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zepelin [54]
2 years ago
5

The two blocks are connected by a light string that passes over a frictionless pulley with a negligible mass. The 2 kg block lie

s on a rough horizontal surface with a constant coefficient of kinetic friction 0.4. This block is connected to a spring with spring constant 4 N/m. The second block has a mass of 8 kg. The system is released from rest when the spring is unstretched, and the 8 kg block falls a distance h before it reaches the lowest point. The acceleration of gravity is 9.8 m/s2 . Note: When the 8 kg block is at the lowest point, its velocity is zero.
a. Calculate the falling distance h where the 8 kg blocks stops.
b. Consider the moment when the 8 kg block has descended by a distance of 21.168 m, where 21.168 m is less than h. At this moment calculate the sum of the kinetic energy for the two blocks. Answer in units of J.
Physics
1 answer:
frozen [14]2 years ago
5 0

Answer: YOU ARE A FAT NEGRO

Explanation:YOU ARE

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A charge moves a distance of 1.8 cm in the direction of a uniform electric field having a magnitude of 214 N/C. The electrical p
Andrei [34K]

Answer:

 13.4 x 10 raise to power -19  C

Explanation:

. The distance moved by a charge in the direction of a uniform electric field is d= 1.8 cm =0.018 m

. The uniform electric field is  E = 214 N/M

, The decrease in electrical potential energy is   d(P.E) = 51.63 x 10 raise to power -19 J

Let the magnitude of the charge of the moving particle be q

which is given by the equation

d(P.E) =qEd

51.63 x 10 power -19 = q(214)(0.018)

51.63 x 10 power -19 =3.852q

by making q the formular,

q = 13.4 x 10 power -19 C  

5 0
3 years ago
Consider the following mass distribution where the x- and y-coordinates are given in meters: 5.0 kg at (0.0, 0.0) m, 2.9 kg at (
Sauron [17]

Answer:

x = -1.20 m

y = -1.12 m

Explanation:

as we know that four masses and their position is given as

5.0 kg (0, 0)

2.9 kg (0, 3.2)

4 kg (2.5, 0)

8.3 kg (x, y)

As we know that the formula of center of gravity is given as

x_{cm} = \frac{m_1 x_1 + m_2x_2 + m_3x_3 + m_4x_4}{m_1 + m_2 + m_3 + m_4}

0 = \frac{5(0) + 2.9(0) + 4(2.5) + 8.3 x}{5 + 2.9 + 4 + 8.3}

10 + 8.3 x = 0

x = -1.20 m

Similarly for y direction we have

y_{cm} = \frac{m_1 y_1 + m_2y_2 + m_3y_3 + m_4y_4}{m_1 + m_2 + m_3 + m_4}

0 = \frac{5(0) + 2.9(3.2) + 4(0) + 8.3 y}{5 + 2.9 + 4 + 8.3}

9.28 + 8.3 x = 0

x = -1.12 m

5 0
3 years ago
Suppose a ball is dropped from shoulder height, falls, makes a perfectly elastic collision with the floor, and rebounds to shoul
Bezzdna [24]

Answer:Same magnitude

Explanation:

When ball is dropped from shoulder height h then velocity at the bottom is given by

v_1=\sqrt{2gh}

if it makes elastic collision then it will acquire the same velocity and riser up to the same height

If m is the mass of ball then impulse imparted is given by

J=m(v_2-v_1)

J=2m\sqrt{2gh}

Thus impulse imparted by gravity and Floor will have same magnitude of impulse but direction will be opposite to each other.

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What is the motion of the particles in this kind of wave? A hand holds the left end of a set of waves. The waves themselves make
Sonja [21]

Answer:

The transverse wave

Explanation:

4 0
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mash [69]

We know, P = F . ΔV

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