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UkoKoshka [18]
2 years ago
5

Relative to the ground, what is the gravitational potential energy of 68 kg

Physics
1 answer:
Kisachek [45]2 years ago
8 0

Answer:

Explanation:

p.E=mgh

P.E=68*10*443

p.E=301240j

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The ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straig
Virty [35]

Answer:

1. The tension in the rope is everywhere the same.

2. The magnitudes of the forces exerted on the two objects by the rope are the same.

3. The forces exerted on the two objects by the rope must be in opposite directions.

Explanation:

"Massless ropes" do not have a<em> "net force"</em> which means that it is able to transmit the force from one end of the rope to the other end, perfectly. It is known for its property of having a total force of zero. In order to attain this property, the magnitude of the forces exerted on the two stationary objects by the rope are the same and in opposite direction. <u>So this explains number 2 & 3 answers.</u>

Since the objects that are held by the rope are stationary, then this means that the tension in the rope is also stationary. This means that the tension in the rope everywhere is the same (provided that the rope is still or in a straight line, as stated in the situation above, and is being held by two points). <u>So, this explains number 1.</u>

6 0
3 years ago
Which best describes the current atomic theory? (1 point) a Since it is only a theory it should not be used in practice. b It ha
Ahat [919]

Answer:scientist try to fix it

Explanation:

8 0
3 years ago
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A uniform ladder is 10 m long and weighs 400 n. it rests with its upper end against a frictionless vertical wall. its lower end
Nikolay [14]
<h3><u>Answer;</u></h3>

200 N

<h3><u>Solution;</u></h3>

Given ;

Weight = 400 N

Length of the ladder = 10 m

Using the equation;

Torque = d x F, where F is the perpendicular force and d is the distance

Calculating torque about the lower end,

Tweight = Twall

400N(5cos30) = F (10sin30)

 F = 200 N

Therefore, the magnitude of force exerted on the peg by the ladder is 200 N

5 0
4 years ago
A block with mass m = 0.2 kg oscillates with amplitude A = 0.3 m at the end of a spring with force constant k = 12 N/m on a fric
Llana [10]

Answer:

Explanation:

Given that,

A block of mass m = 0.2kg

Amplitude of oscillation A = 0.3m

Spring Constant k = 12N/m

We want to rank the period of oscillation in each case from the smallest to the largest.

Period of oscillation can be determine using

T = 2π√m/k

Where,

T is period in seconds

m is mass in kg

k is spring constant in N/m

So, the only things that affect the period is mass and the spring constant

A. Using the above information

Where m = 0.2kg and k = 12 N/m

Then, T = 2π√m/k

T = 2π√(0.2/12)

T = 2π × 0.129

Ta = 0.81 seconds

B. The amplitude is change to 1.6m, A = 1.6m

The period T for a pendulum is nearly independent of amplitude.

Since the period is independent of the amplitude, then, the period does not change

So, Tb = Ta = 0.81seconds

C. If the mass is change to 1.6kg

Now, m=1.6kg

Then, T = 2π√m/k

T = 2π√(1.6/12)

T = 2π × 0.365

Tc = 2.28 seconds

D. If the force constant is change to 30N/m

Now, k = 30

Then, T = 2π√m/k

T = 2π√(0.2/ 30)

T = 2π × 0.0816

Td = 0.513 seconds

E. The small resistive force does not affect the period

So the period remains unchanged

Ta = Te = 0.81 seconds

Ranking the periods

Tc > Ta = Tb = Te > Td

3 0
4 years ago
Suppose that a constant force is applied to an object. Newton's Second Law of Motion states that the acceleration of the object
omeli [17]
<span>(9 kg)(5 m/s^2) = M(3 m/s^2) 
</span><span>that the acceleration of the object varies inversely with its mass.</span>
4 0
3 years ago
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