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2 years ago

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Newton's second law A. describes how an object accelerates when a force is applied B. says that objects eventually stop unless a

force is applied C. objects with mass attract each other D. forces come in action/reaction pairs E. an object will remain in uniform motion unless acted upon by a force F. like charges repel, opposite charges attract

1 answer:

Papessa [141]2 years ago

8 0

Answer:

A. describes how an object accelerates when a force is applied

Explanation:

Newton's second law of motion concerns the behavior of objects that do not have a stability between all established forces. The second law states that an object's acceleration depends on two factors: the net force on the entity and the entity's mass.

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Ella makes this table to organize her notes on whether atoms gain or lose energy during the changes of state.

solong [7]

Answer:

The answer is D.

7 0

2 years ago

Read 2 more answers

Suppose someone pours 0.250 kg of 20.0ºC water (about a cup) into a 0.500-kg aluminum pan with a temperature of 150ºC. Assume th

Troyanec [42]

Answer : The temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of aluminium = $0.90J/g^oC$

$c_2$ = specific heat of water = $4.184J/g^oC$

$m_1$ = mass of aluminum = 0.500 kg = 500 g

$m_2$ = mass of water = 0.250 kg = 250 g

$T_f$ = final temperature of mixture = ?

$T_1$ = initial temperature of aluminum = $150^oC$

$T_2$ = initial temperature of water = $20^oC$

Now put all the given values in the above formula, we get:

$500g\times 0.90J/g^oC\times (T_f-150)^oC=-250g\times 4.184J/g^oC\times (T_f-20)^oC$

$T_f=59.10^oC$

Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, $59.10^oC$

8 0

3 years ago

Suddenly a worker picks up the bag of gravel. Use energy conservation to find the speed of the bucket after it has descended 2.3

fiasKO [112]

Explanation:

A worker picks up the bag of gravel. We need to find the speed of the bucket after it has descended 2.30 m from rest. It is case of conservation of energy. So,

$\dfrac{1}{2}mv^2=mgh\\\\v=\sqrt{2gh}$

h = 2.3 m

$v=\sqrt{2\times 9.8\times 2.3} \\\\v=6.71\ m/s$

So, the speed of the bucket after it has descended 2.30 m from rest is 6.71 m/s.

8 0

2 years ago

A drunken sailor stumbles 550 meters north, 500 meters northeast, then 450 meters northwest. What is the total displacement and

cluponka [151]

Answer:

Resultant displacement = 1222.3 m

Angle is 88.3 degree from +X axis.

Explanation:

A = 550 m north

B = 500 m north east

C = 450 m north west

Write in the vector form

A = 550 j

B = 500 (cos 45 i + sin 45 j ) = 353.6 i + 353.6 j

C = 450 ( - cos 45 i + sin 45 j ) = - 318.2 i + 318.2 j

Net displacement is given by

R = (353.6 - 318.2) i + (550 + 353.6 + 318.2) j

R = 35.4 i + 1221.8 j

The magnitude is

$R = \sqrt{35.4^{2}+1221.8^{2}}R = 1222.3 m$

The direction is given by

$tan\theta =\frac{1221.8}{35.4}\\\\\theta = 88.3^{o}$

3 0

2 years ago

The electric motor in the car is powered by a battery.

Vinil7 [7]

Answer:

I = 30 A.

Explanation:

Given that,

The voltage of the battery, V = 230 V

Power used to charge the battery, P = 6.9 kW

We need to find the current used to charge the battery. The formula for the power is given by :

P = VI

Where

I is current

So,

So, the required current is 30 A.

6 0

2 years ago