1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
4vir4ik [10]
3 years ago
5

HELP!!

Physics
1 answer:
Otrada [13]3 years ago
3 0

Answer:

22 N

Explanation:

The box is moving, so there is kinetic friction.

The speed is constant, so there is no acceleration.

Draw a free body diagram of the box.  There are four forces.

Normal force Fn pushing up.

Weigh force mg pulling down.

Applied force F pushing right.

Friction force Fn μk pushing left.

Sum the forces in the y direction:

∑F = ma

Fn − mg = 0

Fn = mg

Sum the forces in the x direction:

∑F = ma

F − Fn μk = 0

F = Fn μk

F = mg μk

F = (11.2 kg) (9.8 m/s²) (0.200)

F = 22.0 N

You might be interested in
The coefficient of friction on a surface is 0.25. A box requires 100N to slide it across the surface. What is the weight of the
Novosadov [1.4K]
So we know coefficient of f times normal force is friction. So do 100= .25 times x. Now solve for x. You get 400. So 400 is the normal force. And we know normal force equals weight in these types of problems so the answer is 400
5 0
3 years ago
22 points !
nordsb [41]
Evaporation. The glass of water was less full after 10 minuets because the water got evaporated




7 0
3 years ago
a physics student throws a stone horizontally off a cliff. one second later, he throws a second identical stone in exactly the s
Virty [35]

The second stone hits the ground exactly one second after the first.

The distance traveled by each stone down the cliff is calculated using second kinematic equation;

h = v_0_yt + \frac{1}{2} gt^2

where;

  • <em>t is the time of motion </em>
  • <em />v_0_y<em> is the initial vertical velocity of the stone = 0</em>

h = \frac{1}{2} gt^2

The time taken by the first stone to hit the ground is calculated as;

t_1 = \sqrt{\frac{2h}{g} }

When compared to the first stone, the time taken by the second stone to hit the ground after 1 second it was released is calculated as

t_2 = \sqrt{\frac{2h}{g} } + 1

t_2 = t_1 + 1

Thus, we can conclude that the second stone hits the ground exactly one second after the first.

"<em>Your question is not complete, it seems be missing the following information;"</em>

A. The second stone hits the ground exactly one second after the first.

B. The second stone hits the ground less than one second after the first

C. The second stone hits the ground more than one second after the first.

D. The second stone hits the ground at the same time as the first.

Learn more here:brainly.com/question/16793944

8 0
3 years ago
Two thin concentric spherical shells of radii r1 and r2 (r1 &lt; r2) contain uniform surface charge densities V1 and V2, respect
Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a)  0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 \pi r^{2}  = \frac{Q1}{E_{0} }

So,

Rearranging the above equation to get Electric field, we will get:

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   }

Multiply and divide by r1^{2}

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } x \frac{r1^{2} }{r1^{2} }

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x r1^{2}) /(E_{0} x r^{2})

c) r > r2 :

Electric Field = ?

E x 4 \pi r^{2}  = \frac{Q1 + Q2}{E_{0} }

Rearranging the above equation for E:

E = \frac{Q1+Q2}{E_{0} . 4 \pi. r^{2}   }

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

As we know from above, that:

\frac{Q1}{E_{0} . 4 \pi. r^{2}   } =  (σ1 x r1^{2}) /(E_{0} x r^{2})

Then, Similarly,

\frac{Q2}{E_{0} . 4 \pi. r^{2}   } = (σ2 x r2^{2}) /(E_{0} x r^{2})

So,

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

Replacing the above equations to get E:

E = (σ1 x r1^{2}) /(E_{0} x r^{2}) + (σ2 x r2^{2}) /(E_{0} x r^{2})

Now, for

d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

σ1 x r1^{2} = - σ2 x r2^{2}

4 0
3 years ago
What is the <br>newton's first law of motion​
koban [17]

Answer:

Anybody which is in state of rest ,will be in rest if we don't apply any external force ...

8 0
3 years ago
Read 2 more answers
Other questions:
  • How do I find the net force=mass• acceleration with 9kg and 3 acceleration??
    9·2 answers
  • What is the name for a quantity that has both magnitude and direction
    5·1 answer
  • A pile driver lifts a 450 kg weight and then lets it fall onto the end of a steel pipe that needs to be driven into the ground.
    12·2 answers
  • A bigger pushing force does make the brick slide across the table.write down one thing that the sliding brick will do to the sur
    7·2 answers
  • Describe the energy transformation that occurs in a digital clock.
    8·2 answers
  • Circle the letter of the sentence that tells how Bohr' model of the atom differed from Rutherford's model A. Bohr's model focuse
    6·1 answer
  • Cuantos CM son:<br><br>8 newtons
    7·1 answer
  • Which statement does not describe a characteristic of a good outline?
    15·2 answers
  • Give an example of an object that has balanced forces acting on it.
    12·2 answers
  • If you were standing on the edge of a thunderstorm as it begins to form, would the wind be blowing into the storm or out of it?
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!