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Liono4ka [1.6K]
2 years ago
8

According to Newton's Third Law of Motion, what happens when two objects of unequal masses collide?

Physics
1 answer:
maksim [4K]2 years ago
8 0
<span>Newton's third law is that for every action, there is an equal and opposite reaction.
A </span><span>Damage is not a part of physics, at least I don't think it is. </span><span>
B </span>Is just downright wrong. If anything, the smaller mass will be pushed forward with the heavier mass. <span>
C Sounds correct
D It'</span><span>s wrong, because the masses are not equal, or else they would stop.</span>
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Match the following items.
Ray Of Light [21]

Answer:

Je ne Sachez que Qu’est-ce que le

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2 years ago
A roller coaster car of mass m = 300 kg is released from rest at the top of a 60 m high hill (position A), and rolls with a negl
Triss [41]

Answer:

The principle of conservation of energy and angular momentum

Explanation:

At point A, the car experienced maximum of potential energy

As it moves down the hill, the potential energy decreases while the kinetic energy increases.

The maximum kinetic energy of the car is needed for the attainment of enough centripetal force to help the car move through the loop without falling .

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2 years ago
Caleb is filling up water balloons for the Physics Olympics balloon tosscompetition. Caleb sets a 0.50-kg spherical water balloo
Mashcka [7]

a)

• P = F/A

P = pressure = 630 N/m^2

F = force

A = area

F = mg = 0.50 kg x 9.8 m/s^2 = 4.9 N

m= mass

g= gravity

P = F/A

A = F/P

A = 4.9 N / 630 N/m^2 = 7.778 x 10^-3 m^2

b)

• Area of a circle = pi* radius ^2

7.778 x 10^-3 m^2 = pi* radius ^2

√(7.778 x 10^-3 m^2 / pi ) = radius

radius = 0.04976 m

Answers:

a ) 7.778 x 10^-3 m^2

b) 0.04976 m

8 0
1 year ago
Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with
Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

<u>Part A)</u>

Let's use the parabolic motion equation to solve it. Let's define the variables:

  • y(i) is the initial height, it is a constant.
  • y(f) is the final height, in our case is 0
  • v(i) is the initial velocity (v(i)=16 m/s)
  • θ1 is the first angle, 30°
  • θ2 is the first angle, -30°

For the first stone

y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}              

0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}

0=y_{i1}+8t_{1}-4.905*t_{1}^{2} (1)  

For the second stone  

0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}    

0=y_{i2}-8t_{2}-4.905t_{2}^{2} (2)            

 

If we solve the equation (1) we will have:

t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}  

We can do the same procedure for the equation (2)

t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

<u>Part B)</u>

We can use the equation of the horizontal position here.

<u>First stone</u>

x_{f1}=x_{i1}+vcos(30)t_{1}

x_{f1}=0+13.86*t_{1}

x_{f1}=13.86*t_{1}

<u>Second stone</u>

x_{2}=x_{i2}+vcos(-30)t_{2}

x_{1}=0+13.86*t_{1}

x_{1}=13.86*t_{2}

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

3 0
3 years ago
When Karl Kaveman adds chilled grog to his new granite mug, he removes 10.9 kJ of energy from the mug. If it has a mass of 625 g
mrs_skeptik [129]

Answer:

3°C

Explanation:

We can that heat Q=mc_p dT

Where m is the mass c_p = specific heat capacity

dT = Temperature difference

here we have given m=625 g =.625 kg

specific heat of granite =0.79 J/(g-K) = 0.79 KJ/(kg-k)

T_1 =25°C

T_2 we have to find

we have also given Q=10.9 KJ

10.9=0.625×0.79×(25-T_2)

25-T_2 =22

T_2=3°C

7 0
3 years ago
Read 2 more answers
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