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Travka [436]
3 years ago
5

With a force of 5 Newton's Amanda pushes the stacks of books to the right. At the same time Jeremih her little brother pushes th

e stacks of books to the left with a force of 10 Newton's what is the direction and magnitude of the resulting force
Physics
2 answers:
nignag [31]3 years ago
5 0

Answer:

5n to the left

Explanation:

STUISLAND 2020

Fynjy0 [20]3 years ago
4 0

Alright, so solving for the net force is a rather simple and easy step. I want to briefly explain how you determine the net force, but before I do, let's examine the problem together. What has been provided to us? What direction is it going? Well, we can look at the information!

Amanda's Force: 5 N

Jeremiah's Force: 10 N

They're both pushing on the books, but in different directions. Left and right. If Amanda is pushing the books to the right 5 newtons, and Jeremiah is pushing the books to the left with 10 newtons, that means the net force is 5. The book is being pushed to the left! We can say that because Jeremiah's force is much larger than his sisters.

We can determine the net force from the following;

Net Force → 10 N - 5 N = 5 N

In conclusion, your answer should be a total of five (5) newtons

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Elena-2011 [213]

Answer:

There are 4

Explanation:

4 0
2 years ago
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What is the internal energy of 3.00mol of N2 gas at 25c?
Citrus2011 [14]

Answer:

18600j

Explanation:

It is given that,

Number of moles = 3

Temperature, T = 25°C = 25+273 = 298 K

The internal energy of N₂ gas is given by :

U=f\times nRTU=f×nRT

f is degrees of freedom. For diatomic gas, degree of freedom is equal to 5/2. So,

\begin{gathered}U=\dfrac{5}{2}\times 3\times 8.31\times 298\\\\U=18572.85\ J\end{gathered}

U=

2

5×3×8.31×298

U=18572.85 J

or

U = 18600 J

So, the internal energy of the gas is 18,600 J

3 0
2 years ago
A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.
Elenna [48]

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s

a) The vertical speed when it leaves the ground. is 6.95 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s

Time taken to reach the maximum height is 0.71 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

3 0
3 years ago
Assume a satellite shines an unpolarized light on a telescope. The intensity of the light as it reaches the telescope is 1.1*10-
n200080 [17]

Answer:

4.125\times 10^{-11}\ W/m^2

Explanation:

I_0 = Intensity of unpolarized light = 1.1\times 10^{-10}\ W/m^2

\theta = Angle of the filter = 30^{\circ}

Intensity of light is given by

I=\dfrac{I_0}{2}cos^2\theta\\\Rightarrow I=\dfrac{1.1\times 10^{-10}}{2}cos^230\\\Rightarrow I=4.125\times 10^{-11}\ W/m^2

The intensity of light detected by the camera is 4.125\times 10^{-11}\ W/m^2

7 0
3 years ago
Tom is throwing an baseball at an aluminum can,
pishuonlain [190]

Answer:

The question relates to the conservation of energy principle, the conservation of the linear momentum, and Newton's Laws of motion

Part A

1) Tom throwing a baseball at a can

The initial velocity of the baseball = v₂

The initial kinetic energy of the baseball, K.E.₂ = (1/2)·m₂·v₂²

∴ The final kinetic energy of the baseball, K.E.₂' = (1/2)·m₂·v₂'² < (1/2)·m₂·v₂²

Therefore, the energy of the ball before the collision is lesser than the energy of the ball after the collision

2) The evidence that would likely support the claim is that the baseball's height above the ground reduces rapidly immediately after the collision which is due to the reduced velocity, and therefore, the reduced (kinetic) energy

The final velocity of the baseball v₂' < v₂

Part B

1) The argument

The initial velocity of the can = v₁ = 0 (The can is initially  at rest)

The initial kinetic energy of the can, K.E.₁ = (1/2)·m₁·v₁² = 0

The final velocity of the can v₁' > v₁ = 0

∴ The final kinetic energy of the can, K.E.₁ = (1/2)·m₁·v₁² > 0

Given that the velocity of the can increases from zero to a positive value after collision with the baseball, the kinetic energy of the can is increased from zero before the collision to a positive value after the collision

2) An evidence in support of the argument is the motion of the can which was initially at rest which is an indication of increase in energy podded by the can

Explanation:

8 0
2 years ago
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