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Bess [88]
3 years ago
6

A box of unknown weight is 1 m away from the fulcrum. To balance the sides, a student applies a force of 4 N at a distance of 3

m away from the fulcrum. How much force is the box exerting on the balance beam? In other words, what is the weight of the box?
This is due by 4:10
Physics
1 answer:
Solnce55 [7]3 years ago
3 0

Answer: I think A

Explanation:

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C.
The particles move perpendicular to the direction of the wave.
3 0
3 years ago
Suppose two children push horizontally, but in exactly opposite directions, on a third child in a wagon. The first child exerts
Nata [24]

(a) The system of interest if the acceleration of the child in the wagon is to be calculated are the wagon and the children outside the wagon.

(b) The acceleration of the child-wagon system is 0.33 m/s².

(c) Acceleration of the child-wagon system is zero when the frictional force is 21 N.

<h3>Net force on the third child</h3>

Apply Newton's second law of motion;

∑F = ma

where;

  • ∑F is net force
  • m is mass of the third child
  • a is acceleration of the third child

∑F = 96 N - 75 N - 12 N = 9 N

Thus, the system of interest if the acceleration of the child in the wagon is to be calculated are;

  • the wagon
  • the children outside the wagon

<h3>Free body diagram</h3>

           →                 →              Ф                         ←

         1st child      friction       wagon                2nd child

<h3>Acceleration of the  child and wagon system</h3>

a = ∑F/m

a = 9 N / 27 kg

a = 0.33 m/s²

<h3>When the frictional force is 21 N</h3>

∑F = 96 N - 75 N - 21 N = 0 N

a = ∑F/m

a = 0/27 kg

a = 0 m/s²

Learn more about net force here: brainly.com/question/14361879

#SPJ1

7 0
2 years ago
A professional racecar driver buys a car that can accelerate at 5.9 m/s2. The racer decides to race against another driver in a
3241004551 [841]

Answer:

(a) Time will be t = 3.56 sec

(b) Distance traveled by car when they are side by side is 37.38712 m

(b) Velocity of race car = 21.004 m/sec

velocity of stock car = 12.816 m/sec            

Explanation:

We have given acceleration of the car a_1=5.9m/sec^2

Acceleration of the stock car a_2=3.6m/sec^2

When 1st car overtakes the second car then distance traveled by both the car will be same

(a) So s_1=s_2

As both car starts from rest so initial velocity of both car will be 0 m/sec

It is given that stock car leaves 1 sec before

So \frac{1}{2}\times 5.9\times t^2=\frac{1}{2}\times (t+1)^2\times 3.6

After solving t = 3.56 sec

(b) From second equation of motion s=ut+\frac{1}{2}at^2=0\times 3.56+\frac{1}{2}\times 5.9\times 3.56^2=37.38712m

(c) From first equation pf motion v = u+at

So velocity of race car v = 0+5.9×3.56 = 21.004 m/sec

Velocity of stock car v = 0+ 3.6×3.56 = 12.816 m/sec

3 0
3 years ago
You are trying to find the amount of heat transferred between two substances. In order to do this, you plan
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Answer:

do you have kids ?

Explanation:

i am a 14 year old girl

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2 years ago
Which statement does not describe the relationship between genes and
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Answer: C. Inherited traits carry the instructions for individual genes.

Explanation: bejewels I know stuff. ☆ - ~ hope this helps

4 0
3 years ago
Read 2 more answers
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