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zubka84 [21]
3 years ago
10

A plastic rod that has been charged to -15.0nC touches a metal sphere. Afterward, the rod's charge is -10.0nC. How many charged

particles were transferred?
Physics
1 answer:
max2010maxim [7]3 years ago
7 0

Answer:

3.125 X 10^{10}

Explanation:

Net charge= -15.0 nC- -10.0 nC=-5nC

\frac {-5 X 10^{-9}}{ -1.6 X 10^{-19}} = 3.125 X 10^{10} electrons were transferred

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Choose the statement(s) that is/are true about the ratio \frac{C_p}{C_v} C p C v for a gas? (Ii) This ratio is the same for all
Blababa [14]

Answer:

(i) false

(ii) true

(iii) true

(iv) false

Explanation:

(i) The ratio of Cp and Cv is not constant for all the gases. It is because the value of cp and Cv is different for monoatomic, diatomic and polyatomic gases.

So, this is false.

(ii) For monoatomic gas

Cp = 5R/2, Cv = 3R/2

So, thier ratio

Cp / Cv = 5 / 3 = 1.67

This statement is true.

(iii) for diatomic gases

Cp = 7R/2, Cv = 5R/2

Cp / Cv = 7 / 5 = 1.4

This statement is true.

(iv) It is false.

6 0
3 years ago
You have a radioactive sample with a half-life of 1 hour. At t = 1 hour, a Geiger counter measures its radiation at 40 counts/mi
Mama L [17]

Answer:

Explanation:

Given that, .

The half-life of a radioactive element is

t½ = 1 hr.

At the first hour, the radioactive element has 40 counts/minute

After 4 hours it has 5 counts/minute

So,

We want to filled the table.

0 hours, I.e at the start

40 counts/mins × 2 = 80 counts / mis

First half-life (first hour) is

40 counts/mins

Second half-life (second hour)

40 counts/mins × ½ = 20 counts/mins

Third half-life (third hour)

40 counts/mins × ¼ = 10 counts / mins

Fourth half life (fourth hour)

40 counts/mins × ⅛ = 5 counts / mins

So, the table is

Time........................Geiger Counter Rate

0 hours.................... 80 counts / minutes

1 hour........................ 40 counts / minutes

2 hours..................... 20 counts / minutes

3 hours..................... 10counts / minutes

4 hours...................... 5 counts / minute

6 0
3 years ago
A crow is flying horizontally with a constant speed of 2.70 m/s when it releases a clam from its beak. the clam lands on the roc
vovangra [49]

Part a)

in horizontal direction there is no gravity or no other acceleration

so in horizontal direction the speed of clam will remain same

v_x = 2.70 m/s

Part b)

In vertical direction we can use kinematics

v_f = v_i + at

v_f = 0 + 2.1 * 9.8

v_f = 20.6 m/s

part c)

if the speed of crow will be increased then the horizontal speed of the clam will also increase but there is no change in the vertical speed

5 0
2 years ago
You are helping two friends from our class with a physics problem where a cart is pushed up a ramp. In examining the motion of t
stiks02 [169]

Answer: Acceleration will have 2 components, vertical and horizontal.

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Since acceleration is a vector quantity and the cart is being pushed up the ramp, the ramp would be at some angle to the horizontal and hence there will be vertical and horizontal components of acceleration.

<u>For vertical acceleration:</u>

If the magnitude of the upward component of the applied acceleration is greater than the value of the acceleration due to gravity then the net vertical acceleration will be upward because it will overtake the value of acceleration due to gravity.

In case the upward component of the applied acceleration is lesser than the value of the acceleration due to gravity then the net vertical acceleration will be downward.

<u>For horizontal acceleration:</u>

This component remains unaffected and is equal to the horizontal component of the applied acceleration because there is no other acceleration acting in the horizontal direction.

But the net acceleration will not be solely in the vertical or horizontal direction because the block has to move forward on the inclined ramp so there will always exist a horizontal and a vertical component making the net acceleration to parallel to the ramp in upward direction if the body is going up the ramp.

8 0
3 years ago
Read 2 more answers
In an earthquake, a P wave is a longitudinal wave. It moves through soil and rock as a
ElenaW [278]
<span>In an earthquake, a P wave is a longitudinal wave. It moves through soil and rock as a C. series of compressions and rarefactions.</span>
7 0
2 years ago
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