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alexira [117]
3 years ago
9

A student shoots a spitball with a perfectly horizontal velocity of 9.7 m/s from a height of 1.8 meters. How long will it take f

or the spitball to hit the ground?
(ignore air resistance) (include units and correct number of significant figures)
Physics
1 answer:
sergejj [24]3 years ago
3 0

Answer:A student shoots a spitball with a perfectly horizontal velocity of 9.7 m/s from a height of 1.8 meters. How long will it take for the spitball to hit the ground?

(ignore air resistance) (include units and correct number of significant figures)

Explanation:La respuesta es porque esa es la respuesta, la respuesta al número es 9.7 1.8 Divide =53.888

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Determine the kinetic energy of a 2000 kg roller coaster car that is moving at the speed of 10 ms
kondaur [170]

Answer:

\boxed {\boxed {\sf 100,000 \ Joules}}

Explanation:

Kinetic energy is energy due to motion. The formula is half the product of mass and velocity squared.

E_k= \frac{1}{2} mv^2

The mass of the roller coaster car is 2000 kilograms and the car is moving 10 meters per second.

  • m= 2000 kg
  • s= 10 m/s

Substitute these values into the formula.

E_k= \frac{1}{2} (2000 \ kg ) \times (10 \ m/s)^2

Solve the exponent.

  • (10 m/s)²= 10 m/s * 10 m/s= 100 m²/s²

E_k= \frac{1}{2} (2000 \ kg ) \times (100 \ m^2/s^2)

Multiply the first two numbers together.

E_k= 1000 \ kg  \times (100 \ m^2/s^2)

Multiply again.

E_k= 100,000 \ kg*m^2/s^2

  • 1 kilogram square meter per square second is equal to 1 Joule.
  • Our answer of 100,000 kg*m²/s² is equal to 100,000 Joules.

E_k= 100,000 \ J

The roller coaster car has <u>100,000 Joules</u> of kinetic energy.

3 0
3 years ago
A sled is pulled up to the top of a hill. At thetop of the hill the sled is released from rest and allowed to coastdown the hill
Stolb23 [73]

Answer:

a.) the speed at the bottom is greater for the steeperhill

Explanation:

since the energy at the bottom of the steeper hilis greater

mgh =\frac{1}{mv^2}

As we can see from above that v is higher when h ishigher.

8 0
2 years ago
A student in the Biomechanics class has decided that she would like to make her arms
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Answer:

what is heat and transfer

7 0
2 years ago
A force of 10 N acts on an object with a mass of 10 kg. What is its acceleration? A. 1 m/s2 B. 10 m/s2 C. 2 m/s2 D. 5 m/s2
Akimi4 [234]

The acceleration exerted by the object of mass 10 kg is \mathbf{1} m / \boldsymbol{s}^{2}

Answer: Option A

<u>Explanation:</u>

According to Newton’s second law of motion, any external force acting on a body will be directly proportional to the mass of the body as well as acceleration exerted by the body. So, the net external force acting on any object will be equal to the product of mass of the object with acceleration exerted by the object. Thus,

                  Force = Mass \times Acceleration

So,

                 Acceleration=\frac{\text {Force}}{\text {Mass}}

As the force acting on the object is stated as 10 N and the mass of the object is given as 10 kg, then the acceleration will be

                  Acceleration =\frac{10 \mathrm{N}}{10 \mathrm{kg}}=1 \mathrm{m} / \mathrm{s}^{2}

So, the acceleration exerted by the object of mass 10 kg is \mathbf{1} m / \boldsymbol{s}^{2}

4 0
3 years ago
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
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