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dmitriy555 [2]
4 years ago
15

Help!!! What is the kinetic energy of a 0.1kg baseball thrown at 20m/s? With work please

Physics
1 answer:
True [87]4 years ago
4 0

Kinetic Energy = 1/2 * mass * velocity^2

Mass = 0.1 kg

Velocity = 20 m/s

Kinetic energy = 1/2 * 0.1 * 20^2

Kinetic energy = 1/2 * 0.1 * 400

Kinetic energy = 20 J

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WILL MARK BRAINLIESTFor the simple harmonic motion equation d =4sin(8pit) what is the period?
shtirl [24]
The period of any wave is the time it takes for its angle
to go from zero to 2pi .

The 'sin' function is a wave.  The angle of this one is (8pi t).

When t=0, the angle is zero.
Wonderful.
Now, how long does it take for the angle to grow to 2pi ?

I*n other words, when is (8pi t) = 2pi ?

Divide each side by '2pi': . . . . . 4 t = 1

Divide each side by ' 4 ': . . . . . t = 1/4

And there you are.  Every time 't' grows by 1/4, (8pi t) grows by 2pi.
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4 0
4 years ago
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
What is the limitation of relative dating?
astraxan [27]
Your not really supposed to date your relative
7 0
3 years ago
What is the atomic number of this element?
Zina [86]
The atomic number is the number of protons in the nucleus of an atom. The number of protons define the identity of an element (i.e., an element with 6 protons is a carbon


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5 0
4 years ago
Read 2 more answers
Marcy pulls a backpack on wheels down the 100 m hall. The 60N force is applied at an angle of 30° above the horizontal. How much
shutvik [7]

Work= Fcos∆×S

W=60N×cos 30⁰×100

W=60×0.866×100=5196.1J

6 0
3 years ago
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