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Bumek [7]
3 years ago
7

What makes is the difference between speed and velocity ​

Physics
2 answers:
astra-53 [7]3 years ago
5 0
Velocity includes direction
LUCKY_DIMON [66]3 years ago
3 0

Answer:

C. Velocity includes direction.

Explanation:

Velocity is speed with a direction.

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Answer:

The steam represents evaporation.

Explanation:

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Determine the potential difference between two charged parallel plates that are 0.50 cm apart and have an electric field strengt
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E = \frac{V}{r} \\ V = \frac{E}{r} \\ V = \frac{9.0V/cm}{0.5cm} \\ V = 18V
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*A car is going through a dip in the road whose curvature approximates a circle of radius 150m. At what velocity will the occupa
Valentin [98]

Answer:

v= 14.85 m/s

Explanation:

  • When at the bottom of the dip, the only force that keeps the car in the circular trajectory, is the centripetal force.
  • This force is not a new force, is just the net force aiming to the center of the circle.
  • In this case, is just the difference between the normal force (always perpendicular to the surface, pointing upward) and the force that gravity exerts on the car (which is known as the weight), pointing downward.
  • So, we can write the following expression:

       F_{cent} = F_{n} - F_{g}  (1)

  • It can be showed that the centripetal force is related to the speed by the following expression:
  • F_{cent} = m*\frac{v^{2}}{r} (2)
  • The normal force, it is called the apparent weight, because it would be the weight as measured by a scale.
  • Replacing (2) in (1), and solving for Fn, we get:

       F_{n} = m*\frac{v^{2} }{r} + m*g (3)

  • Now, we need to find the value of v that makes Fn, exactly 15% more than the weight m*g, so we can write the following equation:

      F_{n} = 1.15*F_{g} = m*\frac{v^{2}}{r} +F_{g}  (4)

  • Replacing Fg by its value, simplifying, and solving for v, we get:

       v = \sqrt{0.15*g*r} = \sqrt{9.8 m/s2*0.15*150m} = 14.85 m/s (5)

3 0
2 years ago
What property of light shown in the picture?
kupik [55]

Answer:

Explanation:

Reflection

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2 years ago
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A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u
uranmaximum [27]

Answer:

a) t = \sqrt{\frac{h}{2g}}

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}  

Where:

y_{0_{1}}: is the initial height = h

v_{0_{1}}: is the initial speed of ball 1 = 0 (it is dropped from rest)

y = h - \frac{1}{2}gt^{2}     (1)

Now, for ball 2 we have:

y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}    

Where:

y_{0_{2}}: is the initial height of ball 2 = 0

y = v_{0_{2}}t - \frac{1}{2}gt^{2}    (2)

By equating equation (1) and (2) we have:

h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}

t=\frac{h}{v_{0_{2}}}

Where the initial velocity of the ball 2 is:

v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh

Since v_{f_{2}}^{2} = 0 at the maximum height (h):

v_{0_{2}} = \sqrt{2gh}

Hence, the time when they pass each other is:

t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}

b) When they are passing the speed of each one is:

For ball 1:

v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}

The minus sign is because ball 1 is going down.

For ball 2:

v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

For ball 2:

y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!                  

7 0
3 years ago
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