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3 years ago

A boy having 0.2kg toy tied with a sting, having force 2N, angle 60º displaces it 5 m. What is the work done?

Physics

1 answer:

snow_lady [41]3 years ago

3 0

Answer:

solution:

here,

force = 2N

displace=5m

then,

w =f*d

=2*5

=10 joule

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A small but measurable current of 3.8 × 10-10 A exists in a copper wire whose diameter is 2.5 mm. The number of charge carriers

Karolina [17]

Answer:

a) 4.9*10^-6

b) 5.71*10^-15

Explanation:

Given

current, I = 3.8*10^-10A

Diameter, D = 2.5mm

n = 8.49*10^28

The equation for current density and speed drift is

J = I/A = (ne) Vd

A = πD²/4

A = π*0.0025²/4

A = π*6.25*10^-6/4

A = 4.9*10^-6

Now,

J = I/A

J = 3.8*10^-10/4.9*10^-6

J = 7.76*10^-5

Electron drift speed is

J = (ne) Vd

Vd = J/(ne)

Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)

Vd = 7.76*10^-5/1.3584*10^10

Vd = 5.71*10^-15

Therefore, the current density and speed drift are 4.9*10^-6

And 5.71*10^-15 respectively

3 0

3 years ago

8. Three grams of Bismuth-218 decay to 0.375 grams in one hour. What is the half-

Evgen [1.6K]

Answer: 0.333 h

Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula</u>:

$A=A_{o}.2^{\frac{-t}{H}}$ (1)

Where:

$A=0.375 g$ is the final amount of the material

$A_{o}=3 g$ is the initial amount of the material

$t=1 h$ is the time elapsed

$H$ is the half life of the material (the quantity we are asked to find)

Knowing this, let's substitute the values and find $h$ from (1):

$0.375 g=(3 g)2^{\frac{-1h}{H}}$ (2)

$\frac{0.375 g}{3 g}=2^{\frac{-1h}{H}}$ (3)

Applying natural logarithm in both sides:

$ln(\frac{0.375 g}{3 g})=ln(2^{\frac{-1 h}{H}})$ (4)

$-2.079=-\frac{1 h}{H}ln(2)$ (5)

Clearing $H$ :

$H=\frac{-1h}{-2.079}(0.693)$ (6)

Finally:

$h=0.333 h$ This is the half-life of the Bismuth-218 isotope

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3 years ago

Helpp .....meeee pleaseee

Lady bird [3.3K]

Answer:

hydrogen

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Explanation:

join this grop to get instant answer

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3 years ago

Which statement is supported by the information in the diagram and table below?

Natasha_Volkova [10]

Answer:

Explanation:

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3 years ago

An auditorium measures 40.0 m 3 20.0 m 3 12.0 m. The density of air is 1.20 kg/m3. What are (a) the volume of the room in cubic

Burka [1]

We use the formula,

m = V\rho

Here, m is the mass, V is the volume and $\rho$ density

Also

$V = l w h$

Here l is length, w is width and h is height.

(a) The volume of the room,

$V=40.0 \ m \times 3 20.0 \ m \times 3 12.0\ m = 3.99 \times 10^{6} m^3$

The volume of the room in cubic feet,

$V = 3.99 \times 10^{6} m^3 \times(\frac{3.281 \ ft}{m} )^3 = 4.3 \times 10^7 \ ft^3$

(b) Now the mass of the air in room,

$m= (3.99 \times 10^{6} \ m^3) (1.20 \ kg/m^3) = 4.8 \times 10^6 kg$ .

Therefore, the weight of the air in room,

$W = mg= 4.8 \times 10^6 kg \times 9.8 m/s^2 = 46.9 \times 10^6 \ N \\\\ W = 4.69 \times 10^7 \ N$ .

The weight of air in the room in pounds,

$W = 4.69 \times 10^7 \ N ( \frac{1 \ lb}{4.448 \ N} ) = 1.1 \times 10^7 \ lb$

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3 years ago