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3 years ago

A box is pushed horizontally with constant speed across a rough horizontal surface. Which of the following must be true?

1 answer:

5 0

D

Because if an object is moving at a constant speed the force of friction must equal the applied (horizontal) force, and for it to be accelerating or decelerating, the force of friction and the applied force must be unequal

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Suppose that, while lying on a beach near the equator watching the Sun set over a calm ocean, you start a stopwatch just as the

Yuri [45]

Ans: Let d is the distance from height to our eyes.

Applying the Pythagoras theorem, we get,

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3 years ago

The handle of a hammer is 20 cm long and the metal end pulls

artcher [175]

Answer:

Given the experimental setup shown below, where should you put object C, ... applies a force of 150 N. What is the mechanical advantage of the hammer? ... Explain why tin snips, designed for cutting metal, have long handles and short blades. ... advantage for scissors, a first-class lever, is the length of the handle divided ...

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3 years ago

If the mass of a moving object is doubled, which variable will also double? *

iVinArrow [24]

Answer:

potential energy

Explanation:

this is because potential energy= mgh

so when m= 2m

p.e. = 2 times the previous value

Therefore , When mass is doubled , Potential energy gets doubled. When mass is halved , Potential energy decreases by half.

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3 years ago

If a horizontal force of 15N applied to a 50N block towards the east is just enough to cause that block to move

vitfil [10]

Answer:

40 N

Explanation:

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3 years ago

A hose on the ground projects a water current upwards at an angle 40 to the horizontal at velocity 20 m/s find height at which w

irinina [24]

Answer:

The water hits the wall at a height of 5.38 m

Explanation:

Projectile Motion

It's the type of motion that experiences an object projected near the Earth's surface and moves along a curved path exclusively under the action of gravity.

The object describes a parabolic path given by the equation:

$y=\tan(\theta )\cdot x-{\frac {g}{2v_{0}^{2}\cos ^{2}\theta }}\cdot x^{2}$

Where:

y = vertical displacement

x = horizontal displacement

θ = Elevation angle

vo = Initial speed

The hose projects a water current upwards at an angle of θ=40° at a speed vo=20 m/s.

The height at which the water hits a wall located at x=8 m from the hose is:

$y=\tan40^\circ\cdot 8-{\frac {9.8}{2*20^{2}\cos ^{2}40^\circ }}\cdot 8^{2}$

Calculating:

y = 5.38 m

The water hits the wall at a height of 5.38 m

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3 years ago