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2 years ago

6

The value of gravitational acceleration of a body on Earth is 9.8 meters/second2. The gravitational potential energy for a 1.00

kilogram
object is found to be 12.5 joules. How high is this object above the ground level, where PE = 0?
O A. 1.3 meters
OB. 3.5 meters
OC. 2.0 meters
OD. 5.6 meters
O E. 7.9 meters

1 answer:

ANTONII [103]2 years ago

5 0

Answer:

A

Explanation:

i don't know but u think the answer is A

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The block in the diagram below is AT REST. However, the tension in the cable is not the only thing holding the block back. Stati

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Answer:

The tension in the rope is 229.37 N.

Explanation:

Given:

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Coefficient of static friction is, $\mu = 0.214$

Angle of inclination is, $\theta = 31.5$ °

Draw a free body diagram of the block.

From the free body diagram, consider the forces in the vertical direction perpendicular to inclined plane.

Forces acting are $mg\cos \theta$ and normal $N$ . Now, there is no motion in the direction perpendicular to the inclined plane. So,

$N=mg\cos \theta\\N=(33.2)(9.8)\cos (31.5)\\N=277.415\ N$

Consider the direction along the inclined plane.

The forces acting along the plane are $mg\sin \theta$ and frictional force, $f$ , down the plane and tension, $T$ , up the plane.

Now, as the block is at rest, so net force along the plane is also zero.

$T=mg\sin \theta+f\\T=mg\sin \theta +\mu N\\T= (33.2)(9.8)(\sin (31.5)+(0.214\times 277.415)\\T= 170+59.37\\T=229.37\ N$

Therefore, the tension in the rope is 229.37 N.

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You toss a rock of mass m vertically upward. Air resistance can be neglected. The rock reaches a maximum height h above your han

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Answer with Explanation:

We are given that

Mass of rock=m

Maximum height=h

a.At maximum height, velocity,v=0

We know that

$v^2=u^2-2gh$

$0+2gh=u^2$

$u^2=2gh$

Height,h=h/4

Again, $v'^2=u^2-2g\times \frac{h}{4}$

$v'^2=2gh-\frac{gh}{2}=\frac{4gh-gh}{2}=\frac{3gh}{2}$

$v'=\sqrt{\frac{3gh}{2}}=\sqrt{\frac{3\times 9.8 h}{2}}=3.83\sqrt h$

Where $g=9.8 m/s^2$

b.When height,h=3h/4

$v'^2=u^2-2gh$

$v'^2=2gh-2g\times \frac{3h}{4}=2gh-\frac{3gh}{2}=\frac{4gh-3gh}{2}=\frac{gh}{2}$

$v'=\sqrt{\frac{9.8h}{2}}=2.2\sqrt h$

$v'=2.2\sqrt h$

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2 years ago

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Somewhere in the vast flat tundra of planet Tehar, a projectile is launched from the ground at an angle of 60 degrees. It reache

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Answer:

R = 0.0503 m

Explanation:

This is a projectile launching exercise, to find the range we can use the equation

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Let's use trigonometry to find the speed

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A certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of What is the

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Answer:

<em>0.45 mm</em>

Explanation:

The complete question is

a certain fuse "blows" if the current in it exceeds 1.0 A, at which instant the fuse melts with a current density of 620 A/ cm^2. What is the diameter of the wire in the fuse?

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B) 0.63 mm

C.) 0.68 mm

D) 0.91 mm

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Current density of the fuse when it melts is 620 A/cm^2

Area of the wire in the fuse = I/ρ

Where I is the current through the fuse

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Area = 1/620 = 1.613 x 10^-3 cm^2

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1.613 x 10^-3 cm^2 = 1.613 x 10^-7 m^2

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$d^{2}$ = $\sqrt{ 2.05*10^-7}$

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