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3 years ago

Which statement about an atom is correct?

Physics

1 answer:

Alexus [3.1K]3 years ago

7 0

Answer:

$\boxed{ \sf{The \: electron \: has \: a \: negative \: charge \: and \: is \: found \: outside \: of \: the \: nucleus}}$

Option A is correct

Explanation:

Let's check the options:

A: The electron has a negative charge and is found outside of the nucleus.

$\mapsto{}$ Yeah! It's TRUE . An electron is a negatively charged particle and is located outside the nucleus of an atom.

B : The neutron has a negative charge and is found in the nucleus.

$\mapsto{}$ No! It's FALSE . A neutron carries no charge. i.e it is a neutral particle and found inside the nucleus.

C : The proton has no charge and is found in the nucleus.

$\mapsto{}$ No! It's FALSE.A proton is a positively charged particle present inside nucleus of an atom.

D : The neutron has no charge and is found outside of the nucleus.

$\mapsto{}$ I agree that the neutron has no charge. But it is found inside the nucleus not outside . So, this statement is FALSE .

Hence, we found our answer! :D

A. The electron has a negative charge and is found outside of the nucleus is the correct statement about an atom.

Hope I helped!

Best regards! :D

~ $\text{TheAnimeGirl}$

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A diver springs upward from a board that is 4.40 m above the water. At the instant she contacts the water her speed is 13.5 m/s

Yakvenalex [24]

The diver has the initial velocity, both (a) magnitude is 9.8 m/s and (b) direction is 73.5°.

<h3>What is free falling?</h3>

When an object is released from rest in free air considering no friction, the motion is depend only on the acceleration due to gravity, g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy(P.E)

U =mgh

When the object strikes the ground, all the potential energy converted into kinetic energy.

K.E = 1/2mv²

where v is the speed just before hitting the ground.

A diver springs upward from a board that is 4.40 m above the water. At the instant, she contacts the water her speed is 13.5 m/s and her body makes an angle of 78.1 ° with respect to the horizontal surface of the water.

(a)

From energy conservation principle, initial and final mechanical energy are equal.

1/2mu² + mgh = 1/2mv²

where, u is the initial velocity of the diver.

u = sq rt (v² - 2gh)

u = sq rt (13.5² - 2x9.81x4.4)

u = 9.798 m/s or 9.8 m/s

Thus, the velocity of the diver is 9.8 m/s.

(b)

The horizontal component of velocity will remain constant.

The horizontal component of acceleration is zero.

Then,

ucosθ = vcosΦ

θ = cos⁻¹ [ (13.5 x cos 78.1)/9.8 ]

θ = 73.5°

Thus, the direction of velocity is 73.5°.

Learn more about free falling.

brainly.com/question/13299152

#SPJ1

5 0

2 years ago

A hydraulic press has one piston of diameter 4.0 cm and the other piston of diameter 8.0 cm. What force must be applied to the s

yan [13]

Answer:

The force is $F_1 = 400.8 \ N$

Explanation:

From the question we are told that

The first diameter is $d_1 = 4.0 \ cm = 0.04 \ m$

The second diameter is $d_2 = 8.0 \ cm = 0.08 \ m$

Generally the first area is

$A_1 = \pi * \frac{d^2_1 }{4}$

=> $A_1 = 3.142 * \frac{0.04^2}{4}$

=> $A_1 = 0.00126 \ m^2$

The second area is

$A_2 = \pi * \frac{d^2_2 }{4}$

$A_2 = 3.142 * \frac{0.08^2}{4}$

$A_2 = 0.00503 \ m^2$

For a hydraulic press the pressure at both end must be equal .

Generally pressure is mathematically represented as

$P = \frac{F}{A}$

$\frac{F_1}{A_1 } = \frac{F_2}{A_2 }$

=> $F_1 = \frac{1600}{0.00503} * 0.00126$

=> $F_1 = 400.8 \ N$

8 0

3 years ago

A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130

konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

$W=Fd$ (1)

F: applied force = 130N

d: distance = 5.0m

$W=(130N)(5.0m)=650J$

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

$W_f=F_fd=\mu Mgd$ (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

$W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J$

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

$W = 0J$

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

$\Delta K=W_T$ (3)

You calculate the total work:

$W_T=Fd-F_fd=(F-F_f)d$ (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

$F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N$

Next, you replace the values of all parameters in the equation (4):

$W_T=(130N-105.84N)(5.00m)=120.80J$

$\Delta K=120.80J$

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

$W_T=\frac{1}{2}M(v_2^2-v_1^2)$ (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

$v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}$

The final speed of the box is 2.58m/s

5 0

3 years ago

Which of the following force fields will deflect electron beams?

inysia [295]

I'm going to have to go with A Magnetic because it seems to be the only one that makes sense it can't B because electron and Electrical are the same thing its not going deflect it and C it might seem right but gravity has nothing to do with deflection so I pick Magnetic.

So your answer is A.

6 0

4 years ago

Read 2 more answers

A man of mass 81 kg escapes from a burning building by jumping from a window situated 32 m above a catching net. The acceleratio

Dafna1 [17]

Answer:

The speed of man before he hits the ground is 23.35 m/s

Explanation:

We know that:

Weight of Man - Force of Friction = Unbalanced Force

but, from Newton's 2nd Law of Motion:

unbalanced force = ma

Therefore,

W - F = ma

a = (W - F)/m

a = (mg - F)/m

where,

m = 81 kg

g = 9.8 m/s²

F = 103 N

a = [(81 kg)(9.8 m/s²) - 103 N]/81 kg

a = 8.52 m/s²

using 3rd equation of motion:

Vf² - Vi² = 2ah

here,

Vi = initial velocity = 0 m/s

Vf = Final Velocity before he hits ground = ?

Vf² - 0² = 2(8.52 m/s²)(32 m)

Vf = √545.28 m²/s²

Vf = 23.35 m/s

4 0

3 years ago