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kkurt [141]
1 year ago
12

What is the kinetic energy of a 2.0 kg object moving at 5 m/s?

Physics
2 answers:
Zolol [24]1 year ago
3 0

Let's put the given values into the formula below and get the result;

  • K.E=\frac{1}{2}mv^2

We know the numerical value of speed. We also know the mass. We can jump straight to the conclusion.

  • K.E=\frac{1}{2}(2kg)(5m/s)^2
  • =\frac{1}{2}(2kg)(25m^2/s^2)
  • =25m^2/s^2=25J
alexira [117]1 year ago
3 0

The Kinetic Energy is 25J.

If we want to accelerate an object, we must apply force on it, after applying Force some work has to be done in which energy should be transferred to the object. The energy is known as kinetic energy. The kinetic energy always depends on the mass and the velocity. It is denoted as K and the Si unit is Joules (J).

To calculate the Kinetic Energy,

       

           K= 1/2 mv^2

m = mass  

v = velocity

K = kinetic energy

In solving the above equation,

K = 1/2 x2 x 5^2

K = 5x5

K = 25

 

The Kinetic Energy is 25J.

To learn about Kinetic Energy:

brainly.com/question/26472013

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Which event causes tides
Reptile [31]

Answer:

High tides and low tides are caused by the Moon.

Explanation:

The Moon's gravitational pull generates something called the tidal force.

3 0
3 years ago
Read 2 more answers
A trumpet creates a sound wave that has a wave speed of 350m/s and a wavelength of 0.8 m . What is the frequency of the sound wa
ValentinkaMS [17]

Answer:

The frequency of sound wave created by trumpet is 437.5Hz

Explanation:

Given

the speed of sound wave = 350 ms^{-1}

the wavelength of sound wave = 0.8 m

the frequency of sound wave = ?

All the waves have same relationship among wavelength, frequency and speed, which is given by the equation:

v = fλ, where

v is speed of the wave

f is frequency of the wave

λ is wavelength of the wave

therefore frequency of sound wave is given by

f = v/λ

 = 350ms^{-1}/0.8m

 = 437.5s^{-1}

 = 437.5Hz (since 1 s^{-1} = 1 Hz (Hertz)

Hence the frequency of sound wave created by trumpet is 437.5Hz

7 0
3 years ago
Why are strong forces able to hold atomic nuclei together?
Galina-37 [17]
The nucleons(protons and neutrons) are held together by means of this strong force. If this strong never existed, all the nucleus will blow themselves due to strong repulsive force between protons(neutron has no charge).
Thats it!
If I explain beyond, it will surely bounce off your head. Anyways, if you wanna know more bout it, ping me. (:
4 0
3 years ago
What is the power of a machine that pushes with a force of 3 n for a distance of 9 m in 8 s?
dalvyx [7]
The work done by the machine is equal to the product between the force applied and the distance over which the force is applieds, so in this case:
W=Fd=(3 N)(9 m)=27 J

And the power of the machine is equal to the ratio between the work done by the machine and the time taken:
P= \frac{W}{t}= \frac{27 J}{8 s}=3.38 W

4 0
3 years ago
Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.
Artist 52 [7]

Answer:

a) C.M =(\bar x, \bar y)=(0.767,0.7)m

b) (x_4,y_4)=(-1.917,-1.75)m

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}

\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}

Where M represent the sum of all the masses on the system.

And the center of mass C.M =(\bar x, \bar y)

Part a

m_1= 3 kg, m_2=5kg,m_3=7kg represent the masses.

(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5) represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m

C.M =(\bar x, \bar y)=(0.767,0.7)m

Part b

For this case we have an additional mass m_4=6kg and we know that the resulting new center of mass it at the origin C.M =(\bar x, \bar y)=(0,0)m and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m

If we solve for a we got:

(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0

a=-\frac{(5kg*2.3m)}{6kg}=-1.917m

\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m

(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0

And solving for b we got:

b=-\frac{(7kg*1.5m)}{6kg}=-1.75m

So the coordinates for this new particle are:

(x_4,y_4)=(-1.917,-1.75)m

5 0
3 years ago
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