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OLga [1]
2 years ago
11

The back shelf of a car has a physics textbook on it. The coefficient of static friction of 0.44 between the book and the surfac

e. If the car is initially moving at a speed of 16 m/s, in how short a distance can the car be stopped at constant acceleration without causing the book to slide forward (and hit the driver in the back of the head)? (This is why you should never keep packages on the back shelf of your car.)
Physics
1 answer:
Law Incorporation [45]2 years ago
8 0

Answer:29.68 m

Explanation :  

Given

coefficient of friction \mu =0.44

Car initial speed=16 m/s

The maximum acceleration that book can bear without sliding is \mu g

a_{max}=0.44\times 9.8=4.312 m/s^2

using v^2-u^2=2as

Final velocity v=0

-(16)^2=2(-4.31)s

s=\frac{256}{2\times 4.312}

s=29.68 m

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3 years ago
What is the easiest way to increase the magnetic force acting on the rotor in an induction motor?
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Answer:

Explanation:

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This permanents magnets are applicable in loudspeakers, generators, induction motor etc.

To increase the

The following will tend to increase the magnetic force acting on the rotor in an induction motor.

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6 0
3 years ago
A satellite that goes around the earth once every 24 hours (86,400 s) is called a geosynchronous satellite. If a geosynchronous
Olegator [25]

Answer:

42244138.951 m

Explanation:

G = Gravitational constant = 6.667 × 10⁻¹¹ m³/kgs²

r = Radius of orbit from center of earth

M = Mass of Earth = 5.98 × 10²⁴ kg

m = Mass of Satellite

The satellite revolves around the Earth at a constant speed

Speed = Distance / Time

The distance is the perimeter of the orbit

v=\frac{2\pi \times r}{24\times 3600}

The Centripetal force of the satellite is balanced by the universal gravitational force

m\frac{v^2}{r}=\frac{GMm}{r^2}\\\Rightarrow \frac{\left(\frac{2\pi \times r}{24\times 3600}\right)^2}{r}=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{r^2}\\\Rightarrow \left(\frac{2\pi \times r}{24\times 3600}\right)^2=6.667\times 10^{-11}\times 5.98\times 10^{24}\\\Rightarrow r^3=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\\\Rightarrow r=\left(\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\right)^{\frac{1}{3}}\\\Rightarrow r=42244138.951\ m

The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m

6 0
3 years ago
If a sheep is running at 3.0 m/s with a mass of 60 kg what is its kinetic energy​
borishaifa [10]

Answer:270joules

Explanation:KE = 1/2mv^2

KE = 1/2(60kg)(3.0)^2 = 270 Joules

8 0
3 years ago
A catamaran with a mass of 5.44×10^3 kg is moving at 12 knots. How much work is required to increase the speed to 16 knots? (One
Andre45 [30]

The work that is required to increase the speed to 16 knots is 14,176.47 Joules

If a catamaran with a mass of 5.44×10^3 kg is moving at 12 knots, hence;

5.44×10^3 kg = 12 knots

For an increased speed to 16knots, we will have:

x = 16knots

Divide both expressions

\frac{5.44 \times 10^3}{x} = \frac{12}{16}\\12x = 16 \times 5.44 \times 10^3\\x = 7.23\times 10^3kg\\

To get the required work done, we will divide the mass by the speed of one knot to have:

w=\frac{7230}{0.51}\\w= 14,176.47Joules

Hence the work that is required to increase the speed to 16 knots is 14,176.47 Joules

Learn more here: brainly.com/question/25573786

8 0
2 years ago
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