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san4es73 [151]
3 years ago
13

Consider a uniform electric field of 50 N/C directed towards east. if the voltage measured relative to ground at a given point i

s 80 V what is the voltage at a point 1.0 m directly west of the point? assume a constant electric field?
Physics
1 answer:
Dahasolnce [82]3 years ago
8 0

Answer:30 V

Explanation:

Electric Field Strength is 50 N/C due to east

Voltage measured at a point P is 80 V

Let us assume point P is at a distance of x m

therefore

E\cdot x=V

50\cdot x=80

x=1.6 m

Now Potential directly 1 m west of P

therefore distance is 1.6 -1 =0.6 m

Thus potential is

E\cdot 0.6=50\cdot 0.6=30 V

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It means mass can be converted into energy.

I. E. In a nuclear fission, we start of with a greater mass of our rectant, which then decrease due to mass being converted into the (thermal) energy released, in our reaction. Later, being used to produce electricity, for example.

Hope this helps!

6 0
2 years ago
A ball starts from rest. It rolls down a ramp and reaches the ground after 4 seconds. Its final velocity when it reaches the gro
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If it starts at rest the initial velocity is 0.
For an acceleration, a, and time, t, the velocity is v=at. Since at t=4, v=7, then a=7/4=1.75m/s^2
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4 years ago
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n the railroad accident, a boxcar weighting 200 kN and traveling at 3 m/s on horizontal track slams into a stationary caboose we
USPshnik [31]

Answer:

ΔK = -6 10⁴ J

Explanation:

This is a crash problem, let's start by defining a system formed by the two trucks, so that the forces during the crash have been internal and the moment is preserved

initial instant. Before the crash

        p₀ = m v₁ + M 0

final instant. Right after the crash

        p_f = (m + M) v

        p₀ = p_f

        mv₁ = (m + M) v

        v = \frac{m}{m+M} \  v_1

     

we substitute

        v = \frac{20}{20+40}   3

        v = 1.0 m / s

having the initial and final velocities, let's find the kinetic energy

        K₀ = ½ m v₁² + 0

        K₀ = ½ 20 10³ 3²

        K₀ = 9 10⁴ J

        K_f = ½ (m + M) v²

        K_f = ½ (20 +40) 10³  1²

        K_f = 3 10⁴ J

the change in energy is

       ΔK = K_f - K₀

       ΔK = (3 - 9) 10⁴

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The negative sign indicates that the energy is ranked in another type of energy

7 0
3 years ago
A 0.500-mol sample of an ideal monatomic gas at 400 kPa and 300 K, expands quasi-statically until the pressure decreases to 160
Marianna [84]

Answer:

(a). The final volume is 2.5 times the initial volume.

The work done is 1143 J.

(b). The final temperature is 207.9 K

The work done is 574 J.

Explanation:

Given that,

Sample of an ideal gas = 0.500 mol

Initial pressure = 400 kPa

Final pressure = 160 kPa

Temperature = 300 K

(a) for isothermal,

Temperature will be same.

We need to calculate the volume of gas

P_{f}V_{f}=P_{i}V_{i}

V_{f}=(\dfrac{P_{i}}{P_{f}})V_{i}

Put the value into the fomrula

V_{f}=(\dfrac{400}{160})V_{i}

V_{f}=2.5 V_{i}

We need to calculate the work done

Using equation of energy

dQ=dW

dQ=nRTln(\dfrac{P_{in}}{P_{f}})

dQ=0.500\times8.314\times300\times ln(\dfrac{400}{160})

dQ=1143\ J

(b). For adiabatic,

No transfer of heat between system and surroundings

We need to calculate the final temperature

Using formula of gas

P_{f}^{1-\gamma}T_{f}^{\gamma}=P_{i}^{1-\gamma}T^{\gamma}

T_{f}=(\dfrac{P_{i}}{P_{f}})^{\frac{1-\gamma}{\gamma}}T_{i}

Put the value into the formula

T_{f}=(\dfrac{400}{160})^{\frac{1-\frac{5}{3}}{\frac{5}{3}}}\times300

T_{f}=207.9\ K

We need to calculate the wok done in adiabatic

Using formula of work done

W=\dfrac{nR(T_{i}-T_{f})}{\gamma-1}

W=\dfrac{0.500\times8.314\times(300-207.9)}{\dfrac{5}{3}-1}

W=574\ J

Hence, (a). The final volume is 2.5 times the initial volume.

The work done is 1143 J.

(b). The final temperature is 207.9 K

The work done is 574 J.

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