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3 years ago

11

A motorcycle of mass 100 kilograms slowly rolls off the edge of a cliff and falls for three seconds before reaching the bottom o

f a gully. Its momentum upon hitting the ground is

1 answer:

Digiron [165]3 years ago

6 0

Answer:

2940 kg m/s

Explanation:

First of all, we need to find the final velocity of the motorcycle when it hits the ground. The horizontal component of the velocity is zero, so we can just find the vertical component, which is given by:

$v_y = v_0+gt$

where

v0 =0 is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration due to gravity

t = 3 s is the time of the fall

Substituting into the formula, we find

$v_y=0+(9.8 m/s^2)(3 s)=29.4 m/s$

And so, we can now find the momentum of the motorcycle upon hitting the ground, which is given by the product between the mass and the velocity:

$p=mv=(100 kg)(29.4 m/s)=2940 kg m/s$

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In February 1955, a paratrooper fell 365 m from an airplane without being able to open his chute but happened to land in snow, s

DaniilM [7]

Answer:

a. $i=4760 kg*m/s$

b. $D_U= 1.11 m$

Explanation:

a)

F= 120,000N

Kinetic energy @ impact = 120,000*depth

$K_E= (1/2)*85kg*(56m/s)^2$

$K_E=133280 J$

$D_U= \frac{133280J}{120000N} = 1.11 m$

b)

The momentum is equal to the impulse on him from the snow so:

$p=m*v$

$p=85kg*56m/s$

$i=4760 kg*m/s$

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3 years ago

A dentist’s drill starts from rest. After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min. Find the

mylen [45]

Answer:

Angular acceleration, is $708.07\ rad/s^2$

Explanation:

Given that,

Initial speed of the drill, $\omega_i=0$

After 4.28 s of constant angular acceleration it turns at a rate of 28940 rev/min, final angular speed, $\omega_f=28940\ rev/min=3030.58\ rad/s$

We need to find the drill’s angular acceleration. It is given by the rate of change of angular velocity.

$\alpha =\dfrac{\omega_f}{t}\\\\\alpha =\dfrac{3030.58\ rad/s}{4.28\ s}\\\\\alpha =708.07\ rad/s^2$

So, the drill's angular acceleration is $708.07\ rad/s^2$ .

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2 years ago

A uniform, solid, 2000.0 kgkg sphere has a radius of 5.00 mm. Find the gravitational force this sphere exerts on a 2.10 kgkg poi

mars1129 [50]

Answer:

$0.0110284391534\ N$

$0.0653784219002\ N$

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$m_1$ = Mass of sphere = 2000 kg

$m_2$ = Mass of other sphere = 2.1 kg

r = Distance between spheres

Force of gravity is given by

$F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 2000\times 2.1}{(5.04\times 10^{-3})^2}\\\Rightarrow F=0.0110284391534\ N$

The gravitational force is $0.0110284391534\ N$

$F=\dfrac{Gm_1m_2}{r^2}\\\Rightarrow F=\dfrac{6.67\times 10^{-11}\times 2000\times 2.1}{(2.07\times 10^{-3})^2}\\\Rightarrow F=0.0653784219002\ N$

The gravitational force is $0.0653784219002\ N$

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3 years ago

1. What is the name given to the total possibility of sound frequencies

exis [7]

Explanation:

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2. The difference between a musical note and another note at twice the frequency is called an octave.

3. Sound intensity varies with the inverse square of distance.

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2 years ago

Which best describes the electric field created by a positive charge?

Vikentia [17]

Answer:

Which best describes the electric field created by a positive charge? ... Its rays point toward the charge.

Explanation:

thank me later

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2 years ago