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Ierofanga [76]
3 years ago
15

The figure shows a positive charge placed in a uniform electric field. Is the force exerted on the +1 C charge directed up or do

wn?

Physics
2 answers:
Sindrei [870]3 years ago
6 0

Answer:

DOWN

Explanation:

When a positive charge is placed in uniform or nonuniform electric field then due to the Electric field present in that region the charge will experience electrostatic force on itself.

This electrostatic force is given as

F = qE

here we know

q = charge

E = electric field

Now if charge is positive then force on the charge is always in the direction of electric field but if charge is negative then force is exactly opposite to electric field direction.

So here it will have force in the direction of electric field which is DOWN

raketka [301]3 years ago
3 0

Answer:

the answer is down on e2020

Explanation:

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If the earth can travel 1800 km in 1 minute what is it’s speed in km/s
Lady bird [3.3K]

Answer:

30km/s

Explanation:

1800/60=30

4 0
3 years ago
Work & Power Problems
Artyom0805 [142]

Answer:

1. Unit for work is measured in Joules.

Formula for work is power * time taken

2 The unit for force is Newton

A work done in lifting a jar of water

3. The unit for measuring power is Joules.

The formula for measuring power is work done/time taken

6 0
3 years ago
A 7750 kg space probe, moving nose-first toward Jupiter at 179 m/s relative to the Sun, fires its rocket engine, ejecting 72.0 k
Reika [66]

Answer:

179.47m/s

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the final velocity

Substitute

7750(179)+72(230) = (7750+72)v

1,387,250+16560 = 7822v

1,403,810 = 7822v

v = 1,403,810/7822

v= 179.47m/s

Hence the final velocity of the probe is 179.47m/s

7 0
3 years ago
A ball is thrown directly downward with an initial speed of 7.65 m/s froma height of 29.0 m. After what time interval does it st
coldgirl [10]

Answer: 1.77 s

Explanation: In order to solve this problem we have to use the kinematic equation for the position, so we have:

xf= xo+vo*t+(g*t^2)/2  we can consider the origin on the top so the xo=0 and xf=29 m; then

(g*t^2)/2+vo*t-xf=0  vo is the initail velocity, vo=7.65 m/s

then by solving the quadratric equation in t

t=1.77 s

8 0
3 years ago
A tank contains gas at 13.0°C pressurized to 10.0 atm. The temperature of the gas is increased to 95.0°C, and half the gas is re
fomenos

Answer:

The pressure of the remaining gas in the tank is 6.4 atm.

Explanation:

Given that,

Temperature T = 13+273=286 K

Pressure = 10.0 atm

We need to calculate the pressure of the remaining gas

Using equation of ideal gas

PV=nRT

For a gas

P_{1}V_{1}=nRT_{1}

Where, P = pressure

V = volume

T = temperature

Put the value in the equation

10\times V=nR\times286....(I)

When the temperature of the gas is increased

Then,

P_{2}V_{2}=\dfrac{n}{2}RT_{2}....(II)

Divided equation (I) by equation (II)

\dfrac{P_{1}V}{P_{2}V}=\dfrac{nRT_{1}}{\dfrac{n}{2}RT_{2}}

\dfrac{10\times V}{P_{2}V}=\dfrac{nR\times286}{\dfrac{n}{2}R368}

P_{2}=\dfrac{10\times368}{2\times286}

P_{2}= 6.433\ atm

P_{2}=6.4\ atm

Hence, The pressure of the remaining gas in the tank is 6.4 atm.

4 0
3 years ago
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