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Shtirlitz [24]
2 years ago
14

A(n) __________ is a recording of a motion picture, or television program for playing through a television.

Physics
2 answers:
Soloha48 [4]2 years ago
4 0

Answer:

Video

Explanation:

Hope this helps! If it does, drop a 5 star!

lord [1]2 years ago
3 0

Answer:

b video

Explanation:

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The angular velocity of a flywheel obeys the equa tion w(1) A Br2, where t is in seconds and A and B are con stants having numer
makkiz [27]

Answer:

A \to rad/s

B  \to rad/s^3

Explanation:

\omega_z(t)=A + Bt^2

Required

The units of A and B

From the question, we understand that:

\omega_z(t) \to rad/s

This implies that each of A and Bt^2 will have the same unit as \omega_z(t)

So, we have:

A \to rad/s

Bt^2 \to rad/s

The unit of t is (s); So, the expression becomes

B * s^2 \to rad/s

Divide both sides by s^2

B  \to \frac{rad/s}{s^2}

B  \to rad/s^3

5 0
3 years ago
The scientist who found a way to measure the distance between the sun and Venus which
Yakvenalex [24]

Answer:

Astronomer Edmond Halley

Explanation:

The astronomical unit using the transit of venus

The underlying principle behind Halley's method is called parallax

8 0
2 years ago
Select the correct answer.
Natali [406]

its b hoped i helped

8 0
3 years ago
Read 2 more answers
A certain spring stretches 3 cm when a load of 15 n is suspended from it. how much will the spring stretch if 30 n is suspended
Alik [6]
Initially, the spring stretches by 3 cm under a force of 15 N. From these data, we can find the value of the spring constant, given by Hook's law:
k= \frac{F}{\Delta x}
where F is the force applied, and \Delta x is the stretch of the spring with respect to its equilibrium position. Using the data, we find
k= \frac{15 N}{3.0 cm}=5.0 N/cm

Now a force of 30 N is applied to the same spring, with constant k=5.0 N/cm. Using again Hook's law, we can find the new stretch of the spring:
\Delta x =  \frac{F}{k}= \frac{30 N}{5.0 N/cm}=6 cm
4 0
3 years ago
Read 2 more answers
(20) A rocket is launched vertically. At time t = 0 seconds, the rocket’s engine shuts down. At the time, the rocket has reached
Diano4ka-milaya [45]

Answer:

Explanation:

Given that,

h(t) = -9.8t² / 2 + 125t + 500

for t > 0.

At t = 0, the rocket is at height h = 500m, at a velocity of Vo = 125m/s.

We want to find the maximum height reached by rocket

Using mathematics maxima and minima

let find the turning point when dh/dt = 0

dh/dt = -9.8t + 125

dh / dt = 0 = -9.8t + 125

9.8t = 125

t = 125 / 9.8

t = 12.76s

Let find the turning point to know if this time t = 12.76 is maximum or minimum point

Let find d²h / dt²

d²h / dt² = -9.8

Since, d²h/dt² < 0, then, at t = 12.76s is the maximum points.

Then, the maximum height reached is

h =  -9.8t² / 2 + 125t + 500

h =  -9.8(12.76)² / 2 + 125(12.76) + 500

h = -797.80 + 1595 + 500

h = 1297.2 m

The maximum height reached is 1297.2 m

From the attachment, the maximum height is 1297.2m at t = 12.76sec.

Comment, the result are the same for both the analysis aspect and the graphical aspect.

3 0
3 years ago
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