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zlopas [31]
3 years ago
14

A 4 kg and 6 kg bowling ball are dropped from the same height at the same time. The two balls strike the ground at the same time

. Which statement is true about the force due to gravity on the balls?
Physics
2 answers:
Ray Of Light [21]3 years ago
5 0
The force of gravity on objects is proportional to the mass of each object.

(That's a big part of the reason why, when you eat more and your mass
increases, you weigh more.)

The forces of gravity between the Earth and the 6kg ball are 50% greater
than the forces of gravity between the Earth and the 4kg ball.

(The gravitational forces between the 4kg ball and the 6kg ball, or between
both bowling balls and you, are so small that they may be ignored.)

mart [117]3 years ago
3 0

Answer:

B) The 6 kg ball has a greater force of gravity exerted on it.

Explanation: got it from usa tp .

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A 30-cm-diameter, 1.2 kg solid turntable rotates on a 1.2-cm-diameter, 450 g shaft at a constant 33 rpm. When you hit the stop s
skelet666 [1.2K]

Answer:

frictional force = 0.52 N

Explanation:

diameter of turn table (D1) = 30 cm = 0.3 m

mass of turn table (M1) = 1.2 kg

diameter of shaft (D2) = 1.2 cm = 0.012 m

mass of shaft (M2) = 450 g = 0.45 kg

time (t) = 15 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

radius of turn table (R1) = 0.3 / 2 = 0.15 m

radius of shaft (R2) = 0.012 / 2 = 0.006 m

total moment of inertia (I) = moment of inertia of turn table + moment of inertia of shaft

I = 0.5(M1)(R1)^{2} + O.5 (M2)(R2)^{2}

I =  0.5(1.2)(0.15)^{2} + O.5 (0.45)(0.006)^{2}

I = 0.0135 + 0.0000081 = 0.0135081

ω₁ = 33 rpm = 33 x \frac{2π}{60} = 3.5 rad/s

α = -ω₁/t = -3.5 / 15 = -0.23 rad/s^{2}

torque = I x α

torque = 0.0135081 x (-0.23) = - 0.00311 N.m

torque = frictional force x R2

- 0.00311 = frictional force x 0.006

frictional force = 0.52 N

6 0
4 years ago
When light passes through a convex lens, it does this.....
Tpy6a [65]

Explanation:

6. Converge or come together

7. convex

3 0
3 years ago
Read 2 more answers
A system undergoes a two-step process. In the first step, the internal energy of the system increases by 222 J when 150 J of wor
joja [24]

Answer:0 J

Explanation:

Given

For first step

change in internal Energy of the system is \Delta U_1=222 J

Work done on the system W_1=-150 J

For second step

change in internal Energy of the system is \Delta U_2=123 J

Work done on the system W_2=-195 J

Work done on the system is considered as Positive and vice-versa.

and from first law of thermodynamics

Q=\Delta U+W

for first step

Q_1=222-150=72 J

Q_2=123-195=-72 J

overall heat added=Q_1+Q_2

Q_{net}=72-72 =0

For overall Process Heat added is 0 J

8 0
3 years ago
Anyone know the answer to number 4…..please help me ASAP!
Damm [24]

Answer:

All planets have an elliptical orbit

all planets have roughly the same SHAPE of orbit

6 0
3 years ago
Read 2 more answers
A 52 N sled is pulled across a cement sidewalk at constant speed. A horizontal force of 36 N is exerted. What is the coefficient
Andre45 [30]

Answer:

μ = 0.692

Explanation:

In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.

Attached is an image with the respective forces:

A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.

Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.

The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.

The process of solving this problem can be seen in the attached image.

5 0
3 years ago
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