All categories

1 year ago

4. Calculate the velocity of the toy car from 0-3 seconds.

Physics

1 answer:

zvonat [6]1 year ago

8 0

Velocity refers to the change in the position of the object with time.

<h3>What is the velocity?</h3>

The term velocity refers to the change in the position of the object with time. In this case, we have the position time graph that we could use to obtain the velocity of the body at any interval. The velocity is the slope of the position time graph.

4) velocity of the toy car from 0-3 seconds;

v = Δx/Δ t = 3 - 0/ 1.5 - 0 = 2 m/s

5) velocity of the toy car from 5-8 seconds

v = Δx/Δ t = 6 - 3/8 - 5 = 3/3 = 1 m/s

v = Δx/Δ t = 0 m/s

Learn more about velocity:brainly.com/question/18084516

#SPJ1

You might be interested in

The tape in a videotape cassette has a total

xxMikexx [17]

Answer:

$w=19.76 \ rad/s$

Explanation:

<u>Circular Motion </u>

Suppose there is an object describing a circle of radius r around a fixed point. If the relation between the angle of rotation by the time taken is constant, then the angular speed is also constant. If that relation increases or decreases at a constant rate, the angular speed is given by:

$w=w_o+\alpha \ t$

Where $\alpha$ is the angular acceleration and t is the time. If the object was instantly released from the circular path, it would have a tangent speed of:

$\displaystyle v_t=w.r$

We have two reels: one loaded with the tape to play and the other one empty and starting to fill with tape. They both rotate at different angular speeds, one is increasing and the other is decreasing as the tape goes from one to the other. We'll assume the tangent speed is constant for both (so the tape can play correctly). Let's call $w_1$ the angular speed of the loaded reel and $w_2$ that from the empty reel. We have

$w_1=w_{o1}+\alpha_1 \ t$

$w_2=w_{o2}+\alpha_2 \ t$

If $r_f=35\ mm=0.035\ m$ is the radius of the reel when it's full of tape, the angular speed for the loaded reel is computed by

$\displaystyle w_{01}=\frac{v_t}{r_f}$

The tangent speed is computed by knowing the length of the tape and the time needed to fully play it.

$t=1.8\ h=1.8*3600=6480\ sec$

$\displaystyle v_t=\frac{x}{t}=\frac{249\ m}{6480\ sec}=0.0384 \ m/s$

$\displaystyle w_{01}=\frac{0.0384}{0.035}=1,098 \ rad/s$

If $r_e=10\ mm=0.001\ m$ is the radius of the reel when it's empty, the angular speed for the empty reel is computed by

$\displaystyle w_{02}=\frac{0.0384}{0.001}=38.426 \ rad/s$

The full reel goes from $w_{01}$ to $w_{02}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_1=\frac{w_{02}-w_{01}}{6480}$

$\displaystyle \alpha_1=\frac{38.426-1.098}{6480}=0.00576 \ rad/sec^2$

The empty reel goes from $w_{02}$ to $w_{01}$ in 6480 seconds, so we can compute the angular acceleration:

$\displaystyle \alpha_2=\frac{1.098-38.426}{6480}=-0.00576 \ rad/sec^2$

So the equations for both reels are

$w_1=1.098+0.00576 \ t$

$w_2=38.426-0.00576 \ t$

They will be the same when

$1.098+0.00576 \ t=38.426-0.00576 \ t$

Solving for t

$\displaystyle t=\frac{38.426-1.098}{0.0115}$

$t=3240 \ sec$

The common angular speed is

$w_1=1.098+0.00576 \ 3240=19.76 \ rad/s$

$w_2=38.426-0.00576 \ 3240=19.76 \ rad/s$

They both result in the same, as expected

$\boxed{w=19.76 \ rad/s}$

6 0

3 years ago

Convert 68852 millijoules into Calories. (Write your answer in the decimal form. Do not include units in your answer).

alex41 [277]

Answer: 68852 millijoules = 16.46 calories

Explanation:

Given;

Convert 68852 millijoules to calories.

1 calorie = 4.184J = 4184millijoules

Therefore,

1 millijoule = 1/4184 calories

68852 millijoule = 68852 × 1/4184 calories

= 16.46 calories

6 0

3 years ago

Compared to a plane gravitational pull on the ground to 7 miles in sky is what

Citrus2011 [14]

When you're in an airplane that's 7 miles up off the ground, the strength of gravity plunges to only 99.6 percent of its strength all the way down on the ground. A big heavy person, who weighs 200 pounds down at the airport, weighs only 199 pounds 4.7 ounces in a plane at the altitude of 7 miles.

4 0

3 years ago

Suppose that a comet has a very eccentric orbit that brings it quite close to the Sun at closest approach (perihelion) and beyon

Nutka1998 [239]

Answer:

16.63min

Explanation:

The question is about the period of the comet in its orbit.

To find the period you can use one of the Kepler's law:

$T^2=\frac{4\pi}{GM}r^3$

T: period

G: Cavendish constant = 6.67*10^-11 Nm^2 kg^2

r: average distance = 1UA = 1.5*10^11m

M: mass of the sun = 1.99*10^30 kg

By replacing you obtain:

$T=\sqrt{\frac{4\pi}{GM}r^3}=\sqrt{\frac{4\pi^2}{(6.67*10^{-11}Nm^2/kg^2)(1.99*10^{30}kg)}(1.496*10^8m)^3}\\\\T=997.9s\approx16.63min$

the comet takes around 16.63min

8 0

3 years ago

Which of the following could be vector magnitudes?

SOVA2 [1]

Answer:

Both A and D

Explanation:

Vector magnintude contains both speed and direction and so do these answer choices of 15km and 30m/s

5 0

3 years ago

Read 2 more answers