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Bas_tet [7]
3 years ago
8

Is Gasoline for you car a clean source of energy?

Physics
1 answer:
bixtya [17]3 years ago
4 0
No, it's not. It's none renewable source of energy causing carbon dioxide emission and global warming, and climate change.
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When will an object dropped from rest attain a speed of 30 m/s?
stich3 [128]
<h3><u>Answer</u> :</h3>

Initial velocity = zero (i.e., free fall)

Final velocity = 30m/s

Acceleration due to gravity = 10m/s²

For a body falling freely under the action of gravity, g is taken positive.

◈ <u>First equation of kinenatics</u> :

⇒ v = u + gt

⇒ 30 = 0 + 10t

⇒ t = 30/10

⇒ <u>t = 3s</u>

Hence, object will attain a speed of 30m/s after 3s.

8 0
3 years ago
A projectile is launched from ground level at an angle of 30 degrees above the horizontal. Neglect air resistance and consider t
Oduvanchick [21]

Answer:

just before landing the ground

Explanation:

Let the velocity of projection is u and the angle of projection is 30°.

Let T is the time of flight and R is the horizontal distance traveled. As there is no force acting in horizontal direction, so the horizontal velocity remains constant. Let the particle hits the ground with velocity v.

initial horizontal component of velocity, ux = u Cos 30

initial vertical component of velocity, uy = u Sin 30

Time of flight is given by

T = \frac{2u Sin\theta }{g}

Final horizontal component of velocity, vx = ux = u Cos 30

Let vy is teh final vertical component of velocity.

Use first equation of motion

vy = uy - gT

v_{y}=u_{y}- g \times \frac{2u Sin\theta }{g}

v_{y}=u Sin 30 - 2u Sin 30

vy = - u Sin 30

The magnitude of final velocity is given by

v = \sqrt{v_{x}^{2}+v_{y}^{2}}

v = \sqrt{\left (uCos 30  \right )^{2}+\left (uSin 30  \right )^{2}}

v = u

Thus, the velocity is same as it just reaches the ground.

6 0
3 years ago
A rock is sitting at the edge of a flat merry-go-round at a distance of 1.6 meters from the center. The coefficient of static fr
PSYCHO15rus [73]

Answer:

ω = 2.1 rad/sec

Explanation:

  • As the rock is moving along with the merry-go-round, in a circular trajectory, there must be an external force, keeping it on track.
  • This force, that changes the direction of the rock but not its speed, is the centripetal force, and aims always towards the center of the circle.
  • Now, we need to ask ourselves: what supplies this force?
  • In this case, the only force acting on the rock that could do it, is the friction force, more precisely, the static friction force.
  • We know that this force can be expressed as follows:

       f_{frs} = \mu_{s} * F_{n} (1)

      where μs = coefficient of static friction between the rock and the merry-

      go-round surface = 0.7, and Fn = normal force.

  • In this case, as the surface is horizontal, and the rock is not accelerated in the vertical direction, this force in magnitude must be equal to the weight of the rock, as follows:
  • Fn = m*g (2)
  • This static friction force is just the same as the centripetal force.
  • The centripetal force depends on the square of the angular velocity and the radius of the trajectory, as follows:

       F_{c} = m* \omega^{2}*r (3)

  • Since (1) is equal to (3), replacing (2) in (1) and solving for ω, we get:

       \omega = \sqrt{\frac{\mu_{s} * g}{r} } = \sqrt{\frac{0.7*9.8m/s2}{1.6m}} = 2.1 rad/sec

  • This is the minimum angular velocity that would cause the rock to begin sliding off, due to that if it is larger than this value , the centripetal force will be larger that the static friction force, which will become a kinetic friction force, causing the rock to slide off.
4 0
2 years ago
What would be a reasonable measurement of normal force if friction force measured 80 Newtons?
ruslelena [56]
These answers are very good. I made a 95% by using this as my answer key, but unfortunately, one of these is wrong. I don't know which one, all I know is that I got 19/20. Just wanted to throw that out there. :-) Good job, Sadaqasalaam3. And thank you.
4 0
3 years ago
Free body diagram definition <br>in physics​
kirill115 [55]

Answer:

In physics and engineering, a free body diagram (force diagram, or FBD) is a graphical illustration used to visualize the applied forces, moments, and resulting reactions on a body in a given condition.

4 0
3 years ago
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