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Crank
3 years ago
6

Hey pls help thanks a lot

Physics
1 answer:
galben [10]3 years ago
4 0

Answer:

I am not sure about the answer as I don't have a proper calculator besides me now

Explanation:

but I used this equation:

(8.20)sin30(1-d)=10d

Idk whether it is correct or not, I'm just a student too

what is your method of doing this question?

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A block of mass 3 kg slides down a frictionless inclined plane of length 6 m and height 4 m. If the block is released from rest
Ludmilka [50]
(if it's frictionless the length doesn't even matter :) )
It would have the same kinetic energy down as the potential energy up. That is, mgh=\frac{mv^2}{2} or 2gh=v^2 (the mass doesn't even matter). The result is \sqrt{2gh}, so only the height matters really. It is almost 9 (it is \sqrt{80}=4\sqrt{5}).
6 0
4 years ago
Learning Goal: To apply the law of conservation of energy to an object launched upward in the gravitational field of the earth.
marishachu [46]

Answer:

h=\frac{1}{2}\frac{v^2}{g}

Explanation:

Let's assume that an object is launched straight upward in a gravitational field. Its initial kinetic energy is given by

K=\frac{1}{2}mv^2 (1)

where m is the mass and v is the initial speed.

As the object goes higher, its kinetic energy decreases and it is converted into gravitational potential energy, since the total mechanical energy (sum of kinetic and potential energy) must remain constant:

E=K+U=const.

At the highest point of the trajectory, the speed of the object is zero (v=0), so the kinetic energy is also zero (K=0), which means that all the kinetic energy has been converted into potential energy:

U=mgh (2)

where g is the gravitational acceleration and h is the maximum height of the object.

Due to conservation of energy, we can write that (1) and (2) are equal, so:

\frac{1}{2}mv^2=mgh

from which we can derive an expression for the maximum height reached by the object

h=\frac{1}{2}\frac{v^2}{g}

5 0
3 years ago
A railroad car moving at a speed of 3.49 m/s overtakes, collides, and couples with two coupled railroad cars moving in the same
Mandarinka [93]

Answer:

Explanation:

Using conservation law of momentum

let m₁ = mass of the railroad, initial u₁ = 3.49 m/s

let m₂ = mass of one of the coupled car,  u₂= 1.28m/s

let m₃ = mass of the second car u₃ = 1.28 m/s

m₁u₁ + m₂u₂ + m₃u₃ = v ( m₁ + m₂ + m₃)

since the masses are the same

m₁ = m₂ = m₃

m ( 3.49 + 1.28 + 1.28) = 3m v

6.05 m = 3 mv

v = 6.05 m / 3m = 2.0167 m/s

b) kinetic energy lost = energy before collision - energy after collision

= (0.5m₁u₁² + 0.5 ( m₁+m₂) u₂² - 0.5 ( m₁ + m₂ + m₃) v

= (6.58365 + 1.7712) - 6.5951 = 1.76J  

7 0
3 years ago
Four +2 μC charges are placed at the positions (10 cm, 0 cm), (−10 cm, 0 cm), (0 cm, 10 cm), and (0 cm, −10 cm) such that they f
Rufina [12.5K]

Answer:

The force on the charge at the origin is 0 N .

Explanation:

All charges are positive. So, in x axis force exerted by the charge located in the position (10 cm, 0 cm) will be canceled with the force exerted by the charge located in the position (-10 cm, 0 cm). In the same way, in y axis the force exerted by the charge located in the position (0 cm, 10 cm) will be canceled with the force exerted by the charge located in the position (0 cm, -10 cm).

4 0
4 years ago
A uniform electric field, with a magnitude of 370 N/C, is directed parallel to the positive x-axis. If the electric potential at
dem82 [27]

Answer:

\Delta U = 0.2072 J

Explanation:

Potential difference between two points in constant electric field is given by the formula

\Delta V = E.\Delta x

here we know that

E = 370 N/C

also we know that

\Delta x = 2.1 - 1.9 = 0.2 m

now we have

\Delta V = 370 (0.2) = 74 V

now change in potential energy is given as

\Delta U = Q\Delta V

\Delta U = (2.80 \times 10^{-3})(74)

\Delta U = 0.2072 J

3 0
4 years ago
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