Why is doing research helpful if you're creating the procedure for an experiment?

What impulse must be applied to a 25.0-kg cart to cause a velocity change<br> of 12.0 m/s?

klio [65]

Answer:

Impulse of force = 300Ns

Explanation:

Given the following data;

Mass = 25kg

Change in velocity = 12m/s

To find the impulse;

Impulse is given by the formula;

$Impulse \; of \; force = mass * change \; in \; velocity$

Substituting into the equation, we have;

$Impulse \; of \; force = 25 * 12$

Impulse of force = 300Ns

8 0

3 years ago

Two red and two overlapping balls in the center are surrounded by a green, fuzzy, circular cloud with a white line running throu

juin [17]

Kyriakos grizzly is the answer

5 0

2 years ago

A NASCAR driver weighing 75 kg enters a corner with a radius of 155 meters. If his acceleration is 100 m/s2. What is tangential

Ymorist [56]

The tangential velocity of the car is 124.5 m/s

Explanation:

The centripetal acceleration of an object in uniform circular motion is given by:

$a=\frac{v^2}{r}$

where

a is the acceleration

v is the tangential velocity

r is the radius of the circle

For the car in this problem, we have:

$a=100 m/s^2$ is the centripetal acceleration

r = 155 m is the radius of the turn

Solving for v, we find the tangential velocity of the car:

$v=\sqrt{ar}=\sqrt{(100)(155)}=124.5 m/s$

Learn more about centripetal acceleration:

brainly.com/question/2562955

#LearnwithBrainly

3 0

3 years ago

In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that

mars1129 [50]

Answer:

The maximum transverse speed of the bead is 0.4 m/s

Explanation:

As we know that the Amplitude of the travelling wave is

A = 3.65 mm

Now the speed of the travelling wave is

$v_x = 13.5 m/s$

now we know that distance of first antinode from one end is 27.5 cm

so length of the loop of the standing wave is given as

$\frac{\lambda}{4} = 27.5 cm$

$\lambda = 110 cm$

now we have

$N = \frac{2L}{\lambda}$

$N = \frac{2(1.65)}{1.10}$

$N = 3$

now we have

$R = 2A sin(kx)$

$R = 2(3.65) sin(\frac{2\pi}{1.10}x)$

$R = 7.3 sin(1.82 \pi x)$

now at x = 13.8 cm

$R = 7.3 sin(1.82 \pi (0.138))$

$R = 5.18 mm$

now we have

$f = \frac{v}{\lambda}$

$f = \frac{13.5}{1.1}$

$f = 12.27 Hz$

now maximum speed is given as

$v_y = R\omega$

$v_y = (5.18 \times 10^{-3})(2\pi(12.27))$

$v_y = 0.4 m/s$

4 0

3 years ago

A car traveling north slows down from 40 m/s to rest in a distance of 100 m. What was the magnitude of its average acceleration

Leviafan [203]

Answer:

8 m/s² south

Explanation:

Given:

v₀ = 40 m/s

v = 0 m/s

Δx = 100 m

Find: a

v² = v₀² + 2a (x − x₀)

0² = 40² + 2a (100)

a = -8

The acceleration is 8 m/s² south.

6 0

3 years ago