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3 years ago

A current of 1.8 A delivers 2.5 C of charge. How much time was required? 0.70 s 0.72 s 1.4 s 4.5 s

Physics

2 answers:

Natalka [10]3 years ago

5 0

Answer:

1.4 s

Explanation:

Given the following data;

Quantity of charge, Q = 2.5 C

Current = 1.8 A

To find the time required;

Mathematically, the quantity of charge passing through a conductor is given by the formula;

Quantity of charge, Q = current * time

Substituting into the formula, we have;

2.5 = 1.8 * time

Time = 2.5/1.8

Time = 1.4 s

alex41 [277]3 years ago

4 0

Answer:

1.4s

Explanation:

EDGE 2021

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A small convex mirror is placed 60 cm from the pole and on the axis of a large concave mirror of radius of curvature 200 cm. The

umka2103 [35]

Answer:

Calculate 'R' of convex mirror and height of the real image

the radius of the convex mirror is 48cm

Explanation:

Distance between convex and concave mirror is =60cm

Radius of the concave mirror (R) = 200cm

For the concave mirror, u = ∞

V = {R}/{2}=100cm

The object for the convex mirror and the final image is on the pole of the concave mirror, and distance between convex and concave mirror is 60cm

u_1=60-100 =-40cm

Object will be behind the convex mirror

1/f=1/40+1/60

f=24cms

the radius of the convex mirror is 48cm

6 0

3 years ago

A 20-kg child is coasting at 3.3 m/s over flat ground in a 4.0-kg wagon. The child drops a 1.0-kg ball out the back of the wagon

Shalnov [3]

Answer:

The final velocity is $u_f = 3.44 \ m/s$

Explanation:

From the question we are told that

The mass of the child is $m_1 = 20 \ kg$

The initial speed of the child is $u_1 = 3.3 \ m/s$

The mass of the wagon is $m_w = 4.0 \ kg$

The initial speed of the wagon is $u_w = 3.3 \ m/s$

The mass of the ball is $m_2 = 1.0 \ kg$

The initial speed off the ball is $u_2 = 3.3 \ m/s$

Generally the initial speed of the system (i.e the child , wagon , ball) is

$u_1 = u_w = u_2 = u =3.3 \ m/s$

Generally from the law of linear momentum conservation

$p_i = p_f$

Here $p_i$ is the momentum of the system before the ball is dropped which is mathematically represented as

$p_i = ( m_1 + m_2 + m_3 ) * u$

=> $p_i = ( 20 + 4 + 1 ) * 3.3$

=> $p_i = 82.5 \ kg \cdot m/s$

and

$p_f$ is the momentum of the system after the ball is dropped which is mathematically represented as

$p_f = ( m_1 + m_w ) * u_f$

=> $p_i = ( 20 + 4 ) * u_f$

$82.5 = 24 * u_f$

=> $u_f = 3.44 \ m/s$

4 0

3 years ago

If the volume is held constant, what happens to the pressure of a gas as temperature is decreased? Explain.

Lady bird [3.3K]

Answer:Decreases

Explanation:

Given

Volume is held constant that is it is a isochoric process.

We know that

PV=nRT

as n,V& R are constant therefore only variables are

P & T

so $\frac{P_1}{T_1}=\frac{P_2}{T_2}$

$\frac{P_1}{P_2}=\frac{T_1}{T_2}$

As $T_1$ is decreasing therefore Pressure must also decrease so that ratio remains constant.

6 0

3 years ago

Q1=-q2=q3=2.10^-8 đặt tại A,B,C .tìm lực điện tổng hợp tác dụng lên q4=2.10^-8 đặt tại D sao cho ANCD là hình

KiRa [710]

Answer:

Wala ko ka sabot ana oy sorry

5 0

3 years ago

Assuming perfect optics, if the smallest feature you can resolve when observing around 660 nm is of angular size 0.04 arcsec, wh

Dmitriy789 [7]

Answer:

The value is $\theta _2 = 0.08 \ arcsec$

Explanation:

From the question we are told that

The first wavelength is $\lambda_1 = 660 \ nm = 660 *10^{-9 }$

The first angular size is $\theta_1 = 0.04 \ arcsec$

The second wavelength is $\lambda _2 = 1320 \ nm = 1320 *10^{-9 } \ m$

Generally according to Rayleigh Criterion we have that

$\theta= 1.22 * \frac{\lambda }{D}$

given every other thing remains constant we have that

$\theta = k * \lambda$

Here k represented constant so

$k = \frac{\theta }{\lambda}$

=> $\frac{\theta_1}{ \lambda_1} = \frac{\theta_2}{ \lambda_2}$

=> $\frac{\theta_1}{ \theta_2} = \frac{\lambda_1}{ \lambda_2}$

$\frac{ 0.04}{ \theta_2} =0.5$

=> $\theta _2 = 0.08 \ arcsec$

3 0

3 years ago