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2 years ago

Emmett is lifting a box vertically. Which forces are necessary for calculating the total force?

1 answer:

4 0

When Emmett is lifting a box vertically, the forces that must be added to calculate the total force are: the gravitational force, tension force(the force exerted by Emmett to the box and the force exerted by the box to Emmett), and air resistance force.

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Describe three pieces of evidence that support the Big Bang theory

lisov135 [29]

Explanation :

According to astronomers, the whole universe is started with a giant explosion called as Big Bang. Big Bang theory shows that the universe is extended from high density state.

There are some evidence for big bang as :

(1) There are some red shifts of different galaxies which means that the universe is expanding.

(2) Due to the expanding of universe, some of the new elements are created like hydrogen, deuterium etc.

(3) Microwaves are detected by orbiting detectors.

All this parameters shows that big bang theory was correct.

6 0

3 years ago

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A 350-kg roller coaster car starts from rest at point A and slides down a frictionless loop-the-loop (Pig. P7.41). (a) How fast

Leona [35]

Answer:

a)Velocity of car =v=16 m/s

b)Force against the track at point B=1.15* $10^{4}$ N

Explanation:

Given mass of roller coaster=m=350 kg

Position of A=Ha=25 m

Position of B=Hb=12 m

Net potential energy=mg(ha-hb)

Net potential energy=(350)(9.80)(25-12)

Net potential energy=44590 J

Using energy conservation

net kinetic energy=net potential energy

(1/2)mv^2=mg(ha-hb)

m=350

velocity=v=16 m/s

b)There two force acting,centripetal force upward and gravity downward.

Thus net force acting will be

Net force=(mv^2/r)-mg

Net force=14933.33-3430

Net force=1.15* $10^{4}$ N

3 0

2 years ago

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(6) Module 4 Study Guide

velikii [3]

We have the meats Arby’s we beat them kids

3 0

2 years ago

A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur

emmasim [6.3K]

• Net vertical force on the block:

∑ F = n - w = 0

(n = magnitude of normal force, w = weight)

n = w = m g

(m = mass, g = 9.8 m/s²)

n = (4 kg) (9.8 m/s²) = 39.2 N

• Net horizontal force:

∑ F = -f = m a

(f = mag. of friction, a = acceleration)

We have f = µ n = 0.5 (39.2 N) = 19.6 N, so

-19.6 N = (4 kg) a

a = -4.9 m/s²

With this acceleration, the block comes to a rest from an initial speed of 5 m/s, so that it travels a distance ∆x in this time such that

0² - (5 m/s)² = 2 (-4.9 m/s²) ∆x

∆x = (25 m²/s²) / (9.8 m/s²) ≈ 2.55 m ≈ 2.6 m

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2 years ago

What type of plate boundary is shown in this illustration?

Inga [223]

This is a transform Boundrary

8 0

3 years ago

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