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Mrrafil [7]
3 years ago
9

a cart is loaded with a brick and pulled up at a constant speed along an inclined plane to the height of a seattop. if the mass

of the loaded cart is 4.0 kg and the height of the seat top is 0.15 meters, then what is the potential energy of the loaded cart at the height of the seat top?​
Physics
2 answers:
Fudgin [204]3 years ago
8 0

Answer:

6 J

Explanation:

Gravitational Potential Energy can be calculated using the formula:

PE = mgh

where m is mass of object in <em>k</em><em>g</em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>g is the gravitational acceleration of 10 m/s^2

h is the height at which object is thrown in <em>m</em><em>.</em>

Hence, by substituting in the values, we get:

GPE = (4.0)(10)(0.15)

<u>6</u><u> </u><u>J</u>

Hence, the potential energy is <u>6</u><u> </u><u>J</u><u>.</u>

ruslelena [56]3 years ago
4 0

Answer: Stuck in the same one!

Explanation:

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Calcium bromide is the name for compounds with the chemical formula CaBr2(H2O)x.

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A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se
zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, \lambda = 1\ nC = 1\times 10^{- 9}\ C

Charge density of rod 2, \lambda = - 1\ nC = - 1\times 10^{- 9}\ C

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}

Also,

\vec{E} = \frac{2K\lambda }{R}                  (1)

where

K = electrostatic constant = \frac{1}{4\pi \epsilon_{o} R}

R = Distance

\lambda = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C

\vec{E} = 1800\ N/C     (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C

\vec{E'} = 1800\ N/C     (towards)

Now, the total field at the origin is the sum of both the fields:

\vec{E_{net}} = 1800 + 1800 = 3600\ N/C

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A 28 kg mass suspends from a light rope 18 m long &amp; is held to one side by the horizontal force, F, as shown below.
frutty [35]

Answer: 215.15 N

Explanation:

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On the other hand, we can calculate \theta as follows:

cos\theta=\frac{s}{l}

\theta=cos^{-1}(\frac{s}{l})

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Answer:

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Explanation:

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