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Korvikt [17]
3 years ago
6

If Rebecca is traveling at 7 m/s east for 1 hour how much is Rebecca displaced

Physics
1 answer:
Mekhanik [1.2K]3 years ago
3 0

Answer:

25.2 km east

Explanation:

The relationship between velocity and displacement is given by:

v=\frac{d}{t}

where

v is the velocity

d is the displacement

t is the time

In this problem, we have:

v=7 m/s east is Rebecca's velocity

t=1 h=60 min=3600 s is the time taken

Re-arranging the equation, we can find Rebecca's displacement:

d=vt=(7 m/s)(3600 s)=25,200 m=25.2 km

and the direction is same as velocity (east)

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In 1666 at the age of 23, what scientist
scoray [572]

Answer:

<h3>B - Isaac Newton</h3>

Explanation:

He first thought of his system of gravitation which he hit upon by observing an apple fall from a tree,

The incident occurring in the late summer of 1666.

3 0
3 years ago
Based on the graph of kinetic energy given (gray curve in the graphing window), sketch a graph of the baseball's gravitational p
chubhunter [2.5K]

1) Physical principles:

a) Total mechanical energy = kinetic energy + potential energy.

b) Total mechanical energy is conserved (neglecting external forces, like drag and friction)

2) Notation:

a) Total mechanical energy: ME.

b) Kinetic energy: KE

c) Gravitational potential energy: PE

∴ ME = KE + PE = constant

3) Solution:

a) Since, ME is conserved, it is constant and would be represented in the graph by a horizontal line.

b) At start (t = 0), the ball has only KE, so KE =ME = E and PE = 0

c) As the time goes, the ball gains altitude (PE increases) and loses speed (KE decreases).

d) PE increases from 0 to a maximum value. In the graph that happens at t = 2s.

At that point, KE = 0, and PE = ME.

That is the point of highest altitude and where the speed is zero.

d) From t = 2 seg, the ball starts to lose altitude, then the ball loses PE, and gains KE.

Just before reaching the ground, at t = 4s, the ball has the same initial KE and PE as at t = 0: KE = ME and PE = 0.

The PE may be sketched on the same graph along with the KE and the ME.

The graph is attached. The red line is the ME and the blue line is the PE.

Note that at any point in the graph PE + KE = ME.

5 0
4 years ago
Read 2 more answers
Two children, Ahmed and Jacques, ride on a merry-go-round. Ahmed is at a greater distance from the axis of rotation than Jacques
Alex

Answer:

A. Ahmed has a greater tangential speed than Jacques.

D. Jacques and Ahmed have the same angular speed.

Explanation:

Kinematics of the merry-go-round

The tangential speed of the merry-go-round is calculated using the following formula:

v = ω*R

Where:

v is the tangential speed in meters/second (m/s)

ω is the angular speed in radians/second (rad/s)

R is the angular speed in meters (m)

Data

dA = RA  : Ahmed distance to the axis of rotation

dJ = RJ   : Jacques distance to the axis of rotation

Problem development

We apply the formula (1)

v = ω*R

vA= ω*RA : Ahmed  tangential speed

vJ= ω*RJ  :  Jacques  tangential speed

Ahmed is at a greater distance from the axis of rotation than Jacques, then,

RA ˃ RJ and Ahmed and Jacques have the same  speed ω, then:

vA ˃ vJ

6 0
3 years ago
Suppose, we have a parallel plate capacitor and we know the following about it: Area of each plate = $0.0012m^2$ Distance betwee
FromTheMoon [43]

Answer:

udegsgsfsjtajtsiaitskhsigsfjsjgsyoeyodogdoydyosiskgoydoydiydiyddohdohdohyodhodohdoydoydoydosoydogdkgdohdhdohdgodyodoydoydyodoydyodoydohxohdohxohxohxohxogxkx skzjgz a few days

7 0
4 years ago
An airliner lands with a speed of 50.0 m/s. Each wheelof the
Yuki888 [10]

Answer:

0.52378

Explanation:

V = Velocity = 50 m/s

r = Radius = 1.25 m

I = Moment of inertia = 110 kgm²

N = Weight supported by wheels = 14000 N

The angular velocity

\omega=\dfrac{V}{r}\\\Rightarrow \omega=\dfrac{50}{1.25}\\\Rightarrow \omega=40\ rad/s

\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{40-0}{0.48}\\\Rightarrow \alpha=83.33\ rad/s^2

Frictional force is given by

f=\dfrac{I\alpha}{r}\\\Rightarrow f=\dfrac{110\times 83.33}{1.25}\\\Rightarrow f=7333.04\ N

The coefficient of friction is given by

\mu=\dfrac{f}{N}\\\Rightarrow \mu=\dfrac{7333.039}{14000}\\\Rightarrow \mu=0.52378

The coefficient of friction is 0.52378

8 0
3 years ago
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