1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
mylen [45]
3 years ago
11

An 8.00 kg mass moving east at 15.4 m/s on a frictionless horizontal surface collides with a 10.0 kg object that is initially at

rest. After the collision, the 8.00 kg object moves south at 3.90 m/s. (a) What is the velocity of the 10.0 kg object after the collision
Physics
1 answer:
andrew-mc [135]3 years ago
6 0

Answer:

9.3m/s

Explanation:

Based on the law of conservation of momentum

Sum of momentum before collision = sum of momentum after collision

m1u1 +m2u2 = m1v1+m2v2

m1 = 8kg

u1 = 15.4m/s

m2 = 10kg

u2 = 0m/s(at rest)

v1 = 3.9m/s

Required

v2.

Substitute

8(15.4)+10(0) = 8(3.9)+10v2

123.2=31.2+10v2

123.2-31.2 = 10v2

92 = 10v2

v2 = 92/10

v2 = 9.2m/s

Hence the velocity of the 10.0 kg object after the collision is 9.2m/s

You might be interested in
The amount of time required for 2 successive wave crests to pass a fixed point is called wave ________.
mihalych1998 [28]
I believe this is known as wave period.
hope this helps!
7 0
3 years ago
A dad takes his kids to their school just 8.0 miles down the road but with traffic it takes him 30 minutes and the fastest he ca
VARVARA [1.3K]

Answer:C 24 mi/hr

Explanation:

8 0
3 years ago
A centrifuge is a device used to separate materials by their masses. A sample in a centrifuge is rotated at high speeds along a
german

Answer:

 F = 1.047 10⁻² N

Explanation:

Let's use kinematics to find the angular acceleration

             w = w₀ + α t

as for rest w₀ = 0

             w = α t

             α = w / t

let's reduce the magnitudes to the SI system

              w = 1000 rev / min (2π rad/ 1 rev) (1 min/ 60s) = 104.72 rad / s

              m = 1.00 g (1 kg / 1000 g) = 1,000 10⁻³ kg

              r = 10.0 cm (1 m / 100 cm) = 0.100 m

let's calculate

              α = 104.72 / 1

              α = 104.72 rad / s²

angular and linear variables are related

               a = α  r

               a = 104.72 0.100

               a = 10.47 m / s²

finally we substitute in Newton's second law

               F = 1 10⁻³ 10.47

               F = 1.047 10⁻² N

8 0
3 years ago
A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th
svp [43]

a) x(t)=2.0 sin (10 t) [m]

The equation which gives the position of a simple harmonic oscillator is:

x(t)= A sin (\omega t)

where

A is the amplitude

\omega=\sqrt{\frac{k}{m}} is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s

The amplitude, A, can be found from the maximum velocity of the spring:

v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m

So, the equation of motion is

x(t)= 2.0 sin (10 t) [m]

b)  t=0.10 s, t=0.52 s

The potential energy is given by:

U(x)=\frac{1}{2}kx^2

While the kinetic energy is given by:

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

The velocity as a function of time t is:

v(t)=v_{max} cos(\omega t)

The problem asks as the time t at which U=3K, so we have:

\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}

However, \frac{m}{k}=\frac{1}{\omega^2}, so we have

(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\

with two solutions:

\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s

\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s

c) 3 seconds.

When x=0, the equation of motion is:

0=A sin (\omega t)

so, t=0.

When x=1.00 m, the equation of motion is:

1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s

The period of a pendulum is:

T=2 \pi \sqrt{\frac{L}{g}}

where L is the length of the pendulum. By using T=0.628 s, we find

L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m






5 0
3 years ago
The Sun is an enormous ball of gas. Left to itself, a ball of so many atoms should collapse under its own tremendous gravity. Wh
ikadub [295]

Answer:

Nuclear fusion in the Sun's core causes the release of tremendous amounts of energy that leads to very high temperatures and pressure which is much hotter and higher than the temperature and pressure at the exterior surface of the Sun causing the particles in the inner core region to push outwards towards the Sun's surface

Explanation:

3 0
2 years ago
Other questions:
  • Accourding to biopsychologists a temperament ()
    10·1 answer
  • What is the internal energy of 2.00 mol of diatomic hydrogen gas (H2) at 35°C?
    14·1 answer
  • A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.0 ft/s,
    11·2 answers
  • Objects have a tendency to resist changing their motion. This property is called: *
    13·1 answer
  • How much heat per second h (=qδt) flows from the boiling water to the ice-water mixture? express your answer in watts?
    12·1 answer
  • If A bar + B bar = C bar and magnitude of A bar + magnitude of B bar = magnitude of C bar. Then the angle between A bar and B ba
    6·1 answer
  • What are the largest and smallest resistances you can obtain by connecting a 36.0-Ω , a 50.0-Ω , and a 700-Ω resistor together?
    6·1 answer
  • Did the aluminum foil and the paper tent versoriums behave the way you predicted? What did you learn that could help you improve
    8·1 answer
  • 2 Select the correct answer. Javed suffers from a severe food allergy. Which device should Javed always carry? A. inhaler B. epi
    11·1 answer
  • A bag contains lenses with focal lengths 10 cm, 20 cm and 25 cm which are not marked with their focal length. Describe a simple
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!