2.3c. What is the frequency of a wave with a length of 12 m that travels at a speed of 36 m/s?

The total mechanical energy of a basketball is 400 J. If the kinetic energy is 286 J, what must the potential energy be?

ch4aika [34]

Answer:

the potential energy is 114 J.

Explanation:

Given;

total mechanical energy, E = 400 J

kinetic energy, K.E = 286 J

The potential energy is calculated as follows;

E = K.E + P.E

where;

P.E is the potential energy

P.E = E - K.E

P.E = 400 J - 286 J

P.E = 114 J

Therefore, the potential energy is 114 J.

7 0

2 years ago

Two parallel plates of area 2.34*10-3 M2 have 7.07*10-7C of charge placed on them. A6.62*10-5C charge q1 is placed between the p

ira [324]

Answer:

The force exerted on the $q_1$ is $F = 2.25*10^{3} \ N$

Explanation:

From the question we are told that

The area is $A = 2.34*10^{-3} \ m^2$

The magnitude of charge placed on them is $q = 7.07 * 10^{-7} C$

The charge placed between the plate is $q_1 = 6.62 *10^{-5} C$

The electric field generated around the plate is mathematically represented as

$E = \frac{q}{A \epsilon_o}$

Substituting values

$E = \frac{7.07*10^{-7}}{2.34*10^{-3} * 8.85 *10^{-12}}$

$E = 34*10^{6} \ V/m$

The force exerted the charge $q_1$ is mathematically represented as

$F = q_1 * E$

Substituting values

$F = 6.62 *10^{-5} * 34*10^{6}$

$F = 2.25*10^{3} \ N$

8 0

3 years ago

ANSWER FOR BRAINLIEST

Nikitich [7]

A (b) would be 42 (c) from x to Y mark me as brainlist thanks

5 0

2 years ago

Which statement is true about the part of the electromagnetic spectrum that human eyes can detect?

LUCKY_DIMON [66]

Your answer is B. The human eye can only detect color and Tv waves

7 0

2 years ago

Read 2 more answers

A very long solid insulating cylinder has radius R = 0.1 m and uniform charge density rho0= 10-3 C/m3. Find the electric field a

Galina-37 [17]

Answer:

$E = (0.56 \times 10^8 ) r \ \ N/c$

Explanation:

Given that:

$\rho_o = (10^{-3} ) \ c/m^3$

R = (0.1) m

To find the electric field for r < R by using Gauss Law

${\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)$

For r < R

$Q_{enclosed}=(\rho) ( \pi r^2 ) l$

$E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}$

$E= \dfrac{\rho ( r)}{2 \varepsilon_o}$

where;

$\varepsilon_o = 8.85 \times 10^{-12}$

$E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}$

$E = (0.56 \times 10^8 ) r \ \ N/c$

4 0

2 years ago