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Inessa [10]
3 years ago
12

When Aditya pushes on Rachel and her bicycle, they accelerate at 0.22 m/s/s. If Aditya pushes on Rachel and her bicycle with twi

ce as much force, then her
acceleration will be?

A. 1/4 as much
B. 1/2 as much
C. twice as much
D. the same
Physics
1 answer:
lord [1]3 years ago
4 0
I think it’s D but I’m not sure
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A careful photographic survey of Jupiter's moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to
Over [174]

The concept used to solve this problem is that given in the kinematic equations of motion. From theory we know that the change in velocities of a body is equivalent to twice the distance traveled by acceleration, in other words:

v_f^2-v_i^2 = 2ax

Where,

v_{f,i} = Final and initial velocity

a = Acceleration

x = Displacement

For the given case, the displacement is equivalent to the height (x = h) and the acceleration is the same gravitational acceleration (a = g). In turn we do not have initial speed therefore

v_f^2 = 2hg

v_f = \sqrt{2hg}

Our values are given as

h = 70km = 70*10^3m

g = 2m/s^2

Replacing we have that,

v_f = \sqrt{2hg}

v_f = \sqrt{2(70*10^3)(2)}

v_f = 529.15m/s

Therefore the speed with which the liquid sulfur left the volcano is 529.15m/s

6 0
3 years ago
The human eye can change the shape of its crystalline lens, and thus modify the lens strength (a.k.a., lens power) of the eye in
Softa [21]

Answer:

Good Luck!

Explanation:

5 0
2 years ago
How is lightning related to static electricity?
Naily [24]
Lighting is the static electricity stored in the clouds that is disposed to the earth.
8 0
3 years ago
A 4.60 gg coin is placed 19.0 cmcm from the center of a turntable. The coin has static and kinetic coefficients of friction with
Komok [63]

Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\  g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s

\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM

Therefore the speed in RPM will be 62.64 RPM.

5 0
2 years ago
Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at
PSYCHO15rus [73]

Answer:

|\Delta v |=\sqrt{\frac{4Gm}{d} }

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

\mu= \frac{m_1m_2}{m_1+m_2}

now, since both the masses have mass m

therefore,

\mu= \frac{m^2}{2m}

= m/2

The final K.E. of the particles is

KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2

Distance between two particles is d and the gravitational potential energy between them is given by

PE_{Gravitational}= \frac{Gmm}{d}

By law of conservation of energy we have

KE_{initial}+KE_{final}= PE_{gravitaional}

Now plugging the values we get

0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}

|\Delta v |=\sqrt{\frac{4Gm}{d} }

=\sqrt{\frac{Gm}{d} }

This the required relation between G,m and d

5 0
2 years ago
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