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Over [174]
3 years ago
10

calculate the height from which a body is released from rest if its velocity just before hitting the ground is 30 metre per seco

nds​
Physics
1 answer:
Anna71 [15]3 years ago
5 0

Answer:

height is 45 metres

Explanation:

KE=PE

¹/²mv² = mgh

¹/²v² = gh

but v =30m/s

(30)² / 2 =10h

450=10h

h = 45m

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Darya [45]
  • Mass=0.25kg
  • Force=95N

\\ \bull\sf\dashrightarrow F=ma

\\ \bull\sf\dashrightarrow a=\dfrac{F}{m}

\\ \bull\sf\dashrightarrow a=\dfrac{95}{0.25}

\\ \bull\sf\dashrightarrow a=380m/s^2

3 0
3 years ago
Pleaseeee help got this homework tomorrow ; if an object vibrates five times every second it has a frequency of ...?
AysviL [449]
Here's a tip: The unit of frequency ... Hertz or Hz ... means "per second". "Five times PER SECOND" is a frequency of 5 Hz.
8 0
3 years ago
Manganese-52 has a half-life of 6 days. How many days would a scientist have to wait for the radioactivity to be 12.5% the start
sineoko [7]

Answer:

Let N = N0 where N0 is the number of atoms originally present.

In 6 days    N = N0 / 2

In 12 days   N = N0 / 4

In 18 days   N = N0 / 8    = .125 N0

So it would take 18 days.

4 0
3 years ago
Read 2 more answers
Consider a satellite in a circular orbit around the earth. Why is it important to give a satellite a horizontal speed when placi
Andrei [34K]

Answer:

In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero,

Explanation:

When a satellite is in orbit the most important force is the docking of gravity with the Earth

             F = m a

where the acceleration is centripetal and F is the force of universal attraction

centripetal acceleration is

             a = v² / r

             

             F = m v² / r

In this case, the horizontal velocity of the rocket starts from the acceleration, so if its velocity drops to zero, the force also drops to serious and the satellite steels to Earth.

The speed of the satellite is provides the speed, by local for smaller speeds in satellite, it descends in its orbits and when the speed is amate you have the energy to stop an orb to go to a higher orbit.

4 0
3 years ago
Coherent light with wavelength 598 nm passes through two very narrow slits, and the interference pattern is observed on a screen
Stella [2.4K]

Answer:

1.196 μm

Explanation:

D = Screen distance = 3 m

\lambda = Wavelength = 598 m

y = Distance of first-order bright fringe from the center of the central bright fringe = 4.84 mm

d = Slit distance

tan\theta=\frac{y}{D}\\\Rightarrow \theta=tan^{-1}{\frac{y}{D}}\\\Rightarrow \theta=tan^{-1}{\frac{4.84\times 10^{-3}}{3}}\\\Rightarrow \theta=0.09243\ ^{\circ}

sin\theta=\frac{\lambda}{d}\\\Rightarrow d=\frac{\lambda}{sin\theta}\\\Rightarrow d=\frac{598\times 10^{-9}}{sin0.09243}\\\Rightarrow d=0.00037066\ m

For first dark fringe

dsin\theta=\frac{\lambda'}{2}\\\Rightarrow \lambda'=2dsin\theta\\\Rightarrow \lambda'=2\times 0.00037066\times sin0.09243\\\Rightarrow \lambda'=1.196\times 10^{-6}\\\Rightarrow \lambda'=1.196\ \mu m

Wavelength of first-order dark fringe observed at this same point on the screen is 1.196 μm

3 0
3 years ago
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