answer.
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
mixer [17]
2 years ago
6

In the Atwood machine shown below, m1 = 2.00 kg and m2 = 6.05 kg. The masses of the pulley and string are negligible by comparis

on. The pulley turns without friction and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at vi = 3.00 m/s downward.. (a) How far will m1 descend below its initial level?
Physics
1 answer:
Rus_ich [418]2 years ago
5 0
M1 descending
−m1g + T = m1a 

m2 ascending
m2g − T = m2a

this gives :
(m2 − m1)g = (m1 + m2)a 

a = (m2 − m1)g/m1 + m2
   = (5.60 − 2)/(2 + 5.60) x 9.81 
   = = 4.65m/s^2
You might be interested in
During the 440, a runner changes his speed as he comes out of the curve onto the home stretch from 18 ft/sec to 38 ft/sec over a
Sloan [31]

Answer:

6.67ft/s^2

Explanation:

We are given that

Initial velocity=u=18ft/s

Final velocity,v=38ft/s

Time=t=3 s

We have to find the average acceleration over that 3 s period.

We know that

Average acceleration,a=\frac{v-u}{t}{t}

Using the formula

Average acceleration,a=\frac{38-18}{3}ft/s^2

Average acceleration,a=\frac{20}{3}ft/s^2

Average acceleration,a=6.67ft/s^2

Hence, the average acceleration=6.67ft/s^2

5 0
2 years ago
You throw a tennis ball (mass 0.0570 kg) vertically upward. It leaves your hand moving at 15.0 m/s. Air resistance cannot be neg
Deffense [45]

Answer:195 J

Explanation:

Given

mass of ball m=0.0570\ kg

ball leaves the hand with u=15\ m/s

maximum height reached by ball h=8\ m

Initial Mechanical energy when ball just leaves the hand

M.E._1=(P.E.+K.E.)_1

M.E._1=(mgh)_1+(\frac{1}{2}mv^2)_1

considering hand to be datum so h_1=0[/tex]

so Potential energy at ground is zero

M.E._1=\frac{1}{2}\times m\times (15)^2

M.E._1=6.41\ J

Mechanical Energy at highest point

(M.E.)_2=(P.E.+K.E.)_2

at highest Point velocity is zero

(M.E.)_2=mgh_2+0

(M.E.)_2=0.0570\times 9.8\times 8

(M.E.)_2=4.46\ J

Decrease in Mechanical energy

(M.E.)_1-(M.E.)_2=6.41-4.46

(M.E.)_1-(M.E.)_2=1.95\ J

3 0
2 years ago
When a car accelerates from a standing start, the crash test dummy appears to be pressed backward into the seat cushion. Which o
Setler [38]

<u>Answer:</u>

Option: D. Gravity is pulling the crash test dummy in the direction the car is moving.

<u>Explanation: </u>

When a car accelerates from a standing start, the crash test dummy appears to be pressed backward into the seat cushion because the gravity is pulling the crash test dummy in the direction the car is moving.  

Basically when the car is starting, the person inside is in static position and the car is going to move. So it is putting a force on the person to move on the same speed. But as the person is sitting static hence gravity is pulling him behind from moving. Hence, The dummy appears to be pressed backward.

7 0
2 years ago
A 1.15-kg mass oscillates according to the equation x = 0.650 cos(8.40t) where x is in meters and t in seconds. Determine (a) th
zheka24 [161]

Answer:

(a) A = 0.650 m

(b) f = 1.3368 Hz

(c) E = 17.1416 J

(d)  K = 11.8835 J

     U = 5.2581 J

Explanation:

Given

m = 1.15 kg

x = 0.650 cos (8.40t)

(a) the amplitude,

A = 0.650 m

(b) the frequency,

if we know that

ω = 2πf = 8.40    ⇒   f = 8.40 / (2π)

⇒   f = 1.3368 Hz

(c) the total energy,

we use the formula

E = m*ω²*A² / 2

⇒  E = (1.15)(8.40)²(0.650)² / 2

⇒  E = 17.1416 J

(d) the kinetic energy and potential energy when x = 0.360 m.

We use the formulas

K = (1/2)*m*ω²*(A² - x²)       (the kinetic energy)

and

U = (1/2)*m*ω²*x²              (the potential energy)

then

K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)

⇒  K = 11.8835 J

U = (1/2)*(1.15)*(8.40)²*(0.360)²

⇒  U = 5.2581 J

4 0
2 years ago
Based on the article “Will the real atomic model please stand up?,” why did J.J. Thomson experiment with cathode ray tubes? to s
PIT_PIT [208]

Answer:

B.) to determine that electric beams in cathode ray tubes were actually made of particles

Explanation:

This is the right answer i just took the quiz on edge.

3 0
2 years ago
Other questions:
  • Does the rankine degree represent a larger or smaller temperature unit than the kelvin degree
    12·1 answer
  • A straight, nonconducting plastic wire 9.50 cm long carries a charge density of 130 nC/m distributed uniformly along its length.
    5·1 answer
  • Two red blood cells each have a mass of 9.05×10−14 kg and carry a negative charge spread uniformly over their surfaces. The repu
    9·1 answer
  • A 50-kg platform diver hits the water below with a kinetic energy of 5000 Joules. The height (relative to the water) from which
    15·1 answer
  • How does the sun transfer energy to Earth?
    12·2 answers
  • Calculate the true mass (in vacuum) of a piece of aluminum whose apparent mass is 4.5000 kgkg when weighed in air. The density o
    5·1 answer
  • ery large accelerations can injure the body, especially if they last for a considerable length of time. One model used to gauge
    8·1 answer
  • Tyler drives 50km north. Tyler then drives back 30km south. What distance did he cover? What was his displacement?
    11·1 answer
  • A closed, rigid container holding 0.2 moles of a monatomic ideal gas is placed over a Bunsen burner and heated slowly, starting
    12·1 answer
  • The coefficient of linear expansion of lead is 29 x 10 K What change in temperature will cause a 5-m long lead bar to change in
    5·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!