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N76 [4]
3 years ago
5

Suppose that after walking across a carpeted floor you reach for a doorknob and just before you touch it a spark jumps 0.95cm fr

om your finger to the knob. Find the minimum voltage needed between your finger and the doorknob to generate this spark. The minimum field that can produce a spark in air is 3.0x10^6 V/m. V = _______ V
Physics
1 answer:
alexandr402 [8]3 years ago
8 0

To solve this problem we will consider the definition of Minimum Energy, as the change between the minimum voltage and its distance. Mathematically it can be written as

E_{min} = \frac{V_{min}}{r}

Rearranging to find the Voltage,

V_{min} = E_{min} \times r

Replacing,

V_{min} = (3.0) \times (0.95*10^4)

V_{min} = 28500V

Therefore the minimum voltage needed between your finger and the doorknob to generate this spark is 28.5kV

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I'm walking 1.6m/s to 7-11 and it started to rain so I sped up to 2.7m/s in 1.2
olga nikolaevna [1]

Answer:

Explanation:

a = \frac{v_f-v_0}{t} which is the final velocity minus the initial velocity in the numerator, and the change in time in the denominator.  For us:

a=\frac{2.7-1.6}{1.2} so

a = .92 m/s/s (NOT negative because you're speeding up)

5 0
3 years ago
If v=10v, what value of R would limit l to 10 mA? (if you have 10 volts, what resistance would be needed to limit the current to
777dan777 [17]

r=v/i =10/0.01=1000ohms B

7 0
3 years ago
The 5-kg block A has an initial speed of 5 m/s as it slides down the smooth ramp, after which it collides with the stationary bl
Akimi4 [234]

Answer:

The coupled velocity of both the blocks is 1.92 m/s.

Explanation:

Given that,

Mass of block A, m_1=5\ kg

Initial speed of block A, u_1=5\ m/s

Mass of block B, m_2=8\ kg

Initial speed of block B, u_2=0

It is mentioned that if the two blocks couple together after collision. We need to find the common velocity immediately after collision. We know that due to coupling, it becomes the case of inelastic collision. Using the conservation of linear momentum. Let V is the coupled velocity of both the blocks. So,

m_1u_1+m_2u_2=(m_1+m_2)V\\\\V=\dfrac{m_1u_1+m_2u_2}{(m_1+m_2)}\\\\V=\dfrac{5\times 5+0}{(5+8)}\\\\V=1.92\ m/s

So, the coupled velocity of both the blocks is 1.92 m/s. Hence, this is the required solution.

8 0
3 years ago
When three identical bulbs of 60W,200V rating are connected in series at a 200V supply the power drawn by them will be?
Law Incorporation [45]

Answer:

P = 180 [w]

Explanation:

To solve this problem we must use ohm's law, which is defined by the following formula.

V = I*R & P = V*I

where:

V = voltage = 200[volts]

I = current [amp]

R = resistance [ohm]

P = power [watts]

Since the bulbs are connected in series, the powers should be summed

P = 60 + 60 + 60

P = 180 [watts]

Now we can calculate the current

I = 180/200

I = 0.9[amp]

Attached is an image where we see the three bulbs connected in series, in the circuit we see that the current is the same for all the elements connected to the circuit.

And the power is defined by P = V*I

we know that the voltage is equal to 200[V], therefore

P = 200*0.9

P = 180 [w]

8 0
3 years ago
A rabbit runs 28 m toward the left in 9 s; then the rabbit suddenly changes direction and runs 18 m toward the right in the next
Viefleur [7K]

Answer:

The answer is 3.111111.

Explanation:

It runs 28 m in the first 9 s, and 28 divided by 9 equals 3.1 and the one goes on forever.

8 0
3 years ago
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