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andriy [413]
2 years ago
12

A car is traveling with speed v0 when it begins to speed up at a rate of Δv every second. After t1 seconds, the car travels with

zero acceleration for t2 seconds. Which of the following is a correct expression for the displacement of the car during this motion?
Physics
1 answer:
Rainbow [258]2 years ago
8 0

Answer:

d = Δv(t2-t1)

Explanation:

Speed is defined as the change of displacement with respect to time. It is expressed as shown;

Speed = change in displacement/change in time

Δv = d/Δt

d = Δv*Δt

d = ΔvΔt

Δt = t2-t1

d = Δv(t2-t1)

Δv is the change in rate of speed

Δt = change in time

The correct expression for the displacement of the car during this motion is d = Δv(t2-t1)

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A vibrating standing wave on a string radiates a sound wave with intensity proportional to the square of the standing-wave ampli
uysha [10]

Answer:

string's damping is 1.03676

Explanation:

given data

sound level = 9.0 dB

time = 1 sec

to find out

string's damping

solution

we will apply here formula for string damping (b) that is

A(t) = A × e^{-bt}   ...................1

we know here I  ∝ A² so

√I(t) = √I × e^{-bt}  

√I(t) / √I =  e^{-bt}     .....................2

we know sound level decreases 9 dB i.e ΔdB = 9

so we can write

ΔdB = 10 log ( I(t) / I)

9 = 10 log ( I(t) / I)

I(t) / I = 10^{-0.9}

I(t) / I = 0.1258

and

√I(t) / I) = √0.1258 = 0.3546     .......................3

from equation 2 and 3 we get

0.3546 = e^{-bt}

take ln both side

-bt = ln 0.3546

here we know t is 1 sec

so

- b = - 1.03676

b = 1.03676

so here string's damping is 1.03676

8 0
2 years ago
Magnetic fields within a sunspot can be as strong as 0.4T. (By comparison, the earth's magnetic field is about 1/10,000 as stron
WARRIOR [948]

Answer:

The speed of ejection is 2.06\times 10^{4}\ m/s

Solution:

As per the question:

Magnetic field density, B = 0.4 T

Density of the material in the sunspot, \rho = 3\times 10^{4}\ kg/m^{3}

Now,

To calculate the speed of ejection of the material, v:

The magnetic field energy density is given by:

U_{B} = \frac{B^{2}}{2\mu_{o}}

This energy density equals the kinetic energy supplied by the field.

Thus

KE = U_{B}

\frac{1}{2}mv^{2} = \frac{B^{2}}{2\mu_{o}}

where

m = mass of the sunspot in 1\ m^{3} = 3\times 10^{- 4}\ kg/m^{3}

v = \frac{B}{\sqrt{\mu_{o}m}}

v = \frac{0.4}{\sqrt{4\pi \times 10^{- 7}\times 3\times 10^{- 4}}} = 2.06\times 10^{4}\ m/s

3 0
2 years ago
A charge of 5.67 x 10-18 C is placed 3.5 x 10 m away from another charge of - 3.79 x 10 "C
miskamm [114]

Answer:

1. 579 x 10 ^-22N

Explanation:

F = kq1q2/r^2

   = 9.0 x 10^9 x 5.67 x 10^-18 x 3.79 x 10^-18/ (3.5 x 10^-2)^2

    = 1. 579 x 10 ^-22N

6 0
2 years ago
Consider a simple ideal Rankine cycle with fixed turbine inlet conditions. What is the effect of lowering the condenser pressure
mr Goodwill [35]

Answer:

The effect of lowering the condenser pressure on different parameters is explained below.

Explanation:

The simple ideal Rankine cycle is shown in figure.

Effect of lowering the condenser pressure on

(a). Pump work input :- By lowering the condenser pressure the pump work increased.

(b) Turbine work output :- By lowering the condenser pressure the turbine work increased.

(c). Heat supplied :- Heat supplied increases.

(d). Heat rejected :- The heat rejected may increased  or decreased.

(e). Efficiency :- Cycle  efficiency is increased.

(f). Moisture content at turbine exit :- Moisture content increases.

8 0
2 years ago
Several charges in the neighborhood of point P produce an electric potential of 6.0 kV (relative to zero at infinity) and an ele
Julli [10]

Answer:

0.018 J

Explanation:

The work done to bring the charge from infinity to point P is equal to the change in electric potential energy of the charge - so it is given by

W = q \Delta V

where

q=3.0 \mu C = 3.0 \cdot 10^{-6} C is the magnitude of the charge

\Delta V = 6.0 kV = 6000 V is the potential difference between point P and infinity

Substituting into the equation, we find

W=(3.0\cdot 10^{-6}C)(6000 V)=0.018 J

4 0
2 years ago
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