All categories

3 years ago

What impulse is needed to slow a 45kg object from 15m/s to 12m/s please show who to work problem

1 answer:

8 0

'Impulse' is a change in momentum.
Momentum = (mass) x (velocity)

Final momentum = (45 x12) = 540 kilogram-meters/second
Original momentum = (45 x 15) = 675 kg-m/sec
Change in momentum = -135 kg-m/sec

The impulse required is 135 kg-m/sec in the direction
opposite to the way the object is moving.

You might be interested in

light of wavelength 520 nm falls on a slit that is 3.20 um wide. Estimate how far the first bright-sh difrrection fringe is

Westkost [7]

Answer:

because of the gravity of the earth

6 0

2 years ago

Which theory explains that cussing in front of your grandparents or hugging a stranger would be okay to you based on social cont

Ainat [17]

Either concrete or cognitive

8 0

3 years ago

A transverse sinusoidal wave on a string has a period T= 25.0ms and travels in the negative x direction with a speed of 30.0 m/s

aleksandrvk [35]

The wave is equation with the given conditions is y = 0.02 cos ( 0 )

<u>Given data</u>

period T= 25.0ms

speed of 30.0 m/s

t = 0

x = 0

transverse position of 2.00cm

speed of 2.0 m/s = v

<h3>writing the wave function </h3>

frequency f = 1 / T

f = 1 / 25

f = 0.04Hz

Angular velocity = ω = 2 * pi * f

ω = 2 * pi * 0.04

ω = 0.251

wave number K = ω / v

k = 0.251 / 2

k = 0.126

The wave equation

y = A cos ( kx + ωt )

this is equivalent to

y = 0.02 cos ( 0.126 * 0 + 0.251 * 0 )

y = 0.02 cos ( 0 )

Read more on wave equation here: brainly.com/question/28167443

#SPJ4

6 0

1 year ago

Crest : trough :: compression : _____ A. frequency B. amplitude C. rarefaction D. wavelength

9966 [12]

Rarefraction.

Crest- tallest spot on transverse wave.

Trough- shortest point on transverse wave.

Compression - spot on a compressional wave where the waves are closer together.

Rarefraction - spot on a compressional wave where the waves are farther apart.

3 0

4 years ago

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

$\omega = 0\,\frac{rad}{s}$

7 0

4 years ago