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3 years ago

Which statement accurately describes the charge of the nucleus of an atom?

Physics

1 answer:

Aleksandr [31]3 years ago

3 0

Answer:

A. The nucleus can be either positively charged or neutral.

Explanation:

The nucleus of an atom contains protons and neutrons only. Whereas electrons revolve around the nucleus. Protons are positively charged, neutrons have no charge and electrons are negatively charged.

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Sics and Chemistry A (2nd Six Weeks) - SC3200SFA/01

Andreyy89

Force = Mass * Acceleration therefore the red ball with the higher mass will have more force and greater acceleration

3 0

3 years ago

You wrap a wire around a piece of iron. If you slowly increase the strength of an electric current flowing through the wire, you

Maksim231197 [3]

The correct answer is A. the magnet to become stronger

The stronger the electric current in the piece of metal, the stronger the magnetic field will be.

4 0

3 years ago

Read 2 more answers

A government agency estimated that air bags have saved over 14,000 lives as of April 2004 in the United States. (They also state

balu736 [363]

To solve this problem it is necessary to apply the concepts related to momentum, momentum and Force. Mathematically the Impulse can be described as

$I = F*t$

Where,

F= Force

t= time

At the same time the moment can be described as a function of mass and velocity, that is

$P = m\Delta v \rightarrow P=m(v_1-v_2)$

Where,

m = mass

v = Velocity

From equilibrium the impulse is equal to the momentum, therefore

$I = p$

$Ft = m(v_1-v_2)$

PART A) Since the body ends at rest, we have the final speed is zero, so the momentum would be

$p=m(v_1-v_2)$

$p = 75*0.15$

$p = 1125Kg\cdot m/s$

Therefore the magnitude of the person's impulse is 1125Kg.m/s

PART B) From the equation obtained previously we have that the Force would be:

$Ft = m(v_1-v_2)$

$F(0.025)= 1125$

$F= 45000N$

Therefore the magnitude of the average force the airbag exerts on the person is 45000N

6 0

3 years ago

Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E

Kay [80]

Answer:

The true course: $40.29^\circ$ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

$V_w$ = velocity of wind = $30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h$

$V_p$ = velocity of plane in still air = $100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h$

Assume:

$V_r$ = resultant velocity of the plane

$\theta$ = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

$\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h$