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Rus_ich [418]
3 years ago
9

What is formula for time and velocity

Physics
1 answer:
Juliette [100K]3 years ago
4 0
Divide distance by the time it takes to travel that distance
the formula for time is divide distance/speed
You might be interested in
As you climb a mountain, you can expect the air temperature to decrease by 6.5 degrees C for every 1000 meters you ascend. This
Reptile [31]

Answer:

the answer will be C.  4 degrees C

Explanation:

you subtract base meters from peak meters to get 4 meters; then Multiply 6.5 by 4.

Then subtract that total from 30 degrees C

***as altitude increases, temperature decreases***

6 0
3 years ago
a basketball is tossed upwards with a speed of 5.0 m/s What is the maximum height reached by the basketball from its release poi
den301095 [7]

Answer:

1.275 m

Explanation:

Let the maximum height reached be h.

Here initial velocity, u = 5 m/s

Final velocity, V = 0

Use third equation of motion

V^2 = u^2 + 2 g h

0 = 25 - 2 × 9.8 × h

h = 25/19.6 = 1.275 m

3 0
4 years ago
Read 2 more answers
What does this mean?
Alexeev081 [22]
Well, in some systems the atoms melt or burn or change state like from liquid to gases when they reach a certain temperature. This could decrease the energy in the system.
Hope I helped
3 0
4 years ago
A large, metallic, spherical shell has no net charge. It is supported on an insulating stand and has a small hole at the top. A
TiliK225 [7]

Answer:

funkin

Explanation:

8 0
4 years ago
If the spring constant is doubled, what value does the period have for a mass on a spring?
quester [9]

Answer:

The period would decrease by `sqrt(2)`.

Explanation:

The angular frequency omega of an oscillating mass m due to a spring with a constant k is given by

\omega = \sqrt{\frac{k}{m}}

(this is obtained by solving the differential equation m\ddot{{ x}}-kx=0)

If k doubles, i.e., k'=2k, then

\omega'=\sqrt{\frac{k'}{m}}=\sqrt{\frac{2k}{m}}=\sqrt{2}\sqrt{\frac{k}{m}}=\omega\sqrt{2}

Since the angular frequency is \omega = \frac{2\pi}{T}, we can say that

\omega\sqrt{2}=\frac{2\pi}{T}\sqrt{2}=\frac{2\pi}{\frac{T}{\sqrt{2}}}=\frac{2\pi}{T'}

and so it becomes clear that the period T will decrease by sqrt(2) as stated in choice (D).

5 0
3 years ago
Read 2 more answers
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