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3 years ago

A free positive charge released in an electric field will:

Physics

1 answer:

choli [55]3 years ago

7 0

Answer:

accelerate in the direction in which the electric field is pointing.

Explanation:

The positive charge feels a force in the same direction as the electric field

F=Eq

F and E are vectors, q is a scalar

(if it were a negative charge the force would be in the opposite direction)

that force will produce an acceleration in the same direction, that acceleration will cause the particle to move in the same direction, ie the direction of the electric field.

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a closed systems internal energy changes by 178 j as a result of being heated with 658 j of energy. the energy used to do work b

Akimi4 [234]

<h3><u>Answer;</u></h3>

= 480 Joules

<h3><u>Explanation;</u></h3>

We use the formula, Q - W = ΔU

Where, Q = Heat transferred to the system

W = Work done by the system

ΔU = Change of internal energy.

As per the question, Q = 658 J

ΔU = 178 J

Thus, W = Q - ΔU = (658 - 178) J = 480 J.

The energy used to do work by the system is 480 J.

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3 years ago

Why doesn't a transformer work with direct current? why is ac required?

Yuri [45]

The electric field depends not on the magnetic field, but change in magnetic field. if the magnetic field is constant. electric field will be zero.

4 0

3 years ago

U guys are the best I’m surprised I’m getting all of this right

ella [17]

The answer is B, 1/4 of the cake

5 0

3 years ago

Read 2 more answers

One recently discovered extrasolar planet, or exoplanet, orbits a star whose mass is 0.70 times the mass of our sun. This planet

Stels [109]

0.078 times the orbital radius r of the earth around our sun is the exoplanet's orbital radius around its sun.

Answer: Option B

<u>Explanation:</u>

Given that planet is revolving around the earth so from the statement of centrifugal force, we know that any

$\frac{G M m}{r^{2}}=m \omega^{2} r$

The orbit’s period is given by,

$T=\sqrt{\frac{2 \pi}{\omega r^{2}}}=\sqrt{\frac{r^{3}}{G M}}$

Where,

$T_{e}$ = Earth’s period

$T_{p}$ = planet’s period

$M_{s}$ = sun’s mass

$r_{e}$ = earth’s radius

Now,

$T_{e}=\sqrt{\frac{r_{e}^{3}}{G M_{s}}}$

As, planet mass is equal to 0.7 times the sun mass, so

$T_{p}=\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}$

Taking the ratios of both equation, we get,

$\frac{T_{e}}{T_{p}}=\frac{\sqrt{\frac{r_{e}^{3}}{G M_{s}}}}{\sqrt{\frac{r_{p}^{3}}{0.7 G M_{s}}}}$

$\frac{T_{e}}{T_{p}}=\sqrt{\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2}=\frac{0.7 \times r_{e}^{3}}{r_{p}^{3}}$

$\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}=\frac{r_{e}^{3}}{r_{p}^{3}}$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{T_{e}}{T_{p}}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}$

Given $T_{p}=9.5 \text { days }$ and $T_{e}=365 \text { days }$

$\frac{r_{e}}{r_{p}}=\left(\left(\frac{365}{9.5}\right)^{2} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=\left(\frac{133225}{90.25} \times \frac{1}{0.7}\right)^{\frac{1}{3}}=(2108.82)^{\frac{1}{3}}$

$r_{p}=\left(\frac{1}{(2108.82)^{\frac{1}{3}}}\right) r_{e}=\left(\frac{1}{12.82}\right) r_{e}=0.078 r_{e}$

7 0

3 years ago

A fan is to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3 /s. determine the minimum power that must be suppl

ELEN [110]

Given:
ρ = 1.18 kg/m³, density of air
v = 8 m/s, flow velocity
Q = 9 m³/s, volumetric flow rate

The minimum power required (at 100% efficiency) is
$\frac{1}{2} (8 \, \frac{m}{s} )^{2}*(9 \, \frac{m^{3}}{s} )*(1.18 \, \frac{kg}{m^{3}}) = 339.84 \, W$

The actual power will be higher because 100% efficiency is not possible.

Answer: 339.8 W (nearest tenth)

8 0

4 years ago