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3 years ago

1400 kg car has a speed of 27 m/s. If it takes 7 s to stop the car, what is the Impulse and the average force acting on the car?

Physics

1 answer:

mixer [17]3 years ago

6 0

Explanation:

Impulse is change in momentum:

I = Δp

I = (1400 kg) (27 m/s) - (1400 kg) (0 m/s)

I = 37800 kg m/s

Impulse is also average force times time:

I = F ΔT

37800 kg m/s = F (7 s)

F = 5400 N

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3 years ago

in three situations, a briefly applied horizontal force changes the velocity of a hockey puck that slides over frictionless ice.

victus00 [196]

The work done on the Puck by the applied force from the most positive to the most negative is c, b, a respectively.

According to Newton's second law of motion, the force applied to an object is directly proportional to the product of mass and acceleration of the object.

F = ma

$F= \frac{mv}{t}$

The force applied to an object increases with increases in the velocity of the object.

In the given diagram, the resultant velocity of the puck is calculated as follows;

Figure a:

$\Delta v = v_f -v_i\\\\\Delta v = 5 - 6 = - 1 \ m/s$

Figure b:

$v = \sqrt{4^2 + 3^2} \\\\v = 5 \ m/s$

Figure c:

$\Delta v = 4 - (-2)\\\\\Delta v = 6 \ m/s$

Thus, the work done on the Puck by the applied force from the most positive to the most negative is c, b, a respectively.

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4 0

3 years ago

A 75-turn coil with a diameter of 6.00 cm is placed in a constant, uniform magnetic field of 1.00 T directed perpendicular to th

kolezko [41]

Answer:

The induced emf in the coil at the t = 5s is 6.363 mV

Explanation:

Given;

number of turns = 75

diameter of the coil = 6 cm

magnetic field strength = 1 T

new magnetic field strength = 1.30 T at t = 10.0 s

$Area \ of \ coil = \frac{\pi d^2}{4} = \frac{\pi *0.06^2}{4} = 0.002828 \ m^2$

$E.M.F = \frac{NA* \delta B}{\delta t}$

Between 0 to 5 s, Induced emf is given as;

$E.M.F = \frac{75*0.002828*(B_5-1)}{5}$

Between 5 to 10 s, Induced emf is given as;

$E.M.F = \frac{75*0.002828*(1.3-B_5)}{5}$

Since the field increased at a uniform rate until it reaches 1.30 T at t = 10.0 s, the induced emf will also increase in uniform rate. And equal time interval will generate same increase in field strength.

B₅ -1 = 1.3 - B₅

2B₅ = 2.3

B₅ = 1.15 T

Thus, magnetic field at t = 5 is 1.15 T

$E.M.F = \frac{75*0.002828*(1.3-1.15)}{5} = 6.363 \ mV$

Therefore, the induced emf in the coil at the t = 5s is 6.363 mV

7 0

3 years ago