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2 years ago

Explain the centripetal force or Newton's second law...

2 answers:

8 0

Answer: a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton's second law of motion, net force is mass times acceleration: net . For uniform circular motion, the acceleration is the centripetal acceleration

Explanation:

stepan [7]2 years ago

6 0

Newton’s second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

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A crate rests on a flatbed truck which is initially traveling at 17.9 m/s on a level road. The driver applies the brakes and the

Mamont248 [21]

Answer:

The minimum coefficient of friction required is 0.35.

Explanation:

The minimum coefficient of friction required to keep the crate from sliding can be found as follows:

$-F_{f} + F = 0$

$-F_{f} + ma = 0$

$\mu mg = ma$

$\mu = \frac{a}{g}$

Where:

μ: is the coefficient of friction

m: is the mass of the crate

g: is the gravity

a: is the acceleration of the truck

The acceleration of the truck can be found by using the following equation:

$v_{f}^{2} = v_{0}^{2} + 2ad$

$a = \frac{v_{f}^{2} - v_{0}^{2}}{2d}$

Where:

d: is the distance traveled = 46.1 m

$v_{f}$ : is the final speed of the truck = 0 (it stops)

$v_{0}$ : is the initial speed of the truck = 17.9 m/s

$a = \frac{-(17.9 m/s)^{2}}{2*46.1 m} = -3.48 m/s^{2}$

If we take the reference system on the crate, the force will be positive since the crate will feel the movement in the positive direction.

$\mu = \frac{a}{g}$

$\mu = \frac{3.48 m/s^{2}}{9.81 m/s^{2}}$

$\mu = 0.35$

Therefore, the minimum coefficient of friction required is 0.35.

I hope it helps you!

4 0

3 years ago

Determine the mass the mass in Kilograms of a Object that has a weight of

34kurt

1. 20mn that’s the answer

6 0

2 years ago

Read 2 more answers

Water flows over the edge of a waterfall at a rate of 1.2 x 10^6 kg/s. There are 50.0 m between the top and bottom of the waterf

Anit [1.1K]

Answer:

5.88×10⁸ W

Explanation:

Power = change in energy / time

P = mgh / t

P = (m/t) gh

P = (1.2×10⁶ kg/s) (9.8 m/s²) (50.0 m)

P = 5.88×10⁸ W

4 0

3 years ago

Read 2 more answers

1. An object (m = 500 g) with an initial speed of 0.2 m/s collides with another object (m = 1.5 kg) which was at rest before the

Anettt [7]

Answer:

Explanation:

1 )

We shall apply conservation of momentum law to solve the problem.

mv = ( M +m) V , m and M are masses of small and large object , v is the velocity of small object before collision and V is the velocity of both the objects together after collision .

.5 x .2 = (1.5 + .5)V

V = .05 m /s

2 ) We shall use formula for velocity of object after elastic collision as follows

v₁ = $\frac{(m_1-m_2)}{(m_1+m_2)} u_1+\frac{2m_2u_2}{(m_1+m_2)}$

m₁ and m₂ are masses of first and second object u₁ and u₂ are their initial velocity and v₁ and v₂ are their final velocity.

Putting the values

= $\frac{(200-1000)}{(1000+200)} \times 1 +\frac{2 \times1000\times0}{(1000+200)}$

= - .66 m /s

Since the sign is negative so it will be in opposite direction .

4 0

3 years ago

PLZ HELP IM REALLY CONFUSED a boy is playing catch with his friend. He throws the ball straight up. When it leaves his hand, the

erastovalidia [21]

Answer:

K.E = 100 J

Final P.E = 100 J

Explanation:

The kinetic energy of any object can be given by the following formula:

$K.E = (\frac{1}{2})mv^{2}$

where,

K.E = Kinetic Energy

m = mass of ball = 2 kg

v = speed of ball

Initially, v = 10 m/s. Therefore, the initial K.E is given as:

$K.E = (\frac{1}{2})(2\ kg)(10\ m/s)^{2}$

Now, at the highest point the K.E of the ball becomes zero. because the ball stops for a moment at the highest point and its velocity becomes zero. So, from Law of Conservation of energy:

Initial K.E + Initial P.E = Final K.E + Final P.E

Initial P.E is also zero due to zero height initially.

K.E + 0 = 0 + Final P.E

<u>Final P.E = 100 J</u>

3 0

3 years ago