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iogann1982 [59]
4 years ago
9

If a 10 kg object has 200 J of gravitational potential energy, how high was has it been lifted above the ground?

Physics
1 answer:
Fantom [35]4 years ago
6 0
Potential energy is possessed by the body while at rest while the body in motion possesses kinetic energy.
Potential energy is given by mgh where m is the mass of the body, g is the gravitational acceleration(9.8m/s^2) and h is the height.
Therefore, 200J = 10  × 9.8 ×h
                        h = 200/98
                        h = 2.041 m
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Is it possible to have negative velocity but positive acceleration? If so, what would this mean?
dusya [7]

Observe that the object below moves in the negative direction with a changing velocity. An object which moves in the negative direction has a negative velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a positive acceleration). The dot diagram shows that each consecutive dot is not the same distance apart (i.e., a changing velocity). The position-time graph shows that the slope is changing (meaning a changing velocity) and negative (meaning a negative velocity). The velocity-time graph shows a line with a positive (upward) slope (meaning that there is a positive acceleration); the line is located in the negative region of the graph (corresponding to a negative velocity). The acceleration-time graph shows a horizontal line in the positive region of the graph (meaning a positive acceleration).

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4 0
3 years ago
He does whatever a spider can. Spider-Man, who has a mass of 76 [kg], is clinging onto an inclined wall forming an inclination a
Aliun [14]

Answer:570.54 N

Explanation:

Given

mass of man=76 kg

\theta =50^{\circ}

As man is standing over inclined building therefore

its weight has two components i.e. sin and cos component

Force perpendicular to inclined wall

F=mgcos\theta =76\times 9.8\times \sin 50

F=570.54 N

4 0
3 years ago
Where are you on Earth if you experience each of the following? (Refer to the discussion in Observing the Sky: The Birth of Astr
Aloiza [94]

Explanation:

We know that the sky appears to us like a sphere called as celestial sphere which appears to rotate around an imaginary axis because of Earth's rotation. Since the axis cuts the celestial sphere at celestial poles all the object seems to circle around the celestial poles.

Condition 1: The stars rise and set perpendicular to the horizon

The observer is at the equator

Condition 2: The stars circle the sky parallel to the horizon

The observer is at the Pole of the Earth

Condition 3: The celestial equator passes through the zenith

The observer is at the equator

Condition 4: In the course of a year, all stars are visible

The observer is at the equator

Condition 5: The Sun rises on March 21 and does not set until September 21 (ideally)

The observer is at North Pole

7 0
3 years ago
A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri
GREYUIT [131]

Answer:

|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}

Now, substitute this into 'dF':

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)

Now we can integrate dF over the rod.

\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)

4 0
3 years ago
Light in the air is incident at an angle to the surface of (12.0 A) degrees on a piece of glass with an index of refraction of (
Orlov [11]

The question is incomplete. You dis not provide values for A and B. Here is the complete question

Light in the air is incident at an angle to a surface of (12.0 + A) degrees on a piece of glass with an index of refraction of (1.10 + (B/100)). What is the angle between the surface and the light ray once in the glass? Give your answer in degrees and rounded to three significant figures.

A = 12

B = 18

Answer:

18.5⁰

Explanation:

Angle of incidence i = 12.0 + A

A = 12

= 12.0 + 12

= 14

Refractive index u = 1.10 + B/100

= 1.10 + 18/100

= 1.10 + 0.18

= 1.28

We then find the angle of refraction index u

u = sine i / sin r

u = sine24/sinr

1.28 = sine 24 / sine r

1.28Sine r = sin24

1.28 sine r = 0.4067

Sine r = 0.4067/1.28

r = sine^-1(0.317)

r = 18.481

= 18.5⁰

4 0
3 years ago
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