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dsp73
3 years ago
9

A clock which has brass pendulum beats seconds correctly when the temperature of the room is 30° C. how many seconds will it gai

n or lose per day when the temperature of the room falls to 10° C?
Physics
1 answer:
Andrei [34K]3 years ago
7 0

Answer:A::C

Solution :

ΔT=

1

2

 

TαΔθ

rArr Δt=

ΔT

T1t

 

But T′≈T

∴Δt=

1

2

 

αΔθ(t)

=

1

2

 

×0.000012×20×24×3600

=10.37s

At higher temperture, length of pendulum clock will be more. So, time period will be more and it will lose the time.

Explanation:

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An object accelerates 20.8 m/s2 when a force of 8.9 newtons is applied. What is the mass of the object?
Leona [35]

Answer: 0.43kg

Explanation: a = 20.8m/s^2

                      F = 8.9N

                      m = ?

F = ma

8.9 = 20.8m

m = 8.9/20.8

= 0.43kg

Hope this helped..

7 0
3 years ago
A radiographer stands six feet from the x-ray source when performing a portable chest exam and receives an exposure of 2 mGy. If
telo118 [61]

Answer:

  I₂ = 8 mG

Explanation:

The intensity of a beam is

          I = P / A

Where P is the emitted power which is 3) 3

           

Let's use index 1 for the initial position of r₁ = 6 ft and 2 for the second position r₂ = 3 ft

          I₁ A₁  = I₂  A₂

           I₂ = I₁ A₁ / A₂

The area of ​​the beam if we assume that it is distributed either in the form of a sphere is

           A₁ = 4π r²

We substitute

            I₂ = I₁ (r₁ / r₂)²

           I₂ = 2 (6/3)²

           I₂ = 2 4

           I₂ = 8 mG

3 0
3 years ago
Read 2 more answers
At time t=0 a grinding wheel has an angular velocity of 30.0 rad/s . It has a constant angular acceleration of 35.0 rad/s2 until
Rama09 [41]

Answer:

(A) 570 rad

(B) 10 s

(C) 12.5 rad/s²

Explanation:

The equations of motion for circular motions are used.

  • Initial angular velocity,  \omega_0 = 30.0 \text{ rad/s}
  • Angular acceleration, \alpha =35.0 \text{ rad/s}^2

(A)

At <em>t</em> = 2.00 s, the angular displacement, <em>θ</em>, is given by

\theta = \omega_0t+\frac{1}{2}\alpha t^2 = (30\times 2) + \frac{1}{2}\times35\times2^2=60+70 = 130\text{ rad}

After this time, it decelerates through an angular displacement of 440 rad.

Total angular displacement = 130 + 440 rad = 570 rad

(B)

At the time the circuit breaker tips, the angular velocity is given by

\omega = \omega_0+\alpha  t = 30.0+(35.0\times 2) = 30.0+70.0 =100.0\ \text{rad/s}

This becomes the initial angular velocity for the decelerating motion. Because it stops, the final angular velocity is 0 rad/s. The time for this part of the motion is calculated thus:

\theta_2 = \left(\dfrac{\omega_i+\omega_f}{2}\right)t

Here, \theta_2=440 (the angular displacement during deceleration)

The subscripts, <em>i</em> and <em>f</em>, on <em>ω</em> denote the initial and final angular velocities during deceleration.

\omega_i = 100

\omega_f = 0

t = \dfrac{2\theta_2}{\omega_i} = \dfrac{2\times400}{100} = 8\ \text{s}

This is the time for deceleration. The deceleration began at <em>t</em> = 2 s.

Hence, the wheel stops at <em>t</em> = 2 + 8 = 10 s.

(C)

The deceleration is given by

\alpha_R = \dfrac{\omega_f-\omega_i}{t} = \dfrac{0-100}{8} = -12.5\text{ rad/s}^2

The negative sign appears because it is a deceleration.

4 0
3 years ago
A small block with mass 0.0400 kg slides in a vertical circle of radius 0.600 m on the inside of a circular track. During one of
Maurinko [17]

Answer:

Explanation:

Given that

The mass of the body is 0.04kg

M=0.04kg

The radius of the paths is 0.6m

r=0.6m

The normal force exerted at A is 3.9N

Fa=3.9N

The normal force exerted at B is 0.69N

Fb=0.69N

Then work done by friction from point A to B will be the change in K.E

W=∆K.E+P.E

So we need to know the velocity at both point A and B

Then since the centripetal force is given as

Ft=mv²/r

Then,

For point A

Fa=mv²/r

3.9=0.04v²/0.6

3.9=0.0667v²

v²=3.9/0.0667

v²=58.5

v=√58.5

v=7.65m/s

Va=7.65m/s

Now at point B

Fb=mv²/r

0.69=0.04v²/0.6

0.69=0.0667v²

v²=0.69/0.0667

v²=10.35

v=√10.35

v=3.22m/s

Vb=3.22m/s

Then, the work done is

W=∆K.E+P.E

P.E is given as mgh

The height will be 2R =1.2m

P.E=mgh

P.E=0.04×9.81×1.2

P.E=0.471J

Final kinetic energy at B minus initial kinetic energy at A

W=K.Eb-K.Ea

K.E is given as 1/2mv²

W=1/2m(Vb²-Va²) +P.E

W=0.5×0.04(3.22²-7.65²) +0.471

W=0.5×0.04×(-48.1541) +0.471

W=-0.96+0.471

W=-0.49J

work was done on the block by friction during the motion of the block from point A to point B is 0.49J.

Friction opposes motions and that is why the work done is negative

3 0
3 years ago
I have a few questions. How would density change if volume is not calculated correctly?
VikaD [51]
Density equals mass divided by volume so the ball the higher amount of mass has more density than the ball with less mass has less density. (The golf ball has more density than the ping pong ball)
7 0
3 years ago
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