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
BARSIC [14]
3 years ago
5

A frictionless pulley of negligible mass is hung from the ceiling using a rope, also of negligible mass. Two masses, m1 and m2 (

m1 < m2), are connected to the rope over the pulley. The masses are free to drop. The magnitude of the tension Ttop is ____ the sum of the tensions T1 and T2.
Physics
1 answer:
kirza4 [7]3 years ago
4 0

Answer:

Ttop = T1 + T2

Explanation:

The three tension forces act on the pulley. With the assumption that the pulley has a negligible mass, the force of gravity on it is zero. So only forces Ttop, T1 and T2 act on the pulley.

The tension force Ttop acts upwards on the pulley while the tension forces T1 and T2 act on the pulley downwards. The pulley is in equilibrium so the sum of upward forces acting on the pulley and the sum of downward forces acting on the pulley must be equal.

So

Ttop = T1 + T2.

You might be interested in
The graph in the accompanying figure (Figure 1) shows the magnitude of the force exerted by a given spring as a function of the
serious [3.7K]

The elastic potential energy stored in the stretched spring is 1 J.

<h3>What is Hooke's law?</h3>

Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.

Given that;

Force on the spring = 350 Newton

Distance stretched = 7 centimeters or 0.07 m

Hence;

F = ke

k = F/e = 350 Newton/0.07 m = 5000 N/m

Work done in stretching a spring = 1/2ke^2

= 0.5 × 5000 × (2 × 10^-2)^2 =1 J

Learn more about elastic potential energy: brainly.com/question/156316

4 0
2 years ago
Please help answer question​
nika2105 [10]

Answer:

C = 1.01

Explanation:

Given that,

Mass, m = 75 kg

The terminal velocity of the mass, v_t=60\ m/s

Area of cross section, A=0.33\ m^2

We need to find the drag coefficient. At terminal velocity, the weight is balanced by the drag on the object. So,

R = W

or

\dfrac{1}{2}\rho CAv_t^2=mg

Where

\rho is the density of air = 1.225 kg/m³

C is drag coefficient

So,

C=\dfrac{2mg}{\rho Av_t^2}\\\\C=\dfrac{2\times 75\times 9.8}{1.225\times 0.33\times (60)^2}\\\\C=1.01

So, the drag coefficient is 1.01.

4 0
3 years ago
an object of mass 8 kg is whriled round in a vertical circle of radius 2m with a constant speed of 6m/s .Then the maximum and mi
algol13

Answer:

Maximum Tension=224N

Minimum tension= 64N

Explanation:

Given

mass =8 kg

constant speed = 6m/s .

g=10m/s^2

Maximum Tension= [(mv^2/ r) + (mg)]

Minimum tension= [(mv^2/ r) - (mg)]

Then substitute the values,

Maximum Tension= [8 × 6^2)/2 +(8×9.8)] = 224N

Minimum tension= [8 × 6^2)/2 -(8×9.8)]

=64N

Hence, Minimum tension and maximum Tension are =64N and 2224N respectively

5 0
2 years ago
A density triangle is a tool that can be used to find
Anna [14]
The answer is D
Hope this helps
4 0
3 years ago
A bicycle rider pushes a 13kg bicycle up a steep hill. the incline is 24 degree and the road is 275m long. the rider pushes the
Digiron [165]

Answer:

A. W = 6875.0 J.

B. W = -14264.6 J.

Explanation:

A. The work done by the rider can be calculated by using the following equation:

W_{r} = |F_{r}|*|d|*cos(\theta_{1})

Where:                

F_{r}: is the force done by the rider = 25 N

d: is the distance = 275 m

θ: is the angle between the applied force and the distance

Since the applied force is in the same direction of the motion, the angle is zero.

W_{r} = |F_{r}|*|d|*cos(0) = 25 N*275 m = 6875.0 J

Hence, the rider does a work of 6875.0 J on the bike.

B. The work done by the force of gravity on the bike is the following:

W_{g} = |F_{g}|*|d|*cos(\theta_{2})  

The force of gravity is given by the weight of the bike.

F_{g} = -mgsin(24)     

And the angle between the force of gravity and the direction of motion is 180°.

W_{g} = |mgsin(24)|*|d|*cos(\theta_{2})  

W_{g} = 13 kg*9.81 m/s^{2}*sin(24)*275 m*cos(180) = -14264.6 J  

The minus sign is because the force of gravity is in the opposite direction to the motion direction.

Therefore, the magnitude of the work done by the force of gravity on the bike is 14264.6 J.  

I hope it helps you!                                                                                          

3 0
3 years ago
Other questions:
  • What is the momentum of an object that is traveling at 3 m/s and has a mass of 5 kg?
    5·1 answer
  • Which will result in positive buoyancy and cause the object to float?
    11·2 answers
  • The water skier has a mass of 73 kg. Find the magnitude of the
    11·1 answer
  • The region, or sphere, of influence around a magnet is called the magnetic_____.
    7·2 answers
  • 1. A giri rides her bike at 15 m/s for 20 s. How far does she travel in that time?
    6·2 answers
  • The name used to describe the range of EM waves
    10·1 answer
  • The electric motor of a model train accelerates the train from rest to 0.685 m/s in 21.5 ms. The total mass of the train is 875
    8·1 answer
  • A 2.0-kg mass is oscillating about the origin at 24 rad/s. The amplitude of the oscillations is 0.040 m. At what position is the
    9·1 answer
  • What is the acceleration for an object moving with constant speed? Explain your answer
    6·2 answers
  • Physics Question 100 Points No Spam.
    10·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!