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3 years ago

Calculate the number of electrons passing a point in the wire in 1 min when the current is 1 A

Physics

1 answer:

Vera_Pavlovna [14]3 years ago

7 0

Explanation:

When one coulomb charge passes through any cross section of the wire per second,the current passing is one ampere. Charge of electron ,e=1.6X10^-19C. n=1/(1.6X10^-19)=6.25X10^18.Sep 17, 2017

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Suppose the coefficient of static friction between a quarter and the back wall of a rocket car is 0.383. At what minimum rate wo

djverab [1.8K]

Answer:

25.59 m/s²

Explanation:

Using the formula for the force of static friction:

$f_s = \mu_s N$ --- (1)

where;

$f_s =$ static friction force

$\mu_s =$ coefficient of static friction

N = normal force

Also, recall that:

F = mass × acceleration

Similarly, N = mg

here, due to min. acceleration of the car;

$N = ma_{min}$

From equation (1)

$f_s = \mu_s ma_{min}$

However, there is a need to balance the frictional force by using the force due to the car's acceleration between the quarter and the wall of the rocket.

Thus,

$F = f_s$

$mg = \mu_s ma_{min}$

$a_{min} = \dfrac{mg }{ \mu_s m}$

$a_{min} = \dfrac{g }{ \mu_s }$

where;

$\mu_s = 0.383$ and g = 9.8 m/s²

$a_{min} = \dfrac{9.8 \ m/s^2 }{0.383 }$

$\mathbf{a_{min}= 25.59 \ m/s^2}$

3 0

2 years ago

Question 7 of 25

Crank

Answer: 350 ms

Explanation:

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3 years ago

Read 2 more answers

An object is located 10.0 cm from a convex mirror. The magnitude of the

Sunny_sXe [5.5K]

Answer:

B. 6.00 cm

Explanation:

6 0

3 years ago

Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 9.50 cm apart. You see an object jump from sid

Serhud [2]

Answer: 12.67 cm, 8 cm

Explanation:

Given

Normal distance of separation of eyes, d(n) = 6 cm

Distance of separation is your eyes, d(y) = 9.5 cm

Angle created during the jump, θ = 0.75°

To solve this, we use the formula,

θ = d/r, where

θ = angle created during the jump

d = separation between the eyes

r = distance from the object

θ = d/r

0.75 = 9.5 / r

r = 9.5 / 0.75

r = 12.67 cm

θ = d/r

0.75 = 6 / r

r = 6 / 0.75

r = 8 cm

Thus, the object is 12.67 cm far away in your own "unique" eyes, and just 8 cm further away to the normal person eye

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3 years ago

kristine speeds past a parked police car at 32 m/s. The police car starts from rest with a uniform acceleration of 2.5 m/s^2. Ho

Digiron [165]

$32 = 0 \times t + \frac{1}{2} \times 2.5 \times t^{2} \\ 32 = 0 + 1.25 \times t {}^{2} \\ 32 = 1.25t {}^{2} \\ \frac{32}{1.25} = \frac{1.25t {}^{2} }{1.25} \\ t {}^{2} = 25.6 \\ \sqrt{t {}^{2} } = \sqrt{25.6} \\ t = 5.1seconds \\$

7 0

2 years ago