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Digiron [165]
2 years ago
9

When a 100-N force acts horizontally on an 8.0-kg chair, the chair moves at a constant speed across the level floor. Which state

ment has to be true?
The chair accelerates at 12.5 m/s^2
The applied force is greater than the frictional force.
The friction force is 100 N
There is no friction between the floor and chair
Physics
1 answer:
TiliK225 [7]2 years ago
7 0

Answer:

The applied force is greater than the frictional force.

Explanation:

the chair moves at <u>a constant speed</u><u> </u><u>therefore</u><u>,</u><u> </u><u>the</u><u> </u><u>answer</u><u> </u><u>is</u><u> </u><u>not</u><u> </u><u>A</u><u> </u><u>or</u><u> </u><u>C</u><u>.</u>

if there is no friction then the chair <u>would accelerate and it would not be at a constant speed</u><u>.</u>

hence, the only possible answer is B.

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An ideal spring is hung vertically from the ceiling. The spring constant is k = 125 N/m. A block of mass m = 650 g (1000 g = 1 k
Brrunno [24]

Answer:

0.102 m

Explanation:

k = spring constant of the spring = 125 N/m

m = mass of the block attached to the spring = 650 g = 0.650 kg

x = maximum extension of the spring

h = height dropped by the block = x

Using conservation of energy

Spring potential energy gained = Gravitational potential energy lost

(0.5) k x² = mgh

(0.5) k x² = mgx

(0.5) (125) x = (0.650) (9.8)

x = 0.102 m

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3 years ago
Which of the following is an example of kinetic energy?
tresset_1 [31]
B. Sound, because everything else sits still and sound waves move
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In pea plants Round (R) is a dominant trait and wrinkled nose (r) is a recessive trait. Which combination would result in wrinkl
bixtya [17]
The answer is would be number 3 rr
7 0
2 years ago
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Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long
Ksivusya [100]

Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

Battery moved 10,000 coulombs

current rate = 0.5 mA

<u>A) Determine how long the clock run on the battery. use the relation below</u>

q = i * t ----- ( 1 )

q = charge , i = current , t = time

10000 = 0.5 * 10^-3 * t

hence  t = 2 * 10^7 secs

hence the time = 231.48  days

<u>B) Determine how many electrons per second flowed </u>

q = n*e ------ ( 2 )

n = number of electrons

e = 1.6 * 10^-19

q = 0.5 * 10^-3 coulomb ( charge flowing per electron )

back to equation 2

n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

hence : n = 3.125 * 10^15

8 0
2 years ago
In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,
Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B =\ 1.0\ Kg.

Let mass of ball A and B\ is\ m  

Final velocity of ball A\ is\ v_1

Final velocity of ball B\ is\ v_2

initial velocity of ball A\ is\ u_1

Initial velocity of ball B\ is\ u_2

Momentum after collision =mv_1+mv_2

Momentum before collision = mu_1+mu_2

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, mu_1+mu_2=mv_1+mv_2

Plugging each trial in this equation we get,

First Trial

mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3

momentum before collision \neq moment after collision

Second Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1

moment before collision = moment after collision

Third Trial

mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1

momentum before collision \neq moment after collision

Fourth Trial

mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0

momentum before collision \neq moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

4 0
3 years ago
Read 2 more answers
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