Ask question

Ask question

All categories

2 years ago

10

Please guys I want your help, there is 20p point

1 answer:

Stells [14]2 years ago

6 0

Answers:

The acceleration due to gravity on the surface of earth is 9.8 ms^(-2).Time period of a simple pendulum on earth and moon are 3.5 second and 8.4 second respectively. Find the acceleration due to gravity on the moon . <br> Hint : T_(e) = 2pi sqrt((L)/(g_(e))) T_(m)= 2pi sqrt((L)/(g_(m))) <br> (T_(e)^(2))/(T_(m)^(2))= (g_(m))/(g_(e)) <br> g_(m) = (T_(e)^(2))/(T_(m)^(2))g_(e)

You might be interested in

List the planets that have MORE THAN 3 moons.

antiseptic1488 [7]

Answer:

jupiter they have 79 moons,saturn has 62,uranus has 27,neptune has 14,pluto has 5,

Explanation:

6 0

3 years ago

The mass of jupiter is 300 times the mass of the earth. Jupiter orbits the sun with Tjupiter = 11.9 yr in an orbit with Rjupiter

mihalych1998 [28]

Answer:

c) 11.9 yr

Explanation:

The orbital period is proportional to r^(3/2) and does not depend on the satellite's mass. Any object at Jupiter position will have the same orbital period regardless of mass.

By keppler's law we know that

T^2= r^3

T= orbital time period

r= mean distance of the planet from the Sun.

clearly, The orbital period does not depend on the satellite's mass

there, the correct answer will be c= 11.9 yr.

5 0

3 years ago

Two capacitors of capacitances 25 µF and 50 µF are connected in series with a 33-V battery. How much energy is stored in the 25-

torisob [31]

Answer:

6.05×10⁻³ J

Explanation:

Note: Two capacitors connected in series behaves like two resistors connected in parallel.

Using

1/Ct = 1/C1+1/C2

Ct = (C1×C2)/(C1+C2)............................ Equation 1

Where Ct = combined capacitance of the two capacitor, C1 = Capacitance of the first capacitor, C2 = capacitance of the second capacitor.

Given: C1 = 25 µF, C2 = 50 µF

Substitute into equation 1

Ct = (25×50)/(25+50)

Ct = 1250/75

Ct = 16.67 µF.

Using

Q = CV.................... Equation 2

Where Q = Charge, V = Voltage.

Given: V = 33 V, C = 16.67 µF = 16.67×10⁻⁶ F

Substitute into equation 2

Q = 33(16.67×10⁻⁶)

Q = 5.5×10⁻⁴ C.

Since both capacitors are connected in series, the same amount of charge flows through them.

Using,

E = 1/2Q²/C.................. Equation 3

Where E = Energy stored in the 25-µF capacitor

Given: Q =5.5×10⁻⁴ C, C = 25 µF = 25×10⁻⁶ F

Substitute into equation 3

E = 1/2(5.5×10⁻⁴)²/ 25×10⁻⁶

E = 6.05×10⁻³ J.

5 0

3 years ago

A 60-W light bulb radiates electromagnetic waves uniformly in all directions. At a distance of 1.0 mm from the bulb, the light i

slega [8]

Answer:

The appropriate solution is:

(a) $\frac{1}{4}(I_o)$

(b) $\frac{1}{4} (u_o)$

(c) $\frac{1}{2}B_o$

Explanation:

According to the question, the value is:

Power of bulb,

= 60 W

Distance,

= 1.0 mm

Now,

(a)

⇒ $\frac{I}{I_o} =\frac{r_o_2}{r_2}$

On applying cross-multiplication, we get

⇒ $I=I_o\times \frac{1_2}{2^2}$

⇒ $=I_o\times \frac{1}{4}$

⇒ $=\frac{1}{4} (I_o)$

(b)

As we know,

⇒ $\frac{u}{u_o} =\frac{I}{I_o}$

By putting the values, we get

⇒ $u=\frac{1}{4}(u_o)$

(c)

⇒ $\frac{B^2}{B_o^2} =\frac{u}{u_o}$

$=\frac{I}{I_o}$

⇒ $B=B_o\times \sqrt{\frac{1}{4} }$

⇒ $=\frac{1}{2}(B_o)$

4 0

3 years ago

Suppose I have a spherical insulating shell with inner radius r and outer radius R. The shell has a uniformly distributed charge

a_sh-v [17]

Answer:

3Q / 4 pi (R^3 - r^3)

Explanation:

Charge density = charge / volume

volume of a spherical shell = $\frac{4}{3} \pi(R^{3} - r^{3})$

4 0

3 years ago