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NemiM [27]
3 years ago
8

The chart shows data for four moving objects.

Physics
1 answer:
vesna_86 [32]3 years ago
5 0

Answer:

Y

Explanation:

by adding all the numbers

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PLEASE HELP! I'LL GIVE BRAINLEST​
erma4kov [3.2K]

Answer:

8.162

Explanation:

W=mg

  = 2.2 x 3.71

  = 8.162 N

5 0
3 years ago
A hockey puck is traveling to the left with a velocity of v=10 when it is struck by the hockey stick
Lera25 [3.4K]
We have to calculate the impulse of a hockey puck.
Imp = m * ( v 1 - v 2 ) = m * Δ v
v 1 = - 10 i m/s,
v 2 = ( 20 * cos 40° ) i + ( 20 * sin 40° ) j =
= ( 20 * 0.766 ) i + ( 20 * 0.64278 ) j = ( 15.32 i + 12.855 j ) m/s
Δ v = ( 15.32 i + 12.855 j ) - ( - 10 i ) =
= 15.32 i + 12.855 j + 10 i = 25.32 i + 12.855 j
| Δv | = √ ( 25.32² + 12.855²) = √806.35 = 28.4 m/s
Imp = 0.2 kg * 28.4 m/s = 5.68 N-s
Answer: D ) 5.68 N-s. 
 
4 0
3 years ago
What velocity must a 2.25 kg croquet mallet have in order to have the same momentum as 1.25 kg ball that has a momentum 6.25 kg•
Ksenya-84 [330]

abcdefghijklmnopqrstuvwxyz

3 0
3 years ago
Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical
lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83

6 0
3 years ago
Read 2 more answers
Which statement relates most directly to the second law?
alukav5142 [94]
Statement three i do believe

7 0
3 years ago
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