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Lelu [443]
3 years ago
8

Two masses are connected by a string which passes over a pulley with negligible mass and friction. One mass hangs vertically and

one mass slides on a horizontal surface. The horizontal surface has a coefficient of kinetic friction of 0.200. The vertically hanging mass is 3.00 kg and the mass on the horizontal surface is 3.00 kg. The magnitude of the acceleration of the vertically hanging mass is (the initial velocity of the horizontal mass is to the right)
Physics
1 answer:
monitta3 years ago
3 0

Answer:

a=2,5m/s^2

Explanation:

From the question we are told that:

Coefficient of kinetic friction \mu= 0.200

Vertical Mass M_v=3kg

Horizontal mass M_h=3.00kg  

Generally the equation for kinetic force F_k is mathematically given by

F_k=\mu N\\F_k=0.2*3\\F_k=0.6

Generally the equation for T is mathematically given by

For M_v=3kg3g-T=3a

For M_h=3kg

T=M_v V+F_k\\T=3.0a+0.6

Therefore substituting

3-3a-0.6=3a\\2.4g=6a

a=2,5m/s^2

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Can anyone help me? (physics)
Masja [62]

Answer:

The initial velocity of the golf is 15.7 m/s.

The direction of the golf is 57°.

Explanation:

The following data were obtained from the question:

Time of flight (T) = 2.7 secs

Range (R) = 23 m

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =.?

Direction (θ) =.?

T = 2U Sine θ /g

2.7 = 2 × U × Sine θ /9.8

Cross multiply

2.7 × 9.8 = 2 × U × Sine θ

26.46 = 2 × U × Sine θ

Divide both side by 2 × Sine θ

U = 26.46 /2 Sine θ

U = 13.23 / Sine θ ... (1)

R = U² Sine 2θ /g

23 = U² Sine 2θ / 9.8

U = 13.23 / Sine θ

23 = (13.23/ Sine θ)² Sine 2θ / 9.8

23 = (175.0329 / Sine² θ) × Sine 2θ / 9.8

23 = 17.8605/Sine² θ × Sine 2θ

Recall:

Sine 2θ = 2SineθCosθ

23 = 17.8605/ Sine² θ × 2SineθCosθ

23 = 17.8605/ Sine θ × 2Cosθ

23 = 35.721 Cos θ /Sine θ

Cross multiply

23 × Sine θ = 35.721 Cos θ

Divide both side by 23

Sine θ = 35.721 Cos θ /23

Sine θ = 1.5531 × Cos θ

Divide both side by Cos θ

Sine θ /Cos θ = 1.5531

Recall:

Sine θ /Cos θ = Tan θ

Sine θ /Cos θ = 1.5531

Tan θ = 1.5531

Take the inverse of Tan

θ = Tan¯¹ (1.5531)

θ = 57°

Therefore, the direction of the golf is 57°

Thus, the initial velocity can be obtained as follow:

U = 13.23 / Sine θ

θ = 57°

U = 13.23 / Sine 57

U = 13.23/0.8387

U = 15.7 m/s

Therefore, the initial velocity of the golf is 15.7 m/s

8 0
3 years ago
What force is necessary to accelerate a 2500 kg care from rest to 20 m/s over 10 seconds?
EleoNora [17]
Force = mass x acceleration
force = 2500kg x (20m/s / 10m/s)
force = 2500kg x 2m/s^2
force = 5000kg m/s^2 = 5kN

i hope this is right (^^)
4 0
2 years ago
If a force of 19 N is applied to a crate to push it 23 m, how much work is done on the crate by the force?
swat32

Answer:

437 Joules

Explanation:

Use the formula for work directly

(work) = (force) x (displacement)

to get

(work) = (19 N) x (23 m) = 437 Joules

4 0
3 years ago
Earth's gravity acts upon objects with a steady force of __________.
luda_lava [24]
None of the choices is a force. 'a' and 'b' are speeds. 'C' and 'd' are accelerations. ... The steady force of gravity is 9.8 newtons PER KILOGRAM of mass. ... The question is written by someonewho very much wants to discourage anyone interested in Physics.
6 0
3 years ago
Read 2 more answers
a body initially at rest, starts moving with a constant acceleration of 2ms-2 .calculate the velocity acquired and the distance
Marta_Voda [28]

a) 10 m/s

b) 25 m

Explanation:

a)

The body is moving with a constant acceleration, therefore we can solve the problem by using the following suvat equation:

v=u+at

where

u is the initial velocity

v is the final velocity

a is the acceleration

t is the time

For the body in this problem:

u = 0 (the body starts from rest)

a=2 m/s^2 is the acceleration

t = 5 s is the time

So, the final velocity is

v=0+(2)(5)=10 m/s

b)

In this second part, we want to calculate the distance travelled by the body.

We can do it by using another suvat equation:

v^2-u^2=2as

where

u is the initial velocity

v is the final velocity

a is the acceleration

s is the distance travelled

Here we have

u = 0 (the body starts from rest)

a=2 m/s^2 is the acceleration

v = 10 m/s is the final velocity

Solving for s,

s=\frac{10^2-0^2}{2(2)}=25 m

3 0
3 years ago
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