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Sati [7]
3 years ago
11

How would you find the speed of a ball thrown in to the air- it takes 6 seconds to get tho the top. The height is not given.

Physics
1 answer:
Alecsey [184]3 years ago
7 0
You would need to know how fast you threw the ball in the air once that then multiply the seeped and the time to get your answer<span />
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An air craft heads north at 320 km/hr relative to the wind. the wind velocity is 80km/hr from the north. find the relative veloc
Gnoma [55]

Answer:

Relative to the ground, the velocity of the aircraft is 240 km/hr

Explanation:

Relative velocity is different from normal velocity;

When 2 objects are moving in opposite directions towards each other, they will appear to be faster than they actually are;

This is known as the relative velocity;

The information tells us we have the aircraft moving 320 km/hr northwards relative to the wind;

The wind is in the opposite direction at 80 km/hr;

R = relative velocity of the aircraft

v = actual velocity of the aircraft

w = velocity of the wind

R = v + w

Note: if the wind was moving in the same direction, the formula would be R = v - w

320 = v + 80

v = 320 - 80

v = 240

The velocity relative to the ground is simply the actual velocity as the ground doesn't move;

So, relative to the ground, the velocity of the aircraft is simply 240 km/hr

7 0
3 years ago
How would playing a game of soccer, baseball, or basketball be different if inertia didn't exist?
BigorU [14]
Inertia is what keeps everything moving, so if it didn't exist, the balls wouldn't keep going when they are kicked, or thrown.

4 0
4 years ago
Question 3 of 10
AlexFokin [52]
The answer is Jupiter
6 0
3 years ago
Read 2 more answers
Please help me with this question
valkas [14]

Answer: m∠P ≈ 46,42°

because using the law of sines in ΔPQR

=> sin 75°/ 4 = sin P/3

so ur friend is wrong due to confusion between edges

+) we have: sin 75°/4 = sin P/3

=> sin P = sin 75°/4 . 3 = (3√6 + 3√2)/16

=> m∠P ≈ 46,42°

Explanation:

4 0
3 years ago
A box is initially sliding across a frictionless floor toward a spring which is attached to a wall. the box hits the end of the
Serjik [45]
The elastic potential energy of a spring is given by
U= \frac{1}{2}kx^2
where k is the spring's constant and x is the displacement with respect to the relaxed position of the spring.

The work done by the spring is the negative of the potential energy difference between the final and initial condition of the spring:
W=-\Delta U =  \frac{1}{2}kx_i^2 -  \frac{1}{2}kx_f^2

In our problem, initially the spring is uncompressed, so x_i=0. Therefore, the work done by the spring when it is compressed until x_f is
W=- \frac{1}{2}kx_f^2
And this value is actually negative, because the box is responsible for the spring's compression, so the work is done by the box.
8 0
4 years ago
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