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Over [174]
3 years ago
7

A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h

orizontal. At the lowest point in its swing when it is moving horizontally, the ball collides elastically with a 2.0-kg block initially at rest on a horizontal frictionless surface. What is the speed of the block just after the collision
Physics
1 answer:
hram777 [196]3 years ago
8 0

Answer:

v_{2}=3.5 m/s

Explanation:

Using the conservation of energy we have:

\frac{1}{2}mv^{2}=mgh

Let's solve it for v:

v=\sqrt{2gh}

So the speed at the lowest point is v=7 m/s

Now, using the conservation of momentum we have:

m_{1}v_{1}=m_{2}v_{2}

v_{2}=\frac{1*7}{2}

Therefore the speed of the block after the collision is v_{2}=3.5 m/s

I hope it helps you!

       

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A person weighing 0.9 kN rides in an elevator that has a downward acceleration of 1.9 m/s^2. The acceleration of gravity is 9.8
Keith_Richards [23]

Answer:

The magnitude of the force is 0.7255kN

Explanation:

The elevator floor acts on the person with a force that is due to the gravitational acceleration less the downward acceleration of the elevator:

(force of floor F) = (mass of person m) x [ (grav. acceleration g) - (elevator acceleration a) ]

in other words, considering the elevator floor as a reference frame in the Earth's gravitational field, the person's weight decreases due to the downward acceleration, as follows:

F = m\cdot(g-a)

We are given the person's weight at rest, 0.9kN, from which the mass can be determined as:

900 N = m\cdot g \implies m = \frac{900N}{9.8 \frac{m}{s^2}}

So

F = \frac{900N}{9.8 \frac{m}{s^2}}\cdot(9.8-1.9)\frac{m}{s^2}\approx 725.5N=0.7255kN

3 0
2 years ago
PLZZ HELP ASAP!!
Nimfa-mama [501]
Modern space suits augment the basic pressure garment with a complex system of equipment and environmental systems designed to keep the wearer comfortable, and to minimize the effort required to bend the limbs, resisting a soft pressure garment's natural tendency to stiffen against the vacuum. A self-contained oxygen supply and environmental control system is frequently employed to allow complete freedom of movement, independent of the spacecraft.
Three types of spacesuits exist for different purposes: IVA (intravehicular activity), EVA (extravehicular activity), and IEVA (intra/extravehicular activity). IVA suits are meant to be worn inside a pressurized spacecraft, and are therefore lighter and more comfortable. IEVA suits are meant for use inside and outside the spacecraft, such as the Gemini G4C suit. They include more protection from the harsh conditions of space, such as protection from micrometeorites and extreme temperature change. EVA suits, such as the EMU, are used outside spacecraft, for either planetary exploration or spacewalks. They must protect the wearer against all conditions of space, as well as provide mobility and functionality.
8 0
3 years ago
How much force is required to move a sled 5 meters if a person uses 60 j of work? 300N 12N 65N 30N
maksim [4K]
The answer is b 12N because
We know that<span>
W = F × d × c o s(θ)</span>
assuming theta=0 we then solve and have<span>
F=<span>W/d</span></span>
substitute known values to get:<span><span>
F=<span><span>60J/</span><span>5m</span></span>=12N</span></span>
4 0
3 years ago
Besides a reduction in friction, the only way to increase the amount of work output of a machine is to _____ the work input. Dec
Vedmedyk [2.9K]

Answer:

besides a reduction in friction, the only way to increase the amount of work output of a machine is to Increase the work input

i

Explanation:

7 0
2 years ago
If y = 0.02 sin (30x – 200t) (SI units), the frequency of the wave is
leva [86]

Answer:

31.831 Hz.

Explanation:

<u>Given:</u>

  • \rm y = 0.02\sin(30x-200 t).

The vertical displacement of a wave is given in generalized form as

\rm y = A\sin(kx -\omega t).

<em>where</em>,

  • A = amplitude of the displacement of the wave.
  • k = wave number of the wave = \dfrac{2\pi }{\lambda}.
  • \lambda = wavelength of the wave.
  • x = horizontal displacement of the wave.
  • \omega = angular frequency of the wave = \rm 2\pi f.
  • f = frequency of the wave.
  • t = time at which the displacement is calculated.

On comparing the generalized equation with the given equation of the displacement of the wave, we get,

\rm A=0.02.\\k=30.\\\omega =200.\\

therefore,

\rm 2\pi f=200\\\\\Rightarrow f = \dfrac{200}{2\pi}=31.831\ Hz.

It is the required frequency of the wave.

3 0
2 years ago
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