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2 years ago

7

Discharge is not possible at ntp why

1 answer:

erastova [34]2 years ago

7 0

Answer:

bc discharge cannot happen at ntp

Explanation:

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A student claims that any object in motion must experience a force that keeps it in motion. Do you agree or disagree? Explain yo

frosja888 [35]

Answer:

I disagree

Explanation:

I think the students claim is wrong because according to Newton's First Law an object that is in motion stays in motion unless acted upon by an unbalanced force. Which makes the students claim wrong because a object doesn't require another force to keep it moving.

5 0

3 years ago

Read 2 more answers

A 80 ohms resistor, 0.2 H inductance and 0.1 mF capacitor are connected in series across a generator (60 Hz, V rms=120 V). Deter

qaws [65]

Answer:

Impedance = 93.75 ohms

Current = 1.81 A

Explanation:

Resistance = R = 80 ohms

Inductance = L = 0.2 H

Inductive reactance = XL = $X_{L}=$ = ωL = (2πf) L

= 2 (3.14) (60)(0.2) = 75.398 Ohms

Capacitive reactance = 1 / ωC = 1/(2πf)C = 1 / [(2π)(60)(0.1 × 10⁻3)]

= 26.526 Ohms

Impedance = Z = $\sqrt{R^{2} + (X_{L} - X_{C})^{^{2}}} =$

= $\sqrt{8788.511}$ = 93.747 ohms

Voltage = $\sqrt{2}$ × 120 = 169.7056 V

Current = I = V ÷ R = (169.7056) ÷ 93,747 = 1.81 A

8 0

3 years ago

Volcanoes tend to erupt at places where

tigry1 [53]

Answer:

it is 3

Explanation:

because the crack will be open for the magma to come out

4 0

3 years ago

Read 2 more answers

the rock of 10 kg is falling near the Earth's surface.assume that g =10N/kg and the is no air resistance. what is the accelerati

Ivanshal [37]

Answer:

kya faltu sawal h repetitive g=10N/kg

5 0

3 years ago

Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t

Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

$v^2=v_0^2+2ax\\$

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

= 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

$a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}$

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

$F_f=ma=(1490kg)(5.55m/s^2)=8271.15N$

Furthermore, you use the relation between the friction force and the friction coefficient:

$F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56$

hence, the friction coefficient is 0.56

6 0

2 years ago