All categories

3 years ago

A charged ball of mass 10g and charge 2 C is placed at rest in a field of field strength 5NC as shown.

Physics

1 answer:

Studentka2010 [4]3 years ago

7 0

Answer:

1000 m/s

Explanation:

Given that

Mass of the ball, m = 10 g = 0.01 kg

Charge of the ball, q = 2 C

Field strength of the field, E = 5 N/C

Acceleration of the ball, a = ? m/s²

We know the the equation holds,

F = qE, on substituting, we have

F = 2 C * 5 N/C

F = 10 N

Also, we know that

F = ma, and as such,

a = F / m, on substituting we have

a = 10 N / 0.01 kg

a = 1000 m/s²

Therefore, the acceleration of the ball is 1000 m/s²

You might be interested in

In which of these states is the age of consent 18 years old?

Andreas93 [3]

Answer:

all the ones that say

Explanation:

18.....

4 0

2 years ago

what happens to the period of revolution for the planets as they move farther away in position from the sun

Marina CMI [18]

It increases. Mercury takes 88 days to orbit the sun once. The Earth takes a year. Pluto takes 248 years.

3 0

3 years ago

Rahul and Sonia were playing with blocks and each of them made a train out of them. Both of them thought of measuring the length

lozanna [386]

Answer:

They measured their train using their handspan

7 0

3 years ago

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

7 0

3 years ago

Which of the following is true? A. If an equal force is applied to two objects, the object with greater mass will accelerate mor

docker41 [41]

B

I hope this helps and have a wonderful day filled with joy!!

<3

4 0

3 years ago