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3 years ago

Which unit do astronomers use for angular measurement?

Physics

1 answer:

Alex787 [66]3 years ago

8 0

C and D are units of length or distance.
A is a measured angle.
B is a unit of angular measurement.

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13. An object, travelling along a straight path, covers 35 m distance in 4 seconds. In next 6

WARRIOR [948]

Answer:

(4) 8.5 m/s

Explanation:

You add both the meters together and both the seconds together and then divide them both.

5 0

3 years ago

Read 2 more answers

Which conclusion, based on measurements of the ball in the lab in this lesson, is correct? a. The density of the ball was 1.48 g

ser-zykov [4K]

Answer:

u dummy

Explanation:

7 0

3 years ago

To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3

hodyreva [135]

Answer : The correct option is, (c) $3.7\times 10^2J/^oC$

Explanation :

First we have to calculate the energy or heat.

Formula used :

$E=V\times I\times t$

where,

E = energy (in joules)

V = voltage (in volt)

I = current (in ampere)

t = time (in seconds)

Now put all the given values in the above formula, we get:

$E=(3.6V)\times (2.6A)\times (350s)$

$E=3276J$

Now we have to calculate the heat capacity of the calorimeter.

Formula used :

$C=\frac{E}{\Delta T}=\frac{E}{T_{final}-T_{initial}}$

where,

C = heat capacity of the calorimeter

$T_{initial}$ = initial temperature = $20.3^oC$

$T_{final}$ = final temperature = $29.1^oC$

Now put all the given values in this formula, we get:

$C=\frac{3276J}{(29.1-20.3)^oC}$

$C=372.27J/^oC=3.7\times 10^2J/^oC$

Therefore, the heat capacity of the calorimeter is, $3.7\times 10^2J/^oC$

7 0

3 years ago

A soccer player kicks a ball horizontally at 28 m/s from a bridge and hits the ground after 1.5 s. What is the range of the ball

Lelu [443]

Answer:

Explanation:

range of ball = u×t

= 28×1.5

=42

8 0

2 years ago

A ball is thrown vertically downward from the top of a 30.6-m-tall building. The ball passes the top of a window that is 10.7 m

Gekata [30.6K]

Answer:

v = 19.6 m/s

Explanation:

Height of building = 30.6 m.

Height of window from the ground level= 10.7 m.

Acceleration due to gravity = 9.8 $\frac{m}{s^2}$

At initial condition ball at rest condition so u= 0 m/s.

Lets take when passes through the window ,velocity is v.

Here acceleration is constant so we can apply motion equation .

We know that

v= u + a t

So by putting the values

v = 0 +9.8 x 2

v = 19.6 m/s

So the velocity of ball is 19.6 m/s when passes through the window after 2 s.

7 0

3 years ago