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SpyIntel [72]
3 years ago
6

In lab, a radiation detector was used to calculate the background radiation. The

Physics
1 answer:
Keith_Richards [23]3 years ago
8 0
Not sure but try a tutor!
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Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the workdone by Ryan?​
Xelga [282]
<h2><u>Question</u><u>:</u><u>-</u></h2>

Ryan applied a force of 10N and moved a book 30 cm in the direction of the force. How much was the work done by Ryan?

<h2><u>Answer:</u><u>-</u></h2>

<h3>Given,</h3>

=> Force applied by Ryan = 10N

=> Distance covered by the book after applying force = 30 cm

<h3>And,</h3>

30 cm = 0.3 m (distance)

<h3>So,</h3>

=> Work done = Force × Distance

=> 10 × 0.3

=> 3 Joules

\small \boxed{work \: done \:  by \: Ryan \:  = 3 \: Joules}

4 0
3 years ago
What were some benefits to the american television from analog broadcast to digital broadcast?
Elanso [62]

Answer:

nothing no no no

Explanation:

nothing cause no

8 0
2 years ago
Read 2 more answers
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padilas [110]
The answer would be C. Gamma Rays and High Frequency EM waves travel at the speed of light and are transverse waves.
7 0
3 years ago
Read 2 more answers
An acorn falls from a tree and hits the ground in 0.8 s. How far did the acorn fall . Use g = 9.8 m/s^2. Round your final result
mylen [45]

The distance covered by the acorn is 3.136 m.

<u>Explanation:</u>

The time taken for the acorn to hit the ground is 0.8 s. As it is a free fall, the acorn will be completely under the influence of gravity. So the acceleration will be acceleration due to gravity.

Then using the second law of equation,

s=ut+\frac{1}{2}gt^{2}

Since the initial velocity and time is zero, then the time taken to reach the ground is stated as 0.8 s, so

   s=0+\left(\frac{1}{2} \times 9.8 \times 0.8 \times 0.8\right)=\frac{6.272}{2}=3.136 \mathrm{m}

So the distance covered by the acorn is 3.136 m.

8 0
3 years ago
A circular bird feeder 19.0 cm in radius has rotational inertia 0.130 kg·m2. It's suspended by a thin wire and is spinning slowl
soldi70 [24.7K]

Answer:

N₂=20.05 rpm

Explanation:

Given that

R= 19 cm

I=0.13 kg.m²

N₁ = 24.2 rpm

\omega_1=\dfrac{2\pi \times 24.2}{60}\ rda/s

ω₁= 2.5 rad/s

m= 173 g = 0.173 kg

v=1.2 m

Initial angular momentum L₁

L₁ =  Iω₁  - m v r       ( negative sign because bird coming opposite to motion of the wire motion)

Final linear momentum L₂

L₂=  I₂ ω₂

 I₂ = I + m r²

The is no any external torque that is why angular momentum will be conserve

L₁ = L₂

Iω₁  - m v r =  I₂ ω₂

Iω₁  - m v r =  ( I + m r²) ω₂

Now by putting the all values

Iω₁  - m v r =  ( I + m r²) ω₂

0.13 x 2.5 - 0.173 x 1.2 x 0.19 =  ( 0.13 + 0.173 x  0.19²) ω₂

0.325  - 0.0394 = 0.136 ω₂

ω₂ = 2.1 rad/s

\omega_2=\dfrac{2\pi \times N_2}{60}

N₂=20.05 rpm

3 0
3 years ago
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