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3 years ago

10

A 5.0 kilogram rock rolls off the edge of a cliff. If it is traveling at a speed of 24.2m/s when it hits the ground, what is the

height of the cliff?

1 answer:

timurjin [86]3 years ago

7 0

Using the kinematic equation ;

vf²=vi²+2ad just rearrange for d;

vf²-vi²=2ad;

(vf²-vi²)/2a=d

Now plug the numbers in.

vf=24.2

vi=0

a=9.81=g

(24.2²-0²)/(2*9.81)=29.8 m.

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Tyson Gay's best time to run 100.0 meters was 9.69 seconds. What was his average speed during this run, in miles per hour? (3.28

FinnZ [79.3K]

Answer:

23.086 mile/h

Explanation:

Given,

Distance Tyson Gay run = 100 m

time of run, t = 9.69 s

average speed of the in mph = ?

Speed of the Gay = $\dfrac{distance}{time}$

$v = \dfrac{100}{9.69}$

v = 10.32 m/s

1 m = 3.281 ft

10.32 m = 33.86 ft

1 mile = 5280 ft

1 ft = 1.8939 x 10⁻⁴ mile

33.86 ft/s = 6.413 x 10⁻³ miles/s

Speed of Tyson in mile/hr = 6.413 x 10⁻³ x 3600

= 23.086 mile/h

Hence, speed of Tyson Gay's in mile/ hr is equal to 23.086 mph.

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3 years ago

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Determine the kinetic energy of a 2000 kg roller coaster car that is moving at the speed of 10 ms

kondaur [170]

Answer:

$\boxed {\boxed {\sf 100,000 \ Joules}}$

Explanation:

Kinetic energy is energy due to motion. The formula is half the product of mass and velocity squared.

$E_k= \frac{1}{2} mv^2$

The mass of the roller coaster car is 2000 kilograms and the car is moving 10 meters per second.

m= 2000 kg
s= 10 m/s

Substitute these values into the formula.

$E_k= \frac{1}{2} (2000 \ kg ) \times (10 \ m/s)^2$

Solve the exponent.

(10 m/s)²= 10 m/s * 10 m/s= 100 m²/s²

$E_k= \frac{1}{2} (2000 \ kg ) \times (100 \ m^2/s^2)$

Multiply the first two numbers together.

$E_k= 1000 \ kg \times (100 \ m^2/s^2)$

Multiply again.

$E_k= 100,000 \ kg*m^2/s^2$

1 kilogram square meter per square second is equal to 1 Joule.
Our answer of 100,000 kg*m²/s² is equal to 100,000 Joules.

$E_k= 100,000 \ J$

The roller coaster car has <u>100,000 Joules</u> of kinetic energy.

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3 years ago

Describe how charge is transferred from the ruler to the metal rod.

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When the ruler is broughı near the inetal knob, it repels electrons in the metal. Electrons move away froni the ruler and down the metal rod. The knob now has a positive charge. The thin pieces of metal foil at the bottom of the metal rod now have a negative charge.

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3 years ago

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What is the acceleration of a car that starts from rest and attains a final speed of 20 m

OlgaM077 [116]

Answer:

323.9

Explanation:

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4 years ago

A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr

nikklg [1K]

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

a= μ g ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g =ω²X

We know for spring mass system natural frequency given as

$\omega=\sqrt{\dfrac{k}{M+m}}$

By putting the values

$\omega=\sqrt{\dfrac{150}{5.67+1}}$

ω = 4.47 rad/s

μ g =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

$\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2$

$kX^2=(m+M)v^2$

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

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3 years ago

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