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Tcecarenko [31]

2 years ago

14

A(n) 1100 kg car is parked on a 4◦ incline. The acceleration of gravity is 9.8 m/s 2 . Find the force of friction keeping the ca

r from sliding down the incline.

1 answer:

igomit [66]2 years ago

3 0

Answer:

Force of friction, f = 751.97 N

Explanation:

it is given that,

Mass of the car, m = 1100 kg

It is parked on a 4° incline. We need to find the force of friction keeping the car from sliding down the incline.

From the attached figure, it is clear that the normal and its weight is acting on the car. f is the force of friction such that it balances the x component of its weight i.e.

$f=mg\ sin\theta$

$f=1100\ kg\times 9.8\ m/s^2\ sin(4)$

f = 751.97 N

So, the force of friction on the car is 751.97 N. Hence, this is the required solution.

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The acceleration of a baseball after it is hit by a bat depends on the mass of the ball and the net force on the

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Answer:

B. Newton's second law of motion

Explanation:

Newton's Second Law of Motion states that the acceleration of a physical object is directly proportional to the net force acting on the physical object and inversely proportional to its mass.

Mathematically, it is given by the formula;

$Net \; Force = mass * acceleration$

Making acceleration the subject of formula, we have;

$Acceleration = \frac {Net \; Force}{Mass}$

In this scenario, the acceleration of a baseball after it is hit by a bat depends on the mass of the ball and the net force on the ball. Thus, this example best illustrates Newton's second law of motion.

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A transformer has a primary coil with 106 turns and a secondary coil of 340 turns. The AC voltage across the primary coil has a

UkoKoshka [18]

To solve this problem it is necessary to apply the concepts related to transformers, that is to say passive electrical device that transfers electrical energy from one electrical circuit to one or more circuits.

From the mathematical definition we have that the relationship between the voltage of the first coil and the second coil is proportional to the number of loops of the first and second loop, that is:

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

Where

$V_p =$ input voltage on the primary coil.

$V_s=$ input voltage on the secondary coil.

$N_p=$ number of turns of wire on the primary coil.

$N_s =$ number of turns of wire on the secondary coil.

Replacing our values we have:

$V_p = 128V$

$N_p = 106$

$N_s = 340$

Replacing,

$\frac{V_s}{128} = \frac{340}{106}$

$V_s = 410.56V$

From the same relations of number of turns and the voltage of the first and second coil we also have the relation of electricity and voltage whereby:

$V_s I_s = V_p I_p$

Where

$I_p$ = Current Primary Coil

$I_s$ = Current secundary Coil

Therefore:

$I_s = \frac{V_p I_p}{V_s}$

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2 years ago

A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho

d1i1m1o1n [39]

Answer:

Part a)

$KE_r = 8 J$

Part b)

v = 3.64 m/s

Part c)

$KE_f = 12.7 J$

Part d)

$v = 2.9 m/s$

Explanation:

As we know that moment of inertia of hollow sphere is given as

$I = \frac{2}{3}mR^2$

here we know that

$I = 0.0484 kg m^2$

R = 0.200 m

now we have

$0.0484 = \frac{2}{3}m(0.200)^2$

$m = 1.815 kg$

now we know that total Kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

$KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2$

$20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2$

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Part a)

Now initial rotational kinetic energy is given as

$KE_r = \frac{1}{2}I(\frac{v}{R})^2$

$KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2$

$KE_r = 8 J$

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

$mgh = (KE_i) - KE_f$

$1.815(9.8)(0.900sin27.1) = 20- KE_f$

$7.30 = 20 - KE_f$

$KE_f = 12.7 J$

Part d)

Now we know that

$\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7$

$\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7$

$1.5125 v^2 = 12.7$

$v = 2.9 m/s$

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