y_r} \\\lambda_p =\frac{6.30 * 10^{-3} * 632.8 * 10^{-9}}{6 *10^{-3}} \\\lambda_p = 3.986 * 10^{-6}[/tex]m

Answer:

60 w

Explanation:

Given :

V= 120V

I=0.5 A

Now,

power can be calculated as :

P=VI

where,

V is voltage

I is current

Now,

P=(120)(0.5)

P=60 W

Therefore, 60w power is delivered to a light

Work is done when you lift an object to a certain height. If the force exerted is greater than the weight of the object, input work is greater than the output work. Then the extra energy goes in overcoming the gravitational acceleration and heating up of body etc

Answer:

F = 10.78 N

Hence q₂ will move away from the charge q₁ towards right side.

Explanation:

The force between two charged particles can be found by using Colomb's Law:

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

where,

F = Force = ?

k = Colomb Constant = 8.99 x 10⁹ N.m²/C²

q₁ = charge on first particle = 6 μC = 6 x 10⁻⁶ C

q₂ = charge on second particle = 2 μC = 2 x 10⁻⁶ C

r = distance between particles = 0.1 m

Therefore,

[tex]F = \frac{(8.99\ x\ 10^9\ N.m^2/C^2)(6\ x\ 10^{-6}\ C)(2\ x\ 10^{-6}\ C)}{(0.1\ m)^2}[/tex]

F = 10.78 N

Since both particles have a positive charge. Therefore this force will be the force of repulsion.

Hence q₂ will move away from the charge q₁ towards right side.

Answer:

Explanation:

E2020

Answer:

I don't know hhaha ammmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmmm I wish I could help

This question is incomplete, the complete question is;

In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.

Angular Velocity at time 0s = 12 rad/s

Angular Velocity at time 0.15s = 24 rad/s

a) What is the angular acceleration?

b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g

Answer:

a) the angular acceleration is 80 rad/s²

b) the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Explanation:

Given the data in the question;

from the graph below;

Angular Velocity at time 0s [tex]w_o[/tex] = 12 rad/s

Angular Velocity at time 0.15s [tex]w_f[/tex] = 24 rad/s

a) What is the angular acceleration;

Angular acceleration ∝ = ( [tex]w_f[/tex] - [tex]w_o[/tex] ) / dt

we substitute

Angular acceleration ∝ = ( 24 - 12 ) / 0.15

Angular acceleration ∝ = 12 / 0.15

Angular acceleration ∝ = 80 rad/s²

Therefore, the angular acceleration is 80 rad/s²

b)

If the ball is 0.60 m from her shoulder, i.e s = 0.6 m

the tangential acceleration of the ball will be;

a = ∝ × s

we substitute

a = 80 × 0.6

a = 48 m/s²

a = ( 48 / 9.8 )g

a = 4.9 g

Therefore, the tangential acceleration of the ball is;

- a = 48 m/s²

- a = 4.9 g

Answer:

the sound intensity level for actual intensity ( without the earplugs ) is 103.8 dB

Explanation:

Given the data in the question;

sound intensity reduced by the factor, m = 305

the sound intensity level experienced by the crew members wearing protective earplugs, L = 79 dB

Now, using the expression of sound intensity level;

L = 10log( [tex]I_0[/tex] )

where  [tex]I_0[/tex] is the intensity at L level

so we substitute

79 = 10log( [tex]I_0[/tex] )

[tex]I_0[/tex] = [tex]10^{7.9[/tex]

Now, expression for actual intensity;

[tex]I[/tex] = m[tex]I_0[/tex]

where [tex]I[/tex] is the actual intensity

so we substitute

[tex]I[/tex] = 305 × [tex]10^{7.9[/tex]

Next, we write the expression of sound intensity level for reduced intensity;

L' = 10log( [tex]I[/tex] )

So we substitute

L' = 10log( 305 × [tex]10^{7.9[/tex] )

L' = 10log( 24227011159.09058 )

L' = 103.8 dB

Therefore, the sound intensity level for actual intensity ( without the earplugs ) is 103.8 dB

Answer:

efficiency of a machine is less than 100% because some part is energy is utilized to overcome some opposing forces like friction which is wasted as heat ,sound energy etc

Explanation:

Answer:

in this case around the loop the field points downwards on the outside and upwards on the inside

Explanation:

To find the direction of the magnetic field in a wire you must use the right hand rule.

The thumb points in the direction of the current flow and the other created fingers point in the direction of the magnetic field.

