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Drupady [299]
3 years ago
15

You have been hired by a "storm chaser" as an assistant. This individual loves to find locations at which tornadoes and violent

lightning storms are occurring. While there, he takes photographs and makes electromagnetic measurements. You are on a chase with him to a lightning storm in Florida. He explains to you that a cloud layer and the ground can be considered as the plates of a capacitor that stores charge, with the ground being charged negatively. The capacitor continuously leaks charge due to the free charges in the air between the plates. In thunderclouds, however, various processes result in charge distributions that eventually lead to lightning, a phenomenon that delivers negative charge to the ground. Therefore, the lightning recharges the capacitor. Ahead of you is a cloud layer that the storm chaser measures to be of area 1.65 km2 and height 740 km above the ground. He then uses a special apparatus called a field mill to measure that the electric field under the cloud is 4.00 ✕ 106 N/C. He asks you to do a quick calculation of the charge on the cloud-ground capacitor (in C), so you can know what to expect if there is a bolt of lightning.
Physics
1 answer:
Studentka2010 [4]3 years ago
5 0

Answer:

58.44 C

Explanation:

Electric field is found by

E=\frac {\sigma}{\epsilon_o}=\frac {Q}{A\epsilon_o}

Therefore, the charge is

Q=EA\epsilon_o


Q= 4.00 ✕ 10^{6} N/C*1.65 *10^{6} m^{2}*8.854*10^{-12}= 58.4364 C


Therefore, required charge is 58.44 C

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Through what voltage must an αα-particle, with its charge of +2e+2e, be accelerated so that it has just enough energy to reach a
mezya [45]

Answer:5101.35v

Explanation:

Radius of gold nucleus=7.3×10-15m and a charge of +79e

Q= 79e

e=1.6×10^-19

q= +2e

The nucleus is considered as the point charge where the potential energy between the charges are

U = 1/(4×3.142×Eo)×(qQ)/r

Where r is distance between the charges and the nucleus

r=R+d

V=U/q

U= 1/(4×3.142×Eo)×Q/r

V= 1/(4×3.142×Eo)×Q/(r+d)

9.0×10^9 ×(79×10^-19)/(7.3×10^-15)+(1.5×10^-14)

V= 9.0×10^9 ×(1.264×10^-17)/(2.23×10^-14)

V= 9×10^9×(5.67×10^-14)= 5101.35v

3 0
3 years ago
Light bulb is connected to a 110-V source. What is the resistance of this bulb if it is a 100-W bulb
masya89 [10]

Answer:

<h2>121ohms</h2>

Explanation:

Formula used for calculating power P = current * voltage

P = IV

From ohms law, V = IR where R is the resistance. Substituting V = IR into the formula for calculating power, we will have;

P = IV

P =(V/R)V

P = V²/R

Given parameters

Power rating of the bulb P = 100 Watts

Source voltage V = 110V

Required

Resistance of the bulb R

Substituting the given parameters into the formula for calculating power to get Resistance R;

P = V²/R

100 = 110²/R

R = 110²/100

R = 110 * 110/100

R = 12100/100

R = 121 ohms

<em>Hence, the resistance of this bulb is 121 ohms</em>

5 0
3 years ago
Science and math people where you at
PilotLPTM [1.2K]
For the second question you’re solving for resistance. resistance= voltage/ current. 120/0.5= 240. the answer is 240 ohms
for the third question you would do 2*4 since it’s asking for voltage, the answer is 8 volts :)
8 0
3 years ago
A space probe is fired as a projectile from the Earth's surface with an initial speed of 2.05 104 m/s. What will its speed be wh
Elanso [62]

Answer:

The value is  v  =  2.3359 *10^{4} \ m/s

Explanation:

From the question we are told that

  The  initial speed is u =  2.05 *10^{4} \  m/s

 Generally the total energy possessed by the space probe when on earth is mathematically represented as

             T__{E}} =  KE__{i}} +  KE__{e}}

Here  KE_i is the kinetic energy of the space probe due to its initial speed which is mathematically represented as

          KE_i =   \frac{1}{2}  *  m  *  u^2

=>       KE_i =   \frac{1}{2}  *  m  *  (2.05 *10^{4})^2

=>       KE_i =  2.101 *10^{8} \ \ m \ \ J

And  KE_e is the kinetic energy that the space probe requires to escape the Earth's gravitational pull , this is mathematically represented as

       KE_e =  \frac{1}{2}  *  m *  v_e^2

Here v_e is the escape velocity from earth which has a value v_e =  11.2 *10^{3} \  m/s

=>    KE_e =  \frac{1}{2}  *  m *  (11.3 *10^{3})^2

=>    KE_e =  6.272 *10^{7} \  \  m  \ \   J

Generally given that at a position that is very far from the earth that the is Zero, the kinetic energy at that position is mathematically represented as

        KE_p =  \frac{1}{2}  *  m *  v^2

Generally from the law energy conservation we have that

        T__{E}} =  KE_p

So

       2.101 *10^{8}  m  +  6.272 *10^{7}  m  =   \frac{1}{2}  *  m *  v^2

=>     5.4564 *10^{8} =   v^2

=>     v =  \sqrt{5.4564 *10^{8}}

=>     v  =  2.3359 *10^{4} \ m/s

4 0
3 years ago
One end of a 7-cm-long spring is attached to the ceiling. When a 5.4 kg mass is hung from the other end, the spring is stretched
mash [69]

Answer:

2.63 cm

Explanation:

Hooke's law gives that the force F is equal to cy where c is spring constant and x is extension

Making c the subject of the formula then

c=\frac {F}{y}

Since F is gm but taking the given mass to be F

c=\frac {5.4 kg}{4.3 cm}=1.2558139534883720930232558139534883720930

By substitution now considering F to be 3.3 kg

y=\frac {3.3 kg}{1.2558139534883720930232558139534883720930}=2.6277777777777 cm\approx 2.63 cm

8 0
3 years ago
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