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stiv31 [10]
4 years ago
12

A 1.2-kg object moving with a speed of 8.0 m/s collides perpendicularly with a wall and emerges with a speed of 6.0 m/s in the o

pposite direction. if the object is in contact with the wall for 2.0 ms, what is the magnitude of the average force on the object by the wall? ans: 8400 n
Physics
1 answer:
Rainbow [258]4 years ago
7 0
Newton's second law applies. That is,

F=ma, F = Wall force on the object, m = mass of the object, a = Deceleration = Δv/t = (6-8)/2 = -1 m/s

F = 1.2*1 = 1.2 N
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Answer:

The box will be moving at 0.45m/s. The solution to this problem requires the knowledge and application of newtons second law of motion and the knowledge of linear motion. The vertical component of the force Fp acts vertically upwards against the directio of motion. This causes a constant upward force of 23sin45° to act on the box. Fhe frictional force of 13N also acts vertically upwards and so two forces act upwards against rhe force of gravity resulting un a net force of 0.7N acting kn the box. This corresponds to an acceleration of 0.225m/s². So in w.0s after i start to push v = 0.45m/s.

Explanation:

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Voltage needed to raise current to 3.75a using 20,20,200 resistor set
Varvara68 [4.7K]

<u>Answer:</u> The voltage needed is 35.7 V

<u>Explanation:</u>

Assuming that the resistors are arranged in parallel combination.

For the resistors arranged in parallel combination:

\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}

We are given:

R_1=20\Omega\\R_2=20\Omega\\R_3=200\Omega

Using above equation, we get:

\frac{1}{R}=\frac{1}{20}+\frac{1}{20}+\frac{1}{200}\\\\\frac{1}{R}=\frac{10+10+1}{200}\\\\R=\frac{200}{21}=9.52\Omega

Calculating the voltage by using Ohm's law:

V=IR         .....(1)

where,

V = voltage applied

I = Current = 3.75 A

R = Resistance = 9.52\Omega

Putting values in equation 1, we get:

V=3.75\times 9.52\\\\V=35.7V

Hence, the voltage needed is 35.7 V

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3 years ago
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Batman (mass = 96.1 kg) jumps straight down from a bridge into a boat (mass = 458 kg) in which a criminal is fleeing. The veloci
MariettaO [177]

Answer:

The velocity of the boat after the batman lands in it is +9.26 m/s

Explanation:

Applying the law of conservation of momentum,

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Note: The collision between the Batman and the boat is an inelastic collision.

m'u'+mu = V(m+m').................... Equation 1

Where m' = mass of the Batman, u' = initial velcoity of the batman, m = mass of the boat, u = initial velocity of the boat, V = common velocity.

make V the subject of equation 1

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Given: m' = 96.1 kg, u' = 0 m/s, m = 458 kg, u = +11.3 m/s.

Substitute these values into equation 2

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V = 5129.6/554.1

V = +9.26 m/s

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3 years ago
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