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madreJ [45]
3 years ago
5

What is the velocity of a car that has traveled 500 miles northeast in 5 hours

Physics
2 answers:
Naya [18.7K]3 years ago
6 0

<u>Answer</u>

= 100 miles per hour.

<u>Explanation</u>

Velocity is the rate of change of displacement.

Velocity = Distance /time taken

              = 500/5

              = 100 mile per hour.

Oliga [24]3 years ago
5 0

55 mph........................................

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A driver in a 2290-kg car car traveling at 42.7 m/s slams on the brakes and skids to a stop. If the coefficient of friction betw
8_murik_8 [283]

Answer:

126.56 m

Explanation:

Applying,

-F = ma............. Equation 1

Where F = frictional force, m = mass of the car, a = acceleration.

Note: Frictional force is negative because it act in opposite direction to motion

But,

F = mgμ.......... Equation 2

Where g = acceleration due to gravity, μ = coefficient of friction

Substitute equation 2 in equation 1

-mgμ = ma

a = -gμ.............. Equation 3

From the question,

Given: μ = 0.735

Constant: 9.8 m/s²

Substitute these values in equation 3

a = -9.8×0.735

a = -7.203 m/s²

Finally,

Applying

v² = u²+2as.............. Equation 4

Where v = final velocity, u = initial velocity, s = distance

From the question,

Given: u = 42.7 m/s, v = 0 m/s (to a stop), a = -7.203 m/s²

Substitute these values into equation 4

0² = 42.7²+2(-7.203)s

-1823.29 = -14.406s

s = -1823.29/-14.406

s = 126.56 m

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3 years ago
a kickball is struck with a 15.2 m/s velocity at a 63.0 degree angle. it lands on a rooftop 2.40 s later. how high is the roof?​
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Refer to the attachment

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3 years ago
A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N1 turns. A second toroidal solenoid w
lilavasa [31]

Answer:

Mutual inductance, M=2.28\times 10^{-5}\ H

Explanation:

(a) A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N₁ turns. A second thyroidal solenoid with N₂ turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius.

Mutual inductance is given by :

M=\dfrac{\mu_oN_1N_2A}{2\pi r}

(b) It is given that,

N_1=550

N_2=290

Radius, r = 10.6 cm = 0.106 m

Area of toroid, A=0.76\ cm^2=7.6\times 10^{-5}\ m^2

Mutual inductance, M=\dfrac{4\pi \times 10^{-7}\times 550\times 290\times 7.6\times 10^{-5}}{2\pi \times 0.106}

M=0.0000228\ H

or

M=2.28\times 10^{-5}\ H

So, the value of mutual inductance of the toroidal solenoid is 2.28\times 10^{-5}\ H. Hence, this is the required solution.

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