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ikadub [295]
3 years ago
9

A 1500 kg car accelerats from 55.0 m/s to 90.0 m/s. Calculate the impulse experienced by the car​

Physics
1 answer:
ioda3 years ago
7 0

Answer:

The impulse experienced by the car is 52,500 kg.m/s.

Explanation:

Given;

mass of the car, m = 1500 kg

initial velocity of the car, u = 55 m/s

final velocity of the car, v = 90 m/s

The impulse experienced by the car is the change in linear momentum, calculated as follows;

J = ΔP = mv  - mu

ΔP = m(v - u)

ΔP = 1500(90 - 55)

ΔP = 52,500 kg.m/s

Therefore, the impulse experienced by the car is 52,500 kg.m/s.

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Answer: The longer the lever, the greater the force on the load will be.

Explanation:

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3 years ago
Determine the magnitude of the force between two 42 m-long parallel wires separated by 0.03 m, both carrying 6.3 A in the same d
xz_007 [3.2K]

Answer:

The magnitude of the force between the two parallel wires is 0.0111 N.

Explanation:

Given;

length of the two parallel wires, L = 42 m

distance between the two wires, r = 0.03 m

current in both wires, I₁, I₂ = 6.3 A

Therefore, the magnitude of the repulsive force between the two parallel wires is given by;

F = \frac{\mu_0 I_1I_2l}{2\pi r}\\\\where;\\\mu_0 \ is \ permeability \ of \ free \ space = 4\pi *10^{-7} \ T.m/A \\\\F = \frac{(4\pi *10^{-7})(6.3)^2(42)}{2\pi (0.03)}\\\\F =   0.0111 \ N

Therefore, the magnitude of the force between the two parallel wires is 0.0111 N.

6 0
3 years ago
I need help with this please
Firdavs [7]
C is the answer to the question
6 0
3 years ago
Sandy is whirling a ball attached to a string in a horizontal circle over his head. If Sandy doubles the speed of the ball, what
jeka57 [31]

The tension in the string B) It quadruples.

Explanation:

The ball is in uniform circular motion in a horizontal circle, so the tension in the string is providing the centripetal force that keeps the ball in circular motion. So we can write:

T= m\frac{v^2}{r}

where:

T is the tension in the string

m is the mass of the ball

v is the speed of the ball

r is the radius of the circle (the lenght of the string)

In this problem, we are told that the speed of the ball is doubled, so

v' = 2v

Substituting into the previous equation, we find the new tension in the string:

T' = m \frac{(2v)^2}{r}=4(m\frac{v^2}{r})=4T

Therefore, the tension in the string will quadruple.

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

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6 0
3 years ago
a hawk flies in a horizontal arc of radius 10.3 m at a constant speed of 4.8 m/s. find its centripetal acceleration. answer in u
n200080 [17]

The hawk’s centripetal acceleration is 2.23 m/s²

The magnitude of the acceleration under new conditions is 2.316 m/s²

radius of the horizontal arc = 10.3 m

the initial constant speed = 4.8 m/s

we know that the centripetal acceleration is given by

    a_{c}  = \frac{v^{2} }{r}

   a_{c}  = 23.04/10.3

    a_{c}  = 2.23 m/s²

It continues to fly but now with some tangential acceleration

a_{t} = 0.63 m/s²

therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration

so

a_{net}  =  \sqrt{a_{c} ^{2} +a_{t} ^{2}   }

a_{net}  =  \sqrt{4.97 + 0.396}

a_{net}  =  2.316 m/s²

So the magnitude of  net acceleration will become 2.316 m/s².

learn more about acceleration here :

brainly.com/question/11560829

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8 0
1 year ago
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