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2 years ago

Where an electric field line crosses an equipotential surface, the angle between the field line and the equipotential is

Physics

1 answer:

n200080 [17]2 years ago

4 0

Answer:

90 degree

Explanation:

Electric field line is vertical to the electric field line.

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a ball is projected horizontally with a velocity of 5 m per second from the top of a building 19.6 m high how long will the ball

zepelin [54]

Answer:

1.98s

Explanation:

The time taken to hit the ground is given by

h=ut+ 1/2 at^2

but u =0

so we have

h=1/2at^2

making t the subject

t=√2h/g

√2×19.6/10

1.98s

8 0

2 years ago

A person can jump a maximum horizontal distance (by using a 45◦ projectile angle) of 5 m on Earth. The acceleration of gravity i

snow_lady [41]

Answer:30 m

Explanation:

Given

Maximum Horizontal distance is 5 m on earth

launching angle $=45^{\circ}$

Acceleration due to gravity on earth is $9.8 m/s^2$

Acceleration due to gravity on moon is $\frac{9.8}{6}=1.63 m/s^2$

Range of projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

$R_{earth}=\frac{u^2\sin 2\theta }{g}=5$ ----1

$R_{moon}=\frac{u^2\sin 2\theta }{\frac{g}{6}}$ -----2

Divide 1 & 2

$\frac{5}{R_{moon}}=\frac{1}{6}$

$R_{moon}=30 m$

4 0

3 years ago

An electric space heater draws 16.0 amps from a 120 volt source. This device is operated, on average, for 8.0 hours a day. How m

iogann1982 [59]

Answer:

1920watts

Explanation:

P=IV

P = 16 x 120 = 1920 watts

6 0

3 years ago

If we decrease the distance an object moves we will

Scrat [10]

Decrease the amount of work done.

4 0

3 years ago

Read 2 more answers

Three moles of a monatomic ideal gas are heated at a constant volume of 1.20 m3. The amount of heat added is 5.22x10^3 J.(a) Wha

k0ka [10]

Answer:

A) 140 k

b ) 5.22 *10^3 J

c) 2910 Pa

Explanation:

Volume of Monatomic ideal gas = 1.20 m^3

heat added ( Q ) = 5.22*10^3 J

number of moles (n) = 3

A ) calculate the change in temp of the gas

since the volume of gas is constant no work is said to be done

heat capacity of an Ideal monoatomic gas ( Q ) = n.(3/2).RΔT

make ΔT subject of the equation

ΔT = Q / n.(3/2).R

= (5.22*10^3 ) / 3( 3/2 ) * (8.3144 J/mol.k )

= 140 K

B) Calculate the change in its internal energy

ΔU = Q this is because no work is done

therefore the change in internal energy = 5.22 * 10^3 J

C ) calculate the change in pressure

applying ideal gas equation

P = nRT/V

therefore ; Δ P = ( n*R*ΔT/V )

= ( 3 * 8.3144 * 140 ) / 1.20

= 2910 Pa

3 0

3 years ago