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liq [111]
3 years ago
8

Construct a graph of position versus time for the motion of a dog, using the data in the table below. Explain how the graph indi

cates that the dog is moving at a constant speed.

Physics
1 answer:
Lynna [10]3 years ago
6 0

Answer:

The dog is moving at a constant speed

Explanation:

Given that,

Position : 5, 10, 15, 20, 25

Time = 5. 10, 15, 20, 25

We need to draw a position time graph

Using given data

A graph of position and time shows the speed.

According to graph,

The graph indicates that the dog is moving at a constant speed because the graph is straight line.

Hence, The dog is moving at a constant speed

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A block weighing 30kg is moved at a constant speed over a horizontal surface by a force of 100 N applied parallel
DENIUS [597]
30-100=70^3•É+ 30 =3^E
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What type of infant temperament is generally associated with better adjustment in adulthood?
Rashid [163]
Saying no and not throwing fits and manners.
8 0
3 years ago
You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac
dusya [7]

Answer:

20.62361 rad/s

489.81804 J

Explanation:

I_i = Initial moment of inertia = 9.3 kgm²

I_f = Final moment of inertia = 5.1 kgm²

\omega_i = Initial angular speed = 1.8 rev/s

\omega_f = Final angular speed

As the angular momentum of the system is conserved

I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

6 0
3 years ago
A 10,000kg space ship is orbiting the moon with a radius of 25km from the ship to the center of the moon. It's tangential speed
galina1969 [7]

Answer:

False

Explanation:

ac = v^2/r

acceleration is not dependent on the mass of the orbiting object.

3 0
3 years ago
You are removing branches from your roof after a big storm. You throw a branch horizontally from your roof, which is a height 3.
mart [117]

Answer:

The initial velocity in the x-direction with which the branch was thrown is approximately 10.224 m/s

Explanation:

The given parameters of the motion of the branch are;

The height from which the branch is thrown = 3.00 m

The horizontal distance the branch lands from where it was thrown, x = 8.00 m

The direction in which the branch is thrown = Horizontally

Therefore, the initial vertical velocity of the branch, u_y = 0 m/s

The time it takes an object in free fall (zero initial downward vertical velocity) to reach the ground is given as follows;

s = u_y·t + 1/2·g·t²

Where;

u_y = 0 m/s

s = The initial height of the object = 3.00 m

g = The acceleration due to gravity = 9.8 m/s²

∴ s = 0·t + 1/2·g·t² = 0 × t + 1/2·g·t² = 1/2·g·t²

t = √(2·s/g) = √(2 × 3/9.8) = (√30)/7 ≈ 0.78246

The horizontal distance covered before the branch touches the ground, x = 8.00 m

Therefore, the initial velocity in the horizontal, x-direction with which the branch was thrown, 'uₓ', is given as follows;

uₓ = x/t = 8.00 m/((√30)/7 s)

Using a graphing calculator, we get;

uₓ = 8.00 m/((√30)/7 s) = (28/15)·√30 m/s ≈ 10.224 m/s

The initial velocity in the horizontal, x-direction with which the branch was thrown, uₓ ≈ 10.224 m/s.

3 0
3 years ago
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