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bezimeni [28]
2 years ago
8

a horizontal force of 100N is required to push a crate across a factory floor at a constant speed. What is the net force acting

on the crate/ What is the force of friction acting on the crate?
Physics
1 answer:
Bas_tet [7]2 years ago
8 0

If the crate is moving along the floor in the same direction with a constant speed, it is in dynamic equilibrium. Equilibrium means there is no net force acting on the crate.

Since there is no net force on the crate, there must be a friction force on the crate equal in magnitude and opposite in direction to the applied horizontal force. Therefore the force of friction acting on the crate is 100N.

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Which has the greater density, 5 kg of lead or 10 kg of gold?
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Gold is much heavier than lead. It is very dense. ... Therefore gold weighs 19.3 times as lead
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How does a virus make it into a cell and what happens to it as it tries to get to its target “The Nucleus”?
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Measuring the diameter of a specific object might include which pair of SI units
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Well, the diameter of a circle is simply a length, so your measurement will have units of length.  We just have to find an answer that has only units of length.

A).  gram, second . . . mass and time.  That can't be it.

B).  kilogram, ampere . . . mass and current.  That can't be it.

C).  centimeter, meter . . . both lengths. This one is looking good.

D).  candela, mole . . . light intensity and some chemical thing. That can't be it.

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8 0
3 years ago
The Earth's biosphere is consists of
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8 0
2 years ago
Salmon often jump waterfalls to reach their
PilotLPTM [1.2K]

The minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

The given parameters;

  • height of the waterfall, h = 0.432 m
  • distance of the Salmon from the waterfall, s = 3.17 m
  • angle of projection of the Salmon, = 30.8º

The time of motion to fall from 0.432 m is calculated as;

h = v_0_y + \frac{1}{2} gt^2\\\\0.432 = 0 + (0.5\times 9.8)t^2\\\\0.432 = 4.9t^2\\\\t^2 = \frac{0.432}{4.9} \\\\t^2 = 0.088\\\\t = \sqrt{0.088} \\\\t = 0.3 \ s

The minimum velocity of the Salmon jumping at the given angle is calculated as;

X = v_0_x t\\\\3.17 = (v_0\times cos(30.8)) \times 0.3\\\\10.567 = v_0\times cos(30.8)\\\\v_0 = \frac{10.567}{cos(30.8)} \\\\v_0 = 12.3 \ m/s

Thus, the minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.

Learn more here: brainly.com/question/20064545

8 0
1 year ago
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