What is a metric unit for energy?

We can calculate the work done during segment A by using the work-energy theorem, which states that the work done is equal to the gain in kinetic energy of the object:

$W=K_f -K_i$

where Kf is the final kinetic energy and Ki the initial kinetic energy. The initial kinetic energy is zero (because the initial velocity is 0), while the final kinetic energy is

$K_f =\frac{1}{2}mv^2$

The mass is m=1.3 kg, while the final velocity is v=30 m/s, so the work done is:

$W=K_f = \frac{1}{2}(1.3 kg)(30 m/s)^2=585 J$

5 0

4 years ago

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Ratio of resistances of two bulbs is 2:3. If they are connected in series to a supply, then the ratio of voltages across them is

V125BC [204]

Answer:

Explanation:

Given that,

Two resistor has resistance in the ratio 2:3

Then,

R1 : R2 = 2:3

R1 / R2 =⅔

3 •R1 = 2• R2

Let R2 = R

Then,

R1 = ⅔R2 = 2/3 R

So, if the resistor are connected in series

Let know the current that will flow in the circuit

Series connection will have a equivalent resistance of

Req = R1 + R2

Req = R + ⅔ R = 5/3 R

Req = 5R / 3

Let a voltage V be connect across then, the current that flows can be calculated using ohms law

V = iR

I = V/Req

I = V / (5R /3)

I = 3V / 5R

This the current that flows in the two resistors since the same current flows in series connection

Now, using ohms law again to calculated voltage in each resistor

V= iR

For R1 = ⅔R

V1 =i•R1

V1 = 3V / 5R × 2R / 3

V1 = 3V × 2R / 5R × 3

V1 = 2V / 5

For R2 = R

V2 = i•R2

V2 = 3V / 5R × R

V2 = 3V × R / 5R

V2 = 3V / 5

Then,

Ratio of voltage 1 to voltage 2

V1 : V2 = V1 / V2 = 2V / 5 ÷ 3V / 5

V1 : V2 = 2V / 5 × 5 / 3V.

V1 : V2 =2 / 3

V1:V2 = 2:3

The ratio of their voltages is also 2:3

5 0

3 years ago

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in t

Liono4ka [1.6K]

(a) 5.69 N/C, vertically downward

We can calculate the acceleration of the electron by using the SUVAT equation:

$d=ut+\frac{1}{2}at^2$

where

d = 4.50 m is the distance travelled by the electron

u = 0 is the initial velocity of the electron

$t=3.00 \mu s = 3.0 \cdot 10^{-6} s$ is the time of travelling

a is the acceleration

Solving for a,

$a=\frac{2d}{t^2}=\frac{2(4.50)}{(3.0\cdot 10^{-6})^2}=1.0\cdot 10^{12} m/s^2$

Given the mass of the electron,

$m=9.11\cdot 10^{-31} kg$

We can find the electric force acting on the electron:

$F=ma=(9.11\cdot 10^{-31})(1.0\cdot 10^{12})=9.11\cdot 10^{-19}N$

And the electric force can be written as

$F=qE$

where

$q=-1.6\cdot 10^{-19}C$ is the charge of the electron

E is the magnitude of the electric field

Solving for E,

$E=\frac{F}{q}=\frac{9.11\cdot 10^{-19}}{-1.6\cdot 10^{-19}}=-5.69 N/C$

The negative sign means that the direction of the electric field is opposite to the direction of the force (because the charge is negative): since the force has same direction of the acceleration (vertically upward), the electric field must point vertically downward.

(b) Yes

We can answer the question by calculating the magnitude of the gravitational force acting on the electron, to check if it is relevant or not. The gravitational force on the electron is:

$F=mg$

where

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

$g=9.81 m/s^2$ is the acceleration due to gravity

Substituting,

$F=(9.11\cdot 10^{-31})(9.81)=8.93\cdot 10^{-30}N$

We see that the gravitational force is basically negligible compared to the electric force calculated in part (a), therefore we can say it is justified to ignore the effect of gravity in the problem.

7 0

4 years ago