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Tom [10]
3 years ago
9

An electric bell is placed inside a transparent glass jar. The bell can be turned on and off using a switch on the outside of th

e jar. A vacuum is created inside the jar by sucking out the air. Then the bell is rung using the switch. What will we see and hear?
Physics
2 answers:
Art [367]3 years ago
5 0
The answer is :  We’ll see the bell move, but we won’t hear it ring.  This is    because light can travel through vacuum but sound cannot.  Sound waves are vibrations of particles in any media, so sound requires a medium to travel, and it cannot travel in a vacuum as there is no particles to vibrate.  
qwelly [4]3 years ago
5 0
Well see the bell move but won't hear it ring
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When a 5.0-kilogram cart moving with a speed of 2.8 meters per second on a horizontal surface collides with a 2.0 kilogram cart
mixer [17]

Here in all such collision type question we can use momentum conservation as we can see that there is no external force on this system

m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}

as we know that

m_1 = 5 kg

m_2 = 2 kg

v_{1i} = 2.8 m/s

v_{2i} = 0 m/s

now from above equation we have

5(2.8) + 2(0) = m_1v+ m_2v

14 = (5+ 2) v

v = 2 m/s

so the speed of combined system is 2 m/s

8 0
3 years ago
Two lightbulbs work on a 120-V circuit One 50 W and the other is 100 W. Which bulb has a higher resistance? Explain pls!!!!
Sladkaya [172]

Answer:

Bulb 1 has more resistance.

Explanation:

Given that,

Two lightbulbs work on a 120-V circuit.

The power of circuit 1, P₁ = 50 W

The power of circuit 2, P₂ = 100 W

We need to find the bulb that has a higher resistance.

The power of the bulb is given by :

P = \dfrac{V^2}{R}

For bulb 1,

R_1=\dfrac{V^2}{P_1}\\\\R_1=\dfrac{(120)^2}{50}\\\\R_1=288\ \Omega

For bulb 2,

R_2=\dfrac{V^2}{P_2}\\\\R_2=\dfrac{(120)^2}{100}\\\\R_2=144\ \Omega

So, bulb 1 has higher resistance.

8 0
2 years ago
A car starts from rest and acquires a velocity of 50m/s in 3secs. Calculate i) acceleration ii) distance covered.
mafiozo [28]

Answer: 75.02 m

Explanation:

u = 0 ( starts from rest )

v = 50 m/s

t = 3 s

( i ) a = v - u / t

= 50 - 0 /3

= 16.67

( ii ) s = ut + 1/2 at²

= 0 × 3 + 1/2 × 16.67 × 3 × 3

= <u>75.02 m</u>

Hope this helps...

4 0
2 years ago
The period of a pendulum is measured 16 times. The average value of the period over these 16 trials is calculated to be 1.50 sec
marin [14]

Answer:

The additional trials needed is 48 trials

Explanation:

Given;

initial number of trials, n = 16 trials

the standard deviation, σ = 0.24 s

initial standard error, ε = 0.06 s

The standard error is given by;

\epsilon = \frac{\sigma}{\sqrt{n} }

To reduce the standard error to 0.03 s, let the additional number of trials = x

0.03= \frac{0.24}{\sqrt{n+x} } \\\\0.03= \frac{0.24}{\sqrt{16+x} }\\\\0.03\sqrt{16+x} = 0.24\\\\\sqrt{16+x} = \frac{0.24}{0.03} \\\\\sqrt{16+x} = 8\\\\16+x = 8^2\\\\16+x = 64\\\\x = 64 -16\\\\x = 48 \ trials

Therefore, the additional trials needed is 48 trials.

6 0
3 years ago
If a player through a basketball to the target with an initial velocity of 17 m/s making an angle of 30 degrees with the horizon
Svetllana [295]

Answer:

The final position made with the vertical is 2.77 m.

Explanation:

Given;

initial velocity of the ball, V = 17 m/s

angle of projection, θ = 30⁰

time of motion, t = 1.3 s

The vertical component of the velocity is calculated as;

V_y = Vsin \theta\\\\V_y = 17 \times sin(30)\\\\V_y = 8.5 \ m/s

The final position made with the vertical (Yf) after 1.3 seconds is calculated as;

Y_f = V_yt  - \frac{1}{2}g t^2\\\\Y_f = (8.5 \times 1.3 ) - (\frac{1}{2} \times 9.8 \times 1.3^2)\\\\Y_f = 11.05 \ - \ 8.281\\\\Y_f = 2.77 \ m

Therefore, the final position made with the vertical is 2.77 m.

3 0
2 years ago
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