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sdas [7]
2 years ago
8

Peter throws a snowball at his car parked in the driveway. The snowball disintegrates as it hits the car. By Newton’s third law,

you know that both the car and the snowball exert a force on each other. What can you say about the magnitude of the forces exerted by the snowball and the car?
Physics
1 answer:
elena55 [62]2 years ago
3 0
Well, they are equal and opposite, but the force is high enough to break up the snowball.  That is where the energy is dissipated.
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1. A 100-kg crate is pulled across a warehouse floor using a rope with a force of 250 N at an angle of 45o from the horizontal.
harkovskaia [24]

Answer:

(a) The net force is 80.394 N

    The acceleration of the crate is 0.804 m/s²

(b) the final velocity of the crate is 5.02 m/s

Explanation:

Given;

mass of the crate, m = 100 kg

applied force, F = 250 N

angle of inclination, θ = 45°

coefficient of friction, μ = 0.12

Applied force in y-direction, F_y = Fsin \theta = 250sin45 = 176.78 \ N

Applied force in x-direction, F_x = Fcos \theta = 250cos45 = 176.78 \ N

The normal force is calculated as;

N + Fy -W = 0

N = W - Fy

N = (100 x 9.8) - 176.78

N = 980 - 176.78 = 803.22 N

The frictional force is given by;

Fk = μN

Fk = 0.12 x 803.22

Fk = 96.386 N

(a) The net force is given by;

F_{net} = F_x - F_k\\\\F_{net} = 176.78-96.386\\\\F_{net} = 80.394 \ N

Apply Newton's second law of  motion;

F = ma

a = \frac{F_{net}}{m}\\\\ a = \frac{80.394}{100}\\\\ a = 0.804 \ m/s^2

(b) the velocity of the crate after 5.0 s

F = ma= \frac{m(v-u)}{t} \\\\Ft =m(v-u)\\\\v-u = \frac{Ft}{m}\\\\ v = \frac{Ft}{m} + u\\\\v = \frac{F_{net}*t}{m} + u\\\\v = \frac{80.394*5}{100} + 1\\\\v = 5.02 \ m/s

7 0
3 years ago
Explain why squeezing a plastic mustard bottle forces mustard out to top.
Oliga [24]

Answer:

because when squeezing you are increasing pressure within the bottle and there is less pressure on the outside

5 0
3 years ago
Read 2 more answers
Solution A has a specific heat of 2.0 J/g◦C. Solution B has a specific heat of 3.8 J/g◦C. If equal masses of both solutions start
fgiga [73]

Answer: 2. Solution A attains a higher temperature.

Explanation: Specific heat simply means, that amount of heat which is when supplied to a unit mass of a substance will raise its temperature by 1°C.

In the given situation we have equal masses of two solutions A & B, out of which A has lower specific heat which means that a unit mass of solution A requires lesser energy to raise its temperature by 1°C than the solution B.

Since, the masses of both the solutions are same and equal heat is supplied to both, the proportional condition will follow.

<em>We have a formula for such condition,</em>

Q=m.c.\Delta T.....................................(1)

where:

  • \Delta T= temperature difference
  • Q= heat energy
  • m= mass of the body
  • c= specific heat of the body

<u>Proving mathematically:</u>

<em>According to the given conditions</em>

  • we have equal masses of two solutions A & B, i.e. m_A=m_B
  • equal heat is supplied to both the solutions, i.e. Q_A=Q_B
  • specific heat of solution A, c_{A}=2.0 J.g^{-1} .\degree C^{-1}
  • specific heat of solution B, c_{B}=3.8 J.g^{-1} .\degree C^{-1}
  • \Delta T_A & \Delta T_B are the change in temperatures of the respective solutions.

Now, putting the above values

Q_A=Q_B

m_A.c_A. \Delta T_A=m_B.c_B . \Delta T_B\\\\2.0\times \Delta T_A=3.8 \times \Delta T_B\\\\ \Delta T_A=\frac{3.8}{2.0}\times \Delta T_B\\\\\\\frac{\Delta T_{A}}{\Delta T_{B}} = \frac{3.8}{2.0}>1

Which proves that solution A attains a higher temperature than solution B.

7 0
3 years ago
When the activation energy of an exothermic reaction decreases at a given temperature, the reaction rate increases because the _
fgiga [73]
When the activation energy of an exothermic reaction decreases at a given temperature, the reaction rate increases because the <span>number of successful effective collisions is higher. More of the reactants collide and are able to form products. Hope this answers the question. have a nice day.</span>
6 0
3 years ago
What is the maximum acceleration the belt can have without the crate slipping? express your answer using two significant figures
Montano1993 [528]

To prevent the crate from slipping, the maximum force that the belt can exert on the crate must be equal to the static friction force.


Ff = 0.5 * 16 * 9.8 = 78.4 N

a = 4.9 m/s^2


If acceleration of the belt exceeds the value determined in the previous question, what is the acceleration of the crate?


In this situation, the kinetic friction force is causing the crate to decelerate. So the net force on the crate is 78.4 N minus the kinetic friction force.


Ff = 0.28 * 16 * 9.8 = 43.904 N

Net force = 78.4 – 43.904 = 34.496 N

To determine the acceleration, divide by the mass of the crate.

a = 34.496 ÷ 16 = 2.156 m/s^2



8 0
3 years ago
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