And object has a velocity of 9 m/s and a mass of 20 kg. What is the kinetic energy of this object

Answers

Answer 1

Answer:

The kinetic energy of the object which is moving with velocity of 9 m/s and has a mass of 20 kg will be 810 joules.

What is Kinetic Energy ?

The energy possessed by the body by virtue of its motion is called kinetic energy. Mathematically -

E[K] = 1/2mv²

Given is an object that has a velocity of 9 m/s and a mass of 20 kg.

We can calculate the kinetic energy of the object by using the formula discussed above.

Mass [m] = 20 Kg

Velocity [v] = 9 m/s

Substituting the values, we get -

E[K] = 1/2 x 20 x 9 x 9

E[K] = 10 x 9 x 9

E[K] = 81 x 10

E[K] = 810 joules

Therefore, the kinetic energy of the object which is moving with velocity of 9 m/s and has a mass of 20 kg will be 810 joules.

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Answer 2

Answer:

Explanation:

k.e =1/2m(v)²

k.e=m/2(v)²

so......k.e=20kg/2(9m/s)²

k.e=10kg(81m²/s²)

k.e=810kgm²/s²

HOPE ITS CORRECT

ANSWER

k.e=810kgm²/s²

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What are three physical of aluminum foil

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Solid, ductile, and malleable. Ductile means it can conduct heat. Malleable means it can be shaped differently.

What is a perfect bounce in physics?? Is it force? High School Physics, not middle school physics.

Answers

Perfect Bounce in Physics:

In the physics or the scientific world, each and every fact that occurs is connected with the forces of gravity. Gravity is the only thing which makes life possible on earth. Due to the impact of energies and laws of motion, the bounce occurs in physics. The collisions and other gravitational pull comes into consideration.

However, the bouncing is a temporary act and gets nullified after sometime. Work and force plays equally important role here. It is only because of this, everything in the universe exists. The movement of objects and its displacement is also associated with this. All the frictions and gravity comes under this.

Beginning:

You see,

You see:

It is the bisector of the photosynthesis if you want the nucleustransformation to let you know transpirtation is present for the bisector of photosynthesis and then the nucleus would get a membrane which allowed the photosynthesis synthesise into photos thatneeds to be taken by photographers

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6. A golf ball is hit a distance of 300 yards in 10 sec. What is the speed of the golf ballo

Answers

The speed of the ball is 27.4 m/s

Explanation:

The speed of an object is given by:

$speed=(d)/(t)$

where

d is the distance covered

t is the time taken

In this problem, we have:

d = 300 yards is the distance covered by the golf ball

t = 10 s is the time taken

Keeping in mind that

1 yard = 0.914 m

We can convert the distance from yards to meters:

$d = 300 \cdot 0.914 = 274.2 m$

And substituting into the equation, we find the speed of the ball:

$speed=(274.2)/(10)=27.4 m/s$

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TRUE OR FALSE: with time and pressure, peat turns into lignite, or brown coal.

Answers

True

I hope it's help !

Hey There!

The statement is true, with time and pressure peat does turn into brown coal.

Have A Brainly Day :)

Modern dual-processing theory is the idea that our minds have —tracks.
O
A. conscious and unconscious
O
B. emotional and rational
O
C. conscious and preconscious
O
D. subconscious and conscious

Answers

Modern dual