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Elden [556K]
3 years ago
5

/How much kinetic energy does an 80 kg man have while running at 1.5 m/s

Physics
1 answer:
igomit [66]3 years ago
7 0
K.E = 0.5 * m * v^2 
  = 0.5 * 80 * (1.5)^2 
K.E = 90 joule 
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A thin, flat washer is a disk with an outer diameter of 14 cm and a hole in the center with a diameter of 7 cm. The washer has a
Kryger [21]

Answer:

2583.9 N/C

Explanation:

Parameters given:

Outer diameter = 14 cm

Outer radius, R = 7cm = 0.07m

Inner diameter = 7 cm

Inner radius, r = 3.5 cm = 0.035m

Charge of washer = 8 nC = 8 * 10^(-9)C

Distance from washer, z = 33 cm = 0.33m

The electric field due to a washer (hollow disk) is given as:

E = k * σ * 2π [ 1 - z/(√(z² + R²)]

Where σ = charge per unit area

σ = q/π(R² - r²)

σ = 8 * 10^(-9) /(π*(0.07 - 0.035)²)

σ = 2.077 * 10^(-6) C/m²

=> E = 9 * 10^9 * 2.077 * 10^(-6) * 2π * [1 - 0.33/(√(0.33² + 0.07²)]

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E = 2583.9 N/C

6 0
3 years ago
In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the gr
NikAS [45]

A) 2.64t

B) 2.64h

C) 2.64D

Explanation:

A)

The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The time of flight of a projectile can be found from the equations of motion, and it is found to be

t=\frac{2u sin \theta}{g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the time of flight is t.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:

t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t

B)

The maximum height reached by a projectile can be also found using the equations of motion, and it is given by

h=\frac{u^2 sin^2\theta}{2g}

where

u is the initial speed

\theta is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the maximum height is h.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:

h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h

C)

The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by

D=\frac{u^2 sin(2\theta)}{g}

where:

u is the initial speed

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In this problem, when the athlete is on the Earth, the horizontal distance covered is D.

When she is on Mars, the acceleration due to gravity is:

g'=0.379 g

where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:

D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D

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krok68 [10]

Answer:

Think it is C

Explanation:

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the answer is waxing gibbous moon

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