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xxMikexx [17]
2 years ago
6

A student pushes against a wall with a force of 30N. The wall does not move. What amount of force does the wall exert on the stu

dent?
A. 0N
B. Less than 30N
C. 30N
D. More than 30N
Physics
2 answers:
True [87]2 years ago
8 0

Answer:

C

Explanation:

they both have to be the same for both to not move

satela [25.4K]2 years ago
7 0

Answer: C,30N

Explanation:

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Let us say that x is the cut that we will make on the sides to make a box, therefore the new dimensions are:

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It is 2x since we cut on two sides.

 

We know that volume is:

V = l w x

V = (15 – 2x) (8 – 2x) x

V = 120x – 30x^2 – 16x^2 + 4x^3

V = 120x – 46x^2 + 4x^3

 

Taking the 1st derivative:

dV/dx = 120 – 92x + 12x^2

 

Set dV/dx = 0 to get maxima:

120 – 92x + 12x^2 = 0

 

Divide by 12:

x^2 – (92/12)x + 10 = 0

(x – (92/24))^2 = -10 + (92/24)^2

x - 92/24 = ±2.17

x = 1.66, 6

We cannot have x = 6 because that will make our w negative, so:

x = 1.66 inches

 

So the largest volume is:

V = 120x – 46x^2 + 4x^3

V = 120(1.66) – 46(1.66)^2 + 4(1.66)^3

V = 90.74 cubic inches

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3 years ago
If the force of gravity suddenly stopped acting on planets, they would
Orlov [11]
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Suppose a diving board with no one on it bounces up and down in a SHM with a frequency of 4.00 Hz. The board has an effective ma
kirza4 [7]

Answer:

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The electric field due to two point charges is found by: a) finding the stronger field. The net field will just be equal to the
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Answer:

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Explanation:

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7 0
3 years ago
A polar bear runs at a speed of 11 m/s and has a mass of 380.2 kg. How much Kinetic energy does the bear have?
Yanka [14]

Answer:

\boxed{\sf Kinetic \ energy \ of \ the \ bear (KE) = 23002.1 \ J}

Given:

Mass of the polar bear (m) = 6.8 kg

Speed of the polar bear (v) = 5.0 m/s

To Find:

Kinetic energy of the polar bear (KE)

Explanation:

Formula:

\boxed{ \bold{\sf KE =  \frac{1}{2} m {v}^{2} }}

Substituting values of m & v in the equation:

\sf \implies KE =  \frac{1}{2}  \times 380.2 \times  {11}^{2}

\sf \implies KE = \frac{1}{ \cancel{2}}  \times  \cancel{2} \times 190.1 \times 121

\sf \implies KE = 190.1 \times 121

\sf \implies KE = 23002.1 \: J

\therefore

Kinetic energy of the polar bear (KE) = 23002.1 J

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3 years ago
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