The football player throws the football with an initial diagonal velocity of 17 m/s at an

The diagram shows particles that make up an atom. Which label best completes the diagram?

KonstantinChe [14]

The right answer is "Strong nuclear force"

8 0

3 years ago

A 3-kg ball is thrown with a speed of 8 m/s at an unknown angle above the horizontal. The ball attains a maximum height of 2.8 m

Alina [70]

Answer:

13.5 J

Explanation:

mass of ball, m = 3 kg

maximum height, h = 2.8 m

initial speed, u = 8 m/ s

Angle of projection, θ

use the formula of maximum height

$H = \frac{u^{2}Sin^{2}\theta }{2g}$

$2.8 = \frac{8^{2}Sin^{2}\theta }{2\times 9.8}$

Sin θ = 0.926

θ = 67.8°

The velocity at maximum height is u Cosθ = 8 Cos 67.8 = 3 m/s

So, kinetic energy at maximum height

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

K = 0.5 x 3 x 3 x 3

K = 13.5 J

8 0

4 years ago

When an object such as a plastic comb is charged by rubbing it with a cloth, the net charge is typically a few microcoulombs. If

masya89 [10]

The concept required to solve this problem is quantization of charge.

First the number of electrons will be calculated and then the total mass of the charge.

With these data it will be possible to calculate the percentage of load in the mass.

$Q= ne$

Here Q is the charge, n is the number of electrons and e is the charge on the electron

$n = \frac{Q}{e}$

Replacing,

$n = \frac{4*10^{-6}C}{1.6*10^{-19}}$

$n = 2.5 * 10^{13}77$

According to the quantization of charge the charge is defined as product of the number of electron and the charge on the electron

The total mass of the charge is

$m= nm_e$

Here,

m = Mass of the charge

n = Number of electrons

$m_e$ = Mass of the electron

$\text{Percentage change} = \frac{nm_e}{M}*100$

Replacing we have

$\text{Percentage change} = \frac{(2.5*10^13)(9.1*10^{-28})}{33}*100$

$\text{Percentage change} = 6.9*10^{-14} \%$

6 0

4 years ago

Release an electron initially at rest in the presence of an electric field. The electron tends to go to the region of 1. same el

olga nikolaevna [1]

Answer:

The electron tends to go to the region of 4. higher electric potential.

Explanation:

When a charged particle is immersed in an electric field, it experiences a force given by

$F=qE$

where

q is the charge of the particle

E is the electric field

The direction of the force depends on the sign of the charge. In particular:

- The force and the electric field have the same direction if the charge is positive

- The force and the electric field have opposite directions if the charge is negative

Therefore, an electron (negative charge) moves in the direction opposite to the electric field lines.

However, electric field lines go from points at higher potential to points at lower potential: so, electrons move from regions at lower potential to regions of higher potential.

Therefore, the correct answer is

The electron tends to go to the region of 4. higher electric potential.

4 0

4 years ago

A light spring with force constant 3.80 n/m is compressed by 7.20 cm as it is held between a 0.300-kg block on the left and a 0.

jeka94

Some dogs may inherit a susceptibility to epilepsy.

3 0

4 years ago