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Talja [164]
3 years ago
15

ALOT OF POINTS PLZ HURRYQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQQWhat does Newton's third law sa

y about why momentum is conserved in collisions?
A: Equal Forces act in equal times, so the change in momentum for both objects must be equal.
B: Unequal forces act for unequal times, so the change in momentum for both objects must be unequal.
C: Equal forces act for unequal times, so the change in momentum for both objects must be equal.
D: Unequal forces act for equal times, so the change in momentum for both objects must be equal.
Physics
1 answer:
GarryVolchara [31]3 years ago
8 0

Answer:

A.) Equal Forces act in equal times, so the change in momentum for both objects must be equal.

(Hope this helps! Btw, I am the first to answer.)

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3 years ago
A force in the +x -direction with magnitude F(x)=18.0N−(0.530N/m)x is applied to a 7.90 kg box that is sitting on the horizontal
dsp73

Answer:

v\approx 8.570\,\frac{m}{s}

Explanation:

The equation of equlibrium for the box is:

\Sigma F_{x} = 18\,N-(0.530\,\frac{N}{m} )\cdot x = (7.90\,kg)\cdot a

The formula for the acceleration, given in \frac{m}{s^{2}}, is:

a = \frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg}

Velocity can be derived from the following definition of acceleration:

a = v\cdot \frac{dv}{dx}

v\, dv = a\, dx

\frac{1}{2}\cdot v^{2} = \int\limits^{17\,m}_{0\,m} {\frac{18\,N-(0.530\,\frac{N}{m} )\cdot x}{7.90\,kg} } \, dx

\frac{1}{2}\cdot v^{2} =\frac{18\,N}{7.90\,kg}  \int\limits^{17\,m}_{0\,m}\, dx  - \frac{0.530\,\frac{N}{m} }{7.90\,kg} \int\limits^{17\,m}_{0\,m} {x} \, dx

\frac{1}{2}\cdot v^{2} = (2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}

v =\sqrt{2\cdot[(2.278\,\frac{m}{s^{2}})\cdot x |_{0\,m}^{27\,m}-(0.034\,\frac{1}{s^{2}})\cdot x^{2}|_{0\,m}^{27\,m}]  }

The speed after the box has travelled 17 meters is:

v\approx 8.570\,\frac{m}{s}

3 0
3 years ago
sledge (including load) weighs 5000 N. It is pulled on level snow by a dog team exerting ahorizontal force on it. The coefficient
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Answer:

2.5 x 10^{5} J

Explanation:

weight = 5,000 N

coefficient of friction = 0.05

distance = 1000 m

how much work is done by the dogs pulling the sledge

work done = force x coefficient of friction x distance

work done = 5000 x 0.05 x 1000 = 2.5 x 10^{5} J

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A scalar quantity is a measurement of a quantity, like temperature, or mass.

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