Therefore in this case around the loop the field points downwards on the outside and upwards on the inside

Answer:

required distance is 233.35 m

Explanation:

Given the data in the question;

Sound intensity [tex]I[/tex] = 1.62 × 10⁻⁶ W/m²

distance r = 165 m

at what distance from the explosion is the sound intensity half this value?

we know that;

Sound intensity [tex]I[/tex] is proportional to 1/(distance)²

i.e

[tex]I[/tex] ∝ 1/r²

Now, let r² be the distance where sound intensity is half, i.e [tex]I[/tex]₂ = [tex]I[/tex]₁/2

Hence,

[tex]I[/tex]₂/[tex]I[/tex]₁ = r₁²/r₂²

1/2 = (165)²/ r₂²

r₂² = 2 × (165)²

r₂² = 2 × 27225

r₂² = 54450

r₂ = √54450

r₂ = 233.35 m

Therefore, required distance is 233.35 m

Answer:

B is excellent answer..............

The model of the brain that is shown here is the experimental model that is present in Option B, as it is used to study the brain's parts and its function, which is helpful for a better understanding of the brain.

There are various experimental models of the brain that have been developed to better understand its functions and mechanisms, such as Animal models, such as mice, rats, and primates, have been widely used to study the brain due to their similarity to the human brain in terms of structure and function. Computer models can simulate brain function and behavior at various levels of detail, from individual neurons to large-scale brain networks. These models are useful for testing hypotheses and predicting outcomes, as well as for designing new experiments.

Hence, the model of the brain that is shown here is the experimental model that is present in Option B.

Learn more about the experimental model of the brain here.

https://brainly.com/question/23802617

#SPJ7

Answer:

p = 6.64 cm

Explanation:

For this exercise we use the equation of the constructor

          [tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q}[/tex]

where f is the focal length, p and q are the distance to the object and the image, respectively

They tell us the focal length f = 2.2 cm and that the image as far as it can go is q = 3.29 cm, let's find the position of the object that creates this image

          [tex]\frac{1}{p} = \frac{1}{f} - \frac{1}{q}[/tex]

          1 / p = 1 / 2.2 - 1/3.29

           1 / p = 0.15059

           p = 6.64 cm

therefore the farthest distance from the object is 6.64 c

Answer:

We should not try to park the car because its rest length is greater than the space available.

The car seems to be approximately equal to the friend (L = 3.16 m). Due to this reason he is sure to park.

Explanation:

We should not try to park the car because its rest length is greater than the space available.

The friend is sure about parking because the car appears short in length to him. For this, we will solve Einstein's length contraction formula from theory of relativity:

[tex]L = L_o\sqrt{1-\frac{v^2}{c^2}}[/tex]

where,

L = Relative length observed by friend = ?

L₀ = rest length = 6 m

v = relative speed = 85% of speed of light = 0.85c

Therefore,

[tex]L = (6\ m)\sqrt{1-\frac{(0.85c)^2}{c^2}}[/tex]

L = 3.16 m

Hence, the car seems to be approximately equal to the friend. Due to this reason he is sure to park.

Answer:

[tex](b)\ t_1 - t_0[/tex]

[tex](d)\ t_2 - t_1[/tex]

[tex](e)\ \frac{t_2 - t_0}{2}[/tex]

Explanation:

Given

See attachment for complete question

Required

How long to reach the ground from the maximum height

First, calculate the time of flight (T)

[tex]T =t_2 - t_0[/tex]

The time taken (t) from maximum height to the ground is:

[tex]t = \frac{1}{2}T[/tex]

So, we have:

[tex]t = \frac{t_2 - t_0}{2}[/tex]

Another representation is:

At ymax, the time is: t1

On the ground, the time is t2

The difference between these times is the time taken.

So;

[tex]t = t_2 - t_1[/tex]

Since air resistance is to be ignored, then

[tex]t_2 - t_1 = t_1 - t_0[/tex] --- i.e. time to reach the maximum height from the ground equals time to reach the ground from the maximum height

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[tex] \huge\bold{T}[/tex]he speed of sound is maximum in solids. The speed of sound depends on the density of the medium through which it travels.The higher the density of the medium, the faster the propagation of sound. Since the density of solids is greater than that of liquids and gases, sound travels faster in solids. Also, the molecules in the solid medium are closer together than in liquids or gases, which allows sound waves to travel faster through it.

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Answer:

4210 W

Explanation:

We'll begin by calculating the energy required in lifting the water from the depth. This can be obtained as follow:

Mass (m) = 1200 Kg

Height (h) = 10 m

Acceleration due to gravity (g) = 9.8 m/s²

Energy (E₁) =?

E₁ = mgh

E₁ = 1200 × 9.8 × 10

E₁ = 117600 J

Next, we shall determine the energy required to project the water at 15 m/s. This can be obtained as follow:

Mass (m) = 1200 Kg

Velocity (v) = 15 m/s

Energy (E₂) =?

E₂ = ½mv²

E₂ = ½ × 1200 × 15²

E₂ = 600 × 225

E₂ = 135000 J

Finally, we shall determine the power of the motor. This can be obtained as follow:

Total energy (E) = E₁ + E₂

= 117600 + 135000

= 252600 J

Time (t) = 1 min = 60 s

Power (P) =?

P = E/t

P = 252600 / 60

P = 4210 W

Therefore, the power of the motor is 4210 W.