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Dovator [93]
2 years ago
14

a child riding a bicycle at 15 meters per second accelerates at -3,0 meters per second? for 4.0 seconds. What is the child's spe

ed A at the end of this 4.0-second interval? (with work, all steps))
Physics
1 answer:
Vika [28.1K]2 years ago
5 0

Answer:

<h2> 27m/s</h2>

Explanation:

Given data

initital velocity u=15m/s

deceleration a=3m/s^2

time t= 4 seconds

final velocity v= ?

Applying the expression

v=u+at------1

substituting our data into the expression we have

v=15+3*4

v=15+12

v=27m/s

The velocity after 4 seconds is 27m/s

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N experiment is performed in deep space with two uniform spheres, one with mass 27.0 and the other with mass 107.0 . They have e
Reptile [31]

Answer:

Explanation:

Apply the law of conservation of energy

KE_i+PE_i=KE_f+PE_f

Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)

from the law of conservation of the linear momentum

m_1v_1=m_2v_2

Therefore,

Gm_1m_2[\frac{1}{r_f} -\frac{1}{r_1} ]=\frac{1}{2} (m_1v_1^2+m_2v_2^2)

=\frac{1}{2} [m_1v_1^2+m_2[\frac{m_1v_1}{m_2} ]^2]\\\\=\frac{1}{2} [m_1v_1^2+\frac{m_1^2v_1^2}{m_2} ]\\\\=\frac{m_1v_1^2}{2} [\frac{m_1+m_2}{m_2} ]

v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]

Substitute the values in the above result

v_1^2=[\frac{2Gm_2^2}{m_1+m_2} ][\frac{1}{r_f} -\frac{1}{r_1} ]

=[\frac{2(6.67\times 10^-^1^1)(107)^2}{27+107} ][\frac{1}{26} -\frac{1}{41}] \\\\=1.6038\times 10^-^1^0\\\\v_1=\sqrt{1.6038\times 106-^1^0} \\\\=1.2664 \times 10^-^5m/s

B)  the speed of the sphere with mass 107.0 kg is

v_2=\frac{m_1v_1}{m_2}

=[\frac{27}{107} ](1.2664 \times 10^-^5)\\\\=3.195\times 10^-^6m/s

C)  the magnitude of the relative velocity with which one sphere is

v_r=v_1+v_2\\\\=1.2664\times 10^-^5+3.195\times10^-^6\\\\=15.859\times10^-^6m/s

D) the distance of the centre is proportional to the acceleration

\frac{x_1}{x_2} =\frac{a_1}{a_2} \\\\=\frac{m_2}{m_1} \\\\=3.962

Thus,

x_1=3.962x_2

and

x_2=0.252x_1

When the sphere make contact with eachother

Therefore,

x_1+x_2+2r=41\\x_1+0,252x_1+2r=41\\1.252x_1+2r=41\\x_1=32.747-1.597r

And

x_1+x_2+2r=41\\3.962x_2+x_2+2r+41\\4.962x_2+2r=41\\x_2=8.262-0.403r

The point of contact of the sphere is

32.747-1.597r=8.262-0.403r\\\\r=\frac{24.485}{1.194} \\\\=20.506m

3 0
3 years ago
Please help, no links please! I dont understand stand this question and im going to cry
Vaselesa [24]

Answer:

I'm pretty sure its helium

3 0
2 years ago
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What are non-contact forces please include example in your answer
Alina [70]

Answer:

Gravitational force.  Magnetic force.  Electrostatics.  Nuclear force.

Explanation:

Apple falling from a tree

raindrops falling from the sky

4 0
3 years ago
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Every football field has two 300kg field goal posts separated by 110m of football field (that includes the endzones). What is th
Mama L [17]

Answer:

4.96×10¯¹⁰ N

Explanation:

The following data were obtained from the question:

Mass 1 (M1) = 300 Kg

Mass 2 (M2) = 300 Kg

Separating distance (r) = 110 m

Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²

Gravitational force (F) =?

The gravitational force between the two goal posts can be obtained as follow:

F = GM1M2 / r²

F = 6.67×10¯¹¹ × 300 × 300 / 110²

F = 6.003×10¯⁶ / 12100

F = 4.96×10¯¹⁰ N

Therefore the gravitational force between the two goal posts is 4.96×10¯¹⁰ N

3 0
2 years ago
one thing that stringed, wind, and percussion instruments have in common in regard to the sounds they produce is that a. the ins
mario62 [17]

The answer is:

d) the sound originates from a vibration.

The explanation:

The sound waves are generated by a sound source, such as the vibrating diaphragm of a stereo speaker. The sound source creates vibrations in the surrounding medium. As the source continues to vibrate the medium, the vibrations propagate away from the source at the speed of sound, thus forming the sound wave.

3 0
3 years ago
Read 2 more answers
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