50 POINTS PLSS

a ball of mass 0.80 kg moving at a speed of 2.5 m/s along a straight line collided with a mass 2.5 kg which was initially statio

Likurg_2 [28]

The speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.

The given parameters;

mass of the ball, m₁ = 0.8 kg
speed of the ball, u₁ = 2.5 m/s
mass of the object at rest, m₂ = 2.5 kg
final velocity of the object at rest, v₂ = 1 m/s

Let the final velocity of the 0.8 kg ball immediately after collision = v₁

Apply the principle of conservation of linear momentum;

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.8 x 2.5) + (2.5 x 0) = (0.8)v₁ + 2.5(1)

2 = 2.5 + (0.8)v₁

-0.5 = (0.8)v₁

$v_1 = \frac{-0.5}{0.8} \\\\v_1 = -0.625 \ m/s$

Thus, the speed of the 0.8 kg ball immediately after collision is 0.625 m/s in opposite direction to the stationary ball.

Learn more here: brainly.com/question/7694106

5 0

2 years ago

Determine the maximum shearing stress in a solid shaft of 1.5-in. diameter as it transmits 75 hp at a speed of 1800 rpm. (Round

erma4kov [3.2K]

Answer:

$\tau=3.96\ ksi$

Explanation:

Given that

d= 1.5 in ( 1 in = 0.0254 m)

d= 0.0381 m

P= 75 hp ( 1 hp = 745.7 W)

P= 55927.5 W

N= 1800 rpm

We know that power P is given as

$P=\dfrac{2\pi N\ T}{60}$

T=Torque

N=Speed

$55927.5=\dfrac{2\times \pi \times 1800\ T}{60}$

T=296.85 N.m

The maximum shear stress is given as

$\tau=\dfrac{16 T}{\pi d^3}$

$\tau=\dfrac{16\times 296.85}{\pi \times 0.0381^3}$

$\tau=27.35\ MPa$

We know that 1 MPa =0.145 ksi

$\tau=3.96\ ksi$

3 0

3 years ago

Explain you could use a battery, wire and compass to

gladu [14]

Answer:

oh I'm so sorry I can't answer your question it has been a long time since I learned that. so I totally forgot how to do this. sorry!

5 0

3 years ago

A common physics demonstration is to drop a small magnet down a long, vertical aluminum pipe. Describe the motion of the magnet

Rzqust [24]

Answer and Explanation:

This experiment is known as Lenz's tube.

The Lenz tube is an experiment that shows how you can brake a magnetic dipole that goes down a tube that conducts electric current. The magnet, when falling, along with its magnetic field, will generate variations in the magnetic field flux within the tube. These variations create an emf induced according to Faraday's Law:

$\varepsilon =-\frac{d\phi_B}{dt}$

This emf induced on the surface of the tube generates a current within it according to Ohm's Law:

$V=IR$

This emf and current oppose the flux change, therefore a field will be produced in such a direction that the magnet is repelled from below and is attracted from above. The magnitude of the flux at the bottom of the magnet increases from the point of view of the tube, and at the top it decreases. Therefore, two "magnets" are generated under and above the dipole, which repel it below and attract above. Finally, the dipole feels a force in the opposite direction to the direction of fall, therefore it falls with less speed.

7 0

3 years ago

Read 2 more answers

A commuter train passes a passenger platform at a constant speed of 40.4 m/s. The train horn is sounded at its characteristic fr

mihalych1998 [28]

(a) -83.6 Hz

Due to the Doppler effect, the frequency of the sound of the train horn appears shifted to the observer at rest, according to the formula:

$f' = (\frac{v}{v\pm v_s})f$

where

f' is the apparent frequency

v = 343 m/s is the speed of sound

$v_s$ is the velocity of the source of the sound (positive if the source is moving away from the observer, negative if it is moving towards the observer)