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3 years ago

A dog runs after a car. The car is travelling at an average speed of 5 m/s.

Physics

2 answers:

n200080 [17]3 years ago

3 0

Answer:

<h2>The answer is 4 s</h2>

Explanation:

The time taken can be found by using the formula

$t = \frac{d}{v} \\$

where

d is the distance

v is the velocity

From the question we have

$t = \frac{20}{5} \\$

We have the final answer as

Hope this helps you

Lina20 [59]3 years ago

3 0

Answer:

4 secs

Explanation:

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Please help!! I will give brainliest!

grandymaker [24]

Answer:

Depends.

Explanation:

Whether the object is going left or right, the speed will stay the same until friction eventually stops it. <em>However, </em>if, for example, we're talking about an object going straight before veering right, then yes, speed <em>does</em> matter. An object will normally have to speed up or slow down momentarily when changing direction to keep itself sustained on the ground.

So, honestly? It really depends on what we're talking about!

Hope this helped!

Source(s) used: None.

7 0

3 years ago

1) A satellite in orbit around Earth is in uniform circular motion. What is the angle between the velocity vector and accelerati

artcher [175]

Answer:90

Explanation: bc i looked it up

6 0

1 year ago

FIGURE 2 shows a 1.5 kg block is hung by a light string which is wound around a smooth pulley of radius 20 cm. The moment of ine

Sindrei [870]

Answer:

At t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

Explanation:

First, we consider all the forces acting on the pulley.

There is only one force acting on the pulley, and that is due to the 1.5 kg mass attached to it.

Therefore, the torque on the pulley is

$\tau=Fd=mg\cdot R$

where m is the mass of the block, g is the acceleration due to gravity, and R is the radius of the pulley.

Now we also know that the torque is related to angular acceleration α by

$\tau=I\alpha$

therefore, equating this to the above equation gives

$mg\cdot R=I\alpha$

solving for alpha gives

$\alpha=\frac{mgR}{I}$

Now putting in m = 1.5 kg, g = 9.8 m/s^2, R = 20 cm = 0.20 m, and I = 2 kg m^2 gives

$\alpha=\frac{1.5\cdot9.8\cdot0.20}{2}$ $\boxed{\alpha=1.47s^{-2}}$

Now that we have the value of the angular acceleration in hand, we can use the kinematics equations for the rotational motion to find the angular velocity and the number of revolutions at t = 4.2 s.

The first kinematic equation we use is

$\theta=\theta_0+\omega_0t+\frac{1}{2}\alpha t^2$

since the pulley starts from rest ω0 = 0 and theta = 0; therefore, we have

$\theta=\frac{1}{2}\alpha t^2$

Therefore, ar t = 4.2 s, the above gives

$\theta=\frac{1}{2}(1.47)(4.2)^2$

$\boxed{\theta=12.97}$

So how many revolutions is this?

To find out we just divide by 2 pi:

$\#\text{rev}=\frac{\theta}{2\pi}=\frac{12.97}{2\pi}$ $\boxed{\#\text{rev}=2.06}$

Or about 2 revolutions.

Now to find the angular velocity at t = 4.2 s, we use another rotational kinematics equation:

$\omega^2=w^2_0+2\alpha(\Delta\theta)_{}$

Since the pulley starts from rest, ω0 = 0. The change in angle Δθ we calculated above is 12.97. The value of alpha we already know to be 1.47; therefore, the above becomes:

$\omega^2=0+2(1.47)(12.97)$ $w^2=38.12$ $\boxed{\omega=6.17.}$

Hence, the angular velocity at t = 4.2 w is 6. 17 rad / s

To summerise:

at t = 4.2 s

Angular velocity: 6. 17 rad /s

The number of revolutions: 2.06

3 0

1 year ago

What is the force needed to give a .25 kg arrow an acceleration of 196m/s2?

Rzqust [24]

<span>49N is the force needed to give a .25 kg arrow an acceleration of 196m/s2. F =ma â‡’ =( 0.25kg)(196m/s2) = 49N if the arrow is shot horizontally where the applied force is entirely in the x-direction.</span>

5 0

3 years ago

Read 2 more answers

A medieval instrument used to determine the position of the sun:

Stels [109]

Answer: astrolabe

Explanation:

4 0

2 years ago