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snow_tiger [21]
3 years ago
6

Your friend is bragging about his motorcycle. He claims that it can go from a stopped position to 50 miles per hour in three sec

onds. He is describing the motorcycle's
speed.

acceleration.

velocity.

direction.
Physics
2 answers:
nalin [4]3 years ago
8 0
If the motorcycle can go from 0 to 50mph in 3 seconds the person is describing the motorcycle's acceleration. 
tensa zangetsu [6.8K]3 years ago
5 0

Answer:

acceleration

Explanation:

He is describing the acceleration of motorcycle.

Here the initial velocity is zero, final velocity is 50 miles / hr, time taken is 3 sec.

Convert miles per hour into metre per second.

50 miles / hr = 22.352 m/s

using first equation of motion,

v = u + a x t

a = \left ( \frac{v - u}{t} \right )

a = \left ( \frac{22.352 - 0}{3} \right )

a = 7.45 m/s^2

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3 years ago
A hoop and a solid disc are relased from rest
Murljashka [212]

Answer:

1) The hoop and a solid disc rolling without slipping down an incline plane.

Their final velocities are proportional to their moment of inertia.

The condition for moment of inertia: v = ωR

We will use conservation of energy.

<u>For the hoop:</u>

K_1 + U_1 = K_2 + U_2\\0 + m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}I\omega_h^2 + 0

They are released from rest, so their initial kinetic energy is zero. And when they reach the bottom, their final potential energy is also zero.

The moment of inertia of a hoop is

I_h = m_hR^2

Let's continue with the energy equations:

m_h gh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}(m_hR^2)(\frac{v_h^2}{R^2})\\m_hgh = \frac{1}{2}m_hv_h^2 + \frac{1}{2}m_hv_h^2\\m_hgh = m_hv_h^2\\v_h = \sqrt{gh}

Similarly <u>for the solid disk</u> with a moment of inertia of (1/2)mR^2:

K_1 + U_1 = K_2 + U_2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}I_d\omega_d^2\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{2}(\frac{1}{2}m_dR^2)(\frac{v_d^2}{R^2})\\m_dgh = \frac{1}{2}m_dv_d^2 + \frac{1}{4}m_dv_d^2\\m_dgh = \frac{3}{4}m_dv_d^2\\v_d = \sqrt{\frac{4gh}{3}}

Comparing the final velocities, we can conclude that the solid disk reaches the bottom first.

2) The angular acceleration of the pebble is equal to the angular acceleration of the tire, since they stuck together. We can deduce the angular acceleration of the tire from the linear acceleration of the bicycle.

The kinematics equations states that

v = v_0 + at\\4.47 = 0 + 2a\\a = 2.235 ~m/s^2

where a is the linear acceleration.

The relation with the angular and linear acceleration is

a = \alpha R

where R is the radius of the tire. Since it is not given in the question, we will leave it as R.

The angular acceleration of the small pebble is

\alpha = 2.235/R ~m/s^2

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When you breathe in, or inhale, your diaphragm contracts (tightens) and moves downward. This increases the space in your chest cavity, into which your lungs expand. The intercostal muscles between your ribs also help enlarge the chest cavity. They contract to pull your rib cage both upward and outward when you inhale.
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A ladder rests against a vertical wall at a point 12 feet from the floor. The angle formed by the ladder and the floor is 63°. C
GenaCL600 [577]

Answer:

length of the ladder is 13.47 feet

base of wall to latter distance 6.10 feet

angle between ladder and the wall is 26.95°

Explanation:

given data

height h  = 12 feet

angle 63°

to find out

length of the ladder ( L) and length of wall to ladder ( A) and angle between  ladder and the wall

solution

we consider here angle between base of wall and floor is right angle

we apply here trigonometry rule that is

sin63 = h/L

put here value

L = 12 / sin63

L = 13.47

so length of the ladder is 13.47 feet

and

we can say

tan 63 = h / A

put here value

A = 12 / tan63

A = 6.10

so base of wall to latter distance 6.10 feet

and

we say here

tanθ = 6.10 / 12

θ = 26.95°

so angle between ladder and the wall is 26.95°

8 0
3 years ago
A ball is dropped off a very tall canyon ledge.
oksano4ka [1.4K]

Answer:

V = 49.05 [m/s]

Explanation:

We can easily find the result using kinematics equations, first, we will find the distance traveled during the 5 seconds.

y =y_{o}+(v_{o}*t)+(\frac{1}{2}*g*t^{2} )

where:

Yo = initial position = 0

y = final position [m]

Vo = initial velocity = 0

t = time = 5 [s]

g = gravity aceleration = 9.81 [m/s^2]

The initial speed is zero, as the body drops without imparting an initial speed. Therefore:

y = 0 + (0*5) + (0.5*9.81*5^2)

y = 122.625[m]

Now using the following equation we can find the speed it reaches during the 5 seconds.

v_{f} ^{2}= v_{i} ^{2}+(2*g*y)\\v_{f}=\sqrt{2*9.81*122.625} \\v_{f}=49.05 [m/s]

6 0
4 years ago
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