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sladkih [1.3K]
2 years ago
14

Two students push a heavy crate across the floor. John pushes with a force of 185 N due east and Joan pushes with a force of 165

N at 30∘north of east. What is the resultant horizontal force on the crate?

Physics
1 answer:
zubka84 [21]2 years ago
3 0

Answer:

327.894 N

Explanation:

We have given two forces Jhon pushes with a force of 185 N in east direction and Joan pushes with a force of 165 N at 30° north of east which is clearly shown in figure

We fave to find the total horizontal force

So total horizontal force will be F_X=185+165cos30^{\circ} =327.894 N ( total horizontal force will be sum of force in the east direction and horizontal component of the force in north- east direction)

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Two identical metal balls are rolling without slipping along a horizontal surface with speed V. Each ball encounters a hill ('tw
Anastaziya [24]

Answer:

The angular velocity of Ball A will be greater than the angular velocity of Ball B when they reach the top of the hill.

Explanation:

Angular velocity can be defined as how fast an object rotates relative to a given point or frame of reference.

The question said the hill encountered by Ball A is frictionless, so Ball A will continue to rotate at the same rate it started with even when it reached the top of the hill.

Ball B on the other hand rolls without slipping over its hill, i.e there's friction to slow down its rotational motion which thus reduces how fast Ball B will rotate at the top of the hill

3 0
2 years ago
A force of 150 N accelerates a 25 kg wooden chair across a wood floor at 4.3 m/s2 . How big is the frictional force on the block
solniwko [45]
We can first calculate the net force using the given information.

By Newton's second law, F(net) = ma:

F(net) = 25 * 4.3 = 107.5

We can now calculate the frictional force, f, which is working against the applied force, F(app) (this is why the net force is a bit lower):

f = F(net) - F(app) = 150 - 107.5 = 42.5 N

Now we can calculate the coefficient of friction, u, using the normal force, F(N):

f = uF(n) --> u = f/F(N)
u = 42.5/[25(9.8)]
u = 0.17
4 0
2 years ago
A person weighing 0.70 kn rides in an elevator that has an upward acceleration of 1.5 m/s2. what is the magnitude of the normal
creativ13 [48]
First of all, we can find the mass of the person, since we know his weight W:
W=mg=0.70 kN=700 N
And so
m= \frac{W}{g}= \frac{700 N}{9.81 m/s^2}=71.4 kg

We know for Newton's second law that the resultant of the forces acting on the person must be equal to the product between the mass and the acceleration a of the person itself:
\sum F =  ma
There are only two forces acting on the person: his weight W (downward) and the vincular reaction Rv of the floor against the body (upward). So we can rewrite the previous equation as
R_v -W = ma
We know the acceleration of the system, a=1.5 m/s^2 (upward, so with same sign of Rv), so we can solve to find the value of Rv, the normal force exerted by the elevator's floor on the person:
R_v = ma+W=(71.4 kg)(1.5 m/s^2)+700 N =807N
8 0
2 years ago
A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the
Oksana_A [137]

Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

==> a = 900 ÷ 0.24 = 3750 m/sec2

Now we know;

F = ma

F= -Kx

==> ma= -kx

==> 20 × 3750 = -K × 0.12

==> k = 625000 N/ m

5 0
2 years ago
In this problem, you will analyze a system composed of two blocks, 1 and 2, of respective masses m1 and m2. To simplify the anal
Dafna1 [17]

Answer:

a) p = m1 v1 + m2 v2 , b) dp / dt = m1 a1 + m2 a2 , c) It is equivalent to force

dp / dt = 0

Explanation:

In this problem we have two blocks and the system is formed by the two bodies.

Part A. Initially they ask us to find the moment of the whole system

    p = m1 v1 + m2 v2

Part B.

Find the derivative

     dp / dt = m1 dv1dt + m2 dv2 / dt

     dp / dt = m1 a1 + m2 a2

Part C.

Let's analyze the dimensions

     m a = [kg] [m / s2] = [N]

It is equivalent to force

Part d

Acceleration is due to a net force applied

Part e

The acceleration of block 1 is due to the force exerted by block 2 during the moment change

Part f

Force of block 1 on block 2

True f12 = m1a1        f21 = m2a2

Part g

By the law of action and reaction are equal magnitude F12 = f21

Part H

     dp / dt = 0

Isolated system F12 = F21 and the masses are constant. The total moment is only redistributed

7 0
2 years ago
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