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Nostrana [21]
3 years ago
8

Compare the Vf calculated at the point of impact to the horizontal velocity you calculated using Δx and Δy. Were the vf and the

projectile velocity equal? If not what contributed to the difference? Think about the collision.
Physics
1 answer:
34kurt3 years ago
4 0

Answer: Velocities are not equal

Explanation: A projectile usually have a given initial velocity, the velocity must be resolve into components of x and y with respect to the velocity of the projectile in a certain angle. In a projectile trajectory the final velocity for the y- component is always 0m/s since the object reaches its maximum point.

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Kon'nichiwa~<br>please help me with this question!!​
andrezito [222]

Let's check the relationship

\\ \rm\hookrightarrow g=\dfrac{GM}{r^2}

\\ \rm\hookrightarrow g\propto G

So

  • Raindrops will fall faster . .
  • Also walking on ground would become more difficult as g increases.

Option C is wrong by now .Let's check D once

\\ \rm\hookrightarrow T\propto \dfrac{1}{\sqrt{g}}

  • So time period of simple pendulum would decrease.

4 0
2 years ago
Which means for obtaining hydrogen from water would require the most energy??
Flauer [41]
I think the correct answer would be to electrolyze water (run an electric current through it) to decompose it into hydrogen and oxygen. Assuming 100% efficiency, it is said that it needs about 40kWh per kilogram of water to fully decompose it.
8 0
3 years ago
Through the process of blank
maksim [4K]

Answer:

Subduction, Trench, Mantle

Explanation:

8 0
3 years ago
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Can I PLEASE get some help? I REALLY need it!
soldi70 [24.7K]
The answer is C. Hope this helps.
7 0
3 years ago
A polar bear runs at a speed of 11 m/s and has a mass of 380.2 kg. How much Kinetic energy does the bear have?
Yanka [14]

Answer:

\boxed{\sf Kinetic \ energy \ of \ the \ bear (KE) = 23002.1 \ J}

Given:

Mass of the polar bear (m) = 6.8 kg

Speed of the polar bear (v) = 5.0 m/s

To Find:

Kinetic energy of the polar bear (KE)

Explanation:

Formula:

\boxed{ \bold{\sf KE =  \frac{1}{2} m {v}^{2} }}

Substituting values of m & v in the equation:

\sf \implies KE =  \frac{1}{2}  \times 380.2 \times  {11}^{2}

\sf \implies KE = \frac{1}{ \cancel{2}}  \times  \cancel{2} \times 190.1 \times 121

\sf \implies KE = 190.1 \times 121

\sf \implies KE = 23002.1 \: J

\therefore

Kinetic energy of the polar bear (KE) = 23002.1 J

5 0
3 years ago
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