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raketka [301]
3 years ago
7

In an electricity experiment, a 1.10 g plastic ball is suspended on a 56.0 cm long string and given an electric charge. A charge

d rod brought near the ball exerts a horizontal electrical force F elec on it, causing the ball to swing out to a 21.0 degree angle and remain there.
What is the tension in the string?

Physics
1 answer:
Shalnov [3]3 years ago
8 0

Answer:

Tension, T = 0.0115 N                      

Explanation:

Given that,

Mass of the plastic ball, m = 1.1 g

Length of the string, l = 56 cm

A charged rod brought near the ball exerts a horizontal electrical force F on it, causing the ball to swing out to a 21.0 degree angle and remain there. According to attached figure :

T\cos\theta=mg

T is tension in the string

T=\dfrac{mg}{\cos\theta}\\\\T=\dfrac{1.1\times 10^{-3}\times 9.8}{\cos(21)}\\\\T=0.0115\ N

So, the tension in the string is 0.0115 N.

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In the great shopping cart race, two students push on shopping carts. A having twice the mass of B, with the same force applied
Lesechka [4]

K = 1/2 m x v^2

m = mass on the cart

V = velocity imparted to the cart

KA = 1/2 mA x vA^2.......................(1)

KB = 1/2 mB x vB^2........................(2)

Diving equation 1 by equation 2, we get -

KA/KB = mA/mB

= 2

KA = 2 x KB

Option A is correct

6 0
3 years ago
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
You and a friend frequently play a trombone duet in a jazz band. During such performances it is critical that the two instrument
Rashid [163]

Answer:

f₂ = 468.67 Hz

Explanation:

A beat is a sudden increase and decrease of sound. The beats are produced through the interference of two sound waves of slightly different frequencies. Now we have the following data:

The higher frequency tone = f₁ = 470 Hz

No. of beats = n = 4 beats

Time period = t = 3 s

The lower frequency note = Frequency of Friend's Trombone = f₂ = ?

Beat Frequency = fb

So, the formula for beats per second or beat frequency is given as:

fb = n/t

fb = 4 beats/ 3 s

fb = 1.33 Hz

Another formula for beat frequency is:

fb = f₁ - f₂

f₂ =  f₁ - fb

f₂ = 470 Hz - 1.33 Hz

<u>f₂ = 468.67 Hz</u>

6 0
3 years ago
Look at the resistor illustrated in the figure above. Based on your knowledge of the resistor color code, what are the resistanc
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A because of the resistors are four in this options first option is multiplied by 4
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i sorry i thought of geocentric as something else it appears that the earth was the center

8 0
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