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SVETLANKA909090 [29]
2 years ago
12

A 480-kilogram horse runs across a field at a rate of 40 km/hr. What is the magnitude of the horse's momentum?

Physics
2 answers:
kirill115 [55]2 years ago
6 0
P=mv
<span>p=480 (40)= 19 200 kg km/hr
Hope this helped! :3</span>
Andrej [43]2 years ago
4 0
Momentum (p) = mass × velocity

so, 480×40 = 19,200 kg km/hr

so the answer is C !!
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Answer:

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Explanation:

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2 years ago
Use Newton's laws of motion to explain why it is important that baseballs and softballs each have a small acceptable range of ma
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2 years ago
A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag
klio [65]

Answer:

57.94°

Explanation:

we know that the expression of flux

\Phi =E\times S\times COS\Theta

where Ф= flux

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we have Given Ф= 78 \frac{Nm^{2}}{sec}

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4 0
2 years ago
The food calorie, equal to 4186J , is a measure of how much energy is released when food is metabolized by the body. A certain b
vovangra [49]
<h2>The hiker will go up to 850 m on the hill</h2>

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This energy is consumed in the potential energy acquired , while climbing up the hill.

The potential energy P.E = mass of hiker x acceleration due to gravity x height

Thus

140 x 4186 = 69 x 10 x h

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If the 20% of the total energy is used

the height h₀ = \frac{0.2x4186x140}{69x10} = 170 m

5 0
2 years ago
A majorette in a parade is performing some acrobatic twirlings of her baton. Assume that the baton is a uniform rod of mass 0.12
den301095 [7]

Question

Initially, the baton is spinning about a line through its center at angular velocity 3.00 rad/s.  What is its angular momentum? Express your answer in kilogram meters squared per second.

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Explanation:

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L = \frac{1}{{12}}m{l^2}\omega

Here, l is the length of the baton.

Substitute 0.120 kg for m, 3 rads/s for \omega[\tex] and 0.8 m for l [tex]\begin{array}{c}\\L = \frac{1}{{12}}m{l^2}\omega \\\\ = \frac{1}{{12}}\left( {0.120{\rm{ kg}}} \right){\left( {{\rm{80}}{\rm{.0 cm}}} \right)^2}{\left( {\frac{{1 \times {{10}^{ - 2}}{\rm{m}}}}{{1{\rm{ cm}}}}} \right)^2}\left( {{\rm{3}}{\rm{.00 rad/s}}} \right)\\\\ = 0.0192{\rm{ kg}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/s}}\\\end{array}

5 0
2 years ago
Read 2 more answers
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