A box of oranges with a weight of 96 N is being pushed across a horizontal floor. As it moves, it is slowing at a constant rate of 0.90 m/s each second. The push force has a horizontal component of 20 N and a vertical component of 25 N downward.

Required:
Calculate the coefficient of kinetic friction between the box and floor.

Explanation:

The force between charges is called electrostatic force. The force is given by the relation :

[tex]F=\dfrac{kq_1q_2}{d^2}[/tex]

Where

k is the electrostatic constant

d is the distance between charges

It is clear from the above formula that the electrostatic force is directly proportional to the product of charges and inversely proportional to the square of distance between them.

It means the larger the absolute value of the charges, the larger the force between them. Also, the larger the distance between particles with charge, the weaker the force between them.

Answer:

200m

Explanation:

Answer:

C

Explanation:

Given that a liquid at rest in a fixed container exerts a force perpendicular to the wall of the container. Two students make claims about the microscopic cause of this force. Student A says that the force exerted by individual molecules as they bounce off the wall is always perpendicular to the wall. Student B says that the molecules may strike the wall at angles that are not perpendicular. Which student is correct and why?

The correct answer is option C

Student B . The molecules in a liquid are in random motion at the microscopic scale. For every atom that hits at an angle to one side of perpendicular, there is likely to be another atom hitting at the same speed at the same angle on the other side of perpendicular. On average the nonperpendicular components cancel

Answer:

Option B and C are True

Note: The attachment below shows the force diagram

Explanation:

The weight of the two blocks acts downwards.

Let the weight of the two blocks be W. Solving for T₁ and T₂;

w = T₁/cos 60° -----(1)

w = T₂/cos 30° ----(2)

equating (1) and (2)

T₁/cos 60° = T₂/cos 30°

T₁ cos 30° = T₂ cos 60°

T₂/T₁ = cos 30°/cos 60°

T₂/T₁ =1.73

Therefore, option a is false since T₂ > T₁

Option B is true since T₁ cos 30° = T₂ cos 60°

Option C is true because the T₃ is due to the weight of the two blocks while T₄ is only due to one block.

Option D is wrong because T₁ + T₂ > T₃ by simple summation of the two forces, except by vector addition.

Answer:

I’m really not sur but I think it’s b

Explanation:

Answer:

h= 63.5 m

Explanation:

       [tex]\Delta h = \frac{1}2} * g* t^{2} = \frac{1}{2} *(-9.8 m/s2)*(3.6s)^{2} = -63.5 m[/tex]

        y₀ = h = 63.5 m

Answer:

Disease are caused by micro organisms

All objects in the universe are attracted to one another

species evolved through natural selection

The scientific facts from the given list are,

Microorganisms that are too tiny to be seen with the bare eye are the primary cause of many illnesses.

The force of gravity draws all objects in the cosmos toward one another.

Natural selection is the process through which species change.

Science is the methodical, empirically-based pursuit and application of knowledge and understanding of the natural and social worlds.

Microorganisms that are too tiny to be seen with the bare eye are the primary cause of many illnesses.

The force of gravity draws all objects in the cosmos toward one another.

Natural selection is the process through which species change.

Therefore options A, B, and C are scientific facts.

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Answer:

The plasma membrane, or the cell membrane, provides protection for a cell. It also provides a fixed environment inside the cell, and that membrane has several different functions. One is to transport nutrients into the cell and also to transport toxic substances out of the cell.

Answer:

Force = F = 3.63 [N]

Explanation:

To solve this problem we must use Newton's second law, which tells us that the sum of all forces must be equal to the product of mass by acceleration.

where:

m = mass = 125 [g] = 0.125[kg]

a = 29 [m/s^2]

ΣF = m*a

F = (0.125*29)

F = 3.63 [N]

Answer:

the answer is D

Explanation:

Just took the test

The NCAA Sportsmanship Award honors students who demonstrate the ideals of sportsmanship, including fairness, honesty and respect.

NCAA stands for National Collegiate Athletic Association. The NCAA is a organization which will be useful for the well being and the success for life long to the college Athelet.

A person who got graduate from high school and completed a minimum of 16 core courses for Division I or II and to earn a minimum required grade-point average in normal core courses and also must earn a qualifying test score on either the ACT or SAT will be eligible to join NCAA.

The NCAA Sportsmanship Award honors a student-athlete who will takes actions in the competitive arena of intercollegiate athletics and who demonstrated one or more of the ideals of sportsmanship, including fairness, civility, honesty, unselfishness, respect and responsibility.

Hence, Option D is the correct answer.

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Answer:

Vf=78

d=

Explanation:

vf=vi+at

=72+(2)(3)

=78km/hr

d=vit+1/2at^2

=(72)(3)+1/2 (2)(3)^2

=216+9

=225m

1) 26m/s

2) 69 m

Hi there !

1)

a = (v₂ - v₁)/t => at = v₂ - v₁ => v₂ = at + v₁

v₁ = 72km/h = 72000m/3600s = 20m/s

v₂ = 2m/s²ₓ3s + 20m/s

= 6m/s + 20m/s

= 26 m/s

2)

d = v₁t + at²/2

= 20m/sₓ3s + 2m/s²ₓ9s²/2

= 60m + 9m

= 69 m

Good luck !

Answer:

a) 1.082 × 10⁻¹⁹C  ( e = 1.6 × 10⁻¹⁹C)  

b) 3.466 × 10¹¹ N/C

Explanation:

a)

p(r) = -A exp ( - 2r/a₀)

Q = ₀∫^∞ ₀∫^π ₀∫^2xπ p(r)dV  =  -A  ₀∫^∞ ₀∫^π ₀∫^2π   exp ( - 2r/a₀)r² sinθdrdθd∅

Q = -4πA ₀∫^∞ exp ( - 2r/a₀)r²dr = -e

now using integration by parts;

A = e / πa₀³

p(r) =  - (e / πa₀³) exp (-2r/a₀)

Now Net charge inside a sphere of radius a₀ i.e Qnet is;

= e - (e / πa₀³)  ₀∫^a₀ ₀∫^π ₀∫^2π  r² exp (-2r/a₀)dr

= e - e + 5e exp (-2) = 1.082 × 10⁻¹⁹C  ( e = 1.6 × 10⁻¹⁹C)  

b)

Using Gauss's law,

E × 4πa₀ ² = Qnet / ∈₀

E = 4πa₀ ² × Qnet × 1/a₀²

E = 3.466 × 10¹¹ N/C

Answer:

google answers, i'm not 100% on how to type my own answer

Explanation:

Thaw food submerged under running water at a temperature of 70˚F (21˚C) or lower. The water flow must be strong enough to wash food particles into the overflow drain. In a microwave oven. You can safely thaw food in a microwave oven if the food will be cooked immediately.

The reason defrosting in water is preferable is because water conducts heat better than air. And the faster food is thawed, the better it tastes.

Answer:

The reason defrosting in water is preferable is because water conducts heat better than air. And the faster food is thawed, the better it tastes.

Analysing the question:

We are given that the cat chases a rat on a 1m high table, this means that the height of the cat before falling is 1 m

The horizontal velocity of the cat after sliding off the table will not change in mid-air since there is no force to oppose it

We are given:

height of the cat before falling (h) = 1 m

Distance covered by cat after falling off the edge = 2.3 m

acceleration due to gravity (a) = 9.81 m/s²

initial vertical velocity of the cat (u) = 0 m/s

Since the cat will fall at a constant horizontal velocity, the time taken by the cat to reach the ground times the constant speed of the cat will be equal to the horizontal distance covered by the cat  

[ since time taken to reach the ground = time for which the constant horizontal velocity will be applied]

The above can simply be written as:

time taken by the cat to reach the ground * horizontal velocity = horizontal distance covered

Time taken by the cat to reach the ground:

From the second equation of motion

s = ut + 1/2*at²

replacing the variables with vertical components

1 = (0)(t) + 1/2 * (9.8)(t²)

1 = 4.9t²

t² = 1  / 4.9

t = 1 / √4.9

t = 0.45 seconds

Cat's horizontal velocity when it slid off the table:

Speed of cat = distance travelled / time taken

here, the distance is the horizontal distance covered by the cat after falling off the table

replacing the variables

Speed of the cat = 2.3 / 0.45 = 5.11 m/s

Answer:

Explanation:

The object starts at 0

It moves to - 2

It stops

It then moves to + 6

+6 - 2 = + 4

The new position is 4.

Answer: A-p_Ex

Explanation: answer

Answer: The answer is B; Electrons move from areas of high to low electric potential

Explanation:

I had this question on a quiz and got it correct.

The required Electrons move from areas of high to low electric potential best describes Jodi’s error. Option B is correct.

Electric current is the flow of electric charge through a material. It is typically expressed in units of amperes (A), which represent the rate at which charge flows through a conductor.

Here,
Jodi's error is in statement 2. Electrons move from areas of high electric potential to areas of low electric potential, not from low to high electric potential. This is because electric potential is a measure of the energy per unit charge, so electrons will naturally flow from areas of higher potential to areas of lower potential to balance out the potential difference.

Thus, the required, Electrons move from areas of high to low electric potential best describes Jodi’s error. Option B is correct.

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Complete question is missing, so i have attached it.

Answer:

A) I = 3.5A

B) I = 4.5 A

C) I = 3.15A.

D)I = 3.25 A

Explanation:

a) When we connect across ab, the resistance from ac and bc is;

20ohms + 10ohms = 30 ohms.

This will now be parallel with the 15 ohms resistor.

Thus, equivalent resistance is;

R = (30 × 15)/(30 + 15)

R = 450/45

R = 10 ohms

Formula for Current is; I = (E/R)

Thus;

I = 35/10

I = 3.5A.

b) similarly, when we connect across bc, resistance from ac and ab is;

15 ohms + 20 ohms = 35 ohms.

This will now be parallel with the 10 ohms resistor.

Thus, equivalent resistance is;

R = (35 × 10)/(35 + 10)

R = 350/45

R = 7.778 ohms.

Current is;

I = 35/7.778

I = 4.5 A.

c) similarly, when we connect across ac, resistance from ab and bc is;

15 ohms + 10 ohms = 25 ohms.

This will now be parallel with the 20 ohms resistor.

Thus, equivalent resistance is;

R = (25 × 20)/(25 + 20)

R = 11.11 ohms.

Current is;

I = 35/11.11

I = 3.15A.

d) since connected across bc, the internal resistance will be added to the circuit resistance, which was seen to be 7.778 ohms in step b above

Thus, equivalent resistance = 7.778 + 3 ohms = 10.778

Current is;

I = 35/10.778

I = 3.25 A

(A) The current when connected across ab is 3.5A

(B) The current when connected across bc is 4.5 A

(C) The current when connected across ac is 3.15A

(D) The current when connected across bc with internal resistance is 3.25 A

(a) When connected across ab, the resistance between ac and bc is;

20Ω + 10Ω = 30Ω

It is parallel with 15Ω resistor.

Thus, equivalent resistance is;

R = (30 × 15)/(30 + 15)

R = 450/45

R = 10Ω

Now current is given by

I = (V/R)

I = 35/10

I = 3.5A

(b) When connected across bc, the resistance between ab and ac is;

15Ω + 20Ω = 35Ω

It is in parallel combination with the 10Ω resistor

Thus, equivalent resistance is;

R = (35 × 10)/(35 + 10)

R = 350/45

R = 7.778Ω

Now the current is:

I = 35/7.778

I = 4.5 A

(c) When connected across ac, the resistance between ab and bc is;

15Ω + 10Ω = 25Ω

it is now parallel with the 20Ω resistor.

Thus, - is;

R = (25 × 20)/(25 + 20)

R = 11.11Ω

Current is;

I = 35/11.11

I = 3.15A

(d) when connected across bc, the internal resistance will be added to the circuit resistance across bc.

So, equivalent resistance = 7.778 + 3 = 10.778Ω

Current is;

I = 35/10.778

I = 3.25 A

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Answer: 3.5 × 10^3

Explanation:

Answer:

Let's see what to do buddy...

Explanation:

_________________________________

[tex]a = \frac{60}{ ({ \frac{1}{5}) }^{2} } = \frac{60}{ \frac{1}{25} } = 60 \times 25 \\ \\ a = 15 \times 4 \times 25 = 15 \times 100 \\ a = 1500 \: \frac{m}{ {s}^{2} } [/tex]

_________________________________

The body start from rest which means :

[tex]v(0) = 0[/tex]

_________________________________

In third second means :

t = 2 -----¢ t = 3 -----¢ ∆t = 1

_________________________________

We have this equation to find the distance.

[tex] ∆x = \frac{1}{2} \: a \: {t}^{2} + v(0) \: t \\ [/tex]

_________________________________

[tex]∆x = \frac{1}{2} \times 1500 \times 1 + 0 \times 1 \\ \\ ∆x = 750 + 0 = 750[/tex]

_________________________________

And we're done.

Thanks for watching buddy good luck.

♥️♥️♥️♥️♥️

Answer:

(a) 9.52

(b) 3.81

Explanation:

Draw a free body diagram of block m.  There are two forces:

Weight mg pulling down, and tension T₁ pulling up.

Draw a free body diagram of block M.  There are two forces:

Weight Mg pulling down, and tension T₂ pulling up.

Draw a free body diagram of the double pulley.  There are two torques:

Tension T₂ pulling counterclockwise at r, and tension T₁ pulling clockwise at R.

Let's assume that M moves down, m moves up, and the pulley rotates counterclockwise.

Sum of forces on block m in the +y direction:

∑F = ma

T₁ − mg = ma₁

T₁ = mg + ma₁

Sum of forces on block M in the -y direction:

∑F = ma

Mg − T₂ = Ma₂

T₂ = Mg − Ma₂

Sum of the torques about the pulley's center in the counterclockwise direction:

∑τ = Iα

T₂ r − T₁ R = (½ m₀ r² + ½ m₀ R²) α

T₂ r − T₁ R = ½ m₀ (r² + R²) α

Substitute:

(Mg − Ma₂) r − (mg + ma₁) R = ½ m₀ (r² + R²) α

The strings don't slip on the pulley, so a₁ = αR and a₂ = αr.

(Mg − Mαr) r − (mg + mαR) R = ½ m₀ (r² + R²) α

Mgr − Mαr² − mgR − mαR² = ½ m₀ (r² + R²) α

g (Mr − mR) − α (Mr² + mR²) = ½ m₀ (r² + R²) α

g (Mr − mR) = [½ m₀ (r² + R²) + Mr² + mR²] α

α = g (Mr − mR) / [½ m₀ (r² + R²) + Mr² + mR²]

α = (10) (9×0.2 − 1×0.5) / (½ (0.5) (0.2² + 0.5²) + 9(0.2)² + 1(0.5)²]

α = 19.0 rad/s²

Therefore:

a₁ = αR = (19.0 rad/s²) (0.5 m) = 9.52 m/s²

a₂ = αr = (19.0 rad/s²) (0.2 m) = 3.81 m/s²

Answer:

this question requires us to use our knowledge of newton's 1st 2nd and 3rd laws to fill in the blank spaces. the words in bold are the answers to the question

Explanation:

1. You push a box horizontally along level ground but friction is causing the box to slow down. The magnitude of the friction force from the ground on the box is greater than the magnitude of your force because of newtons second law

2. You push a box horizontally along level ground at constant velocity. The magnitude of the friction force from the ground on the box is equal to the magnitude of your force because of newtons second law

3. You are standing on a hill where the ground is sloped. The magnitude of the normal force from the ground on you is equal to  the magnitude of the normal force from you on the ground because of newtons third law.

4. You are standing on a hill where the ground is sloped. The magnitude of the normal force from the ground on you is less than the magnitude of your weight because of newtons second law

5. You are facing your (very large and very strong) friend, Commander Worf. Worf gets very angry and pushes you to the ground. Worf remains stationary (a = 0). The magnitude of the force from Worf on you is equal to the magnitude of the force from you on Worf because of newtons third law

6. You are standing at a place where the ground is level. The magnitude of your weight is equal to the magnitude of the normal force from the ground on you newtons second law

Answer:

It would be B. The Rope

Explanation:

I say this because the rope is transferring energy from one location to another. Now, I could be totally wrong on this but I think this is right lol.

Answer:

The answer is B the rope.

Hope this helps <3 ;)

Answer:

Ice hockeyyyyyy

Explanation:

Answer:

Using pythagoras theorem

Answer:

use the pythagorean theorem

Explanation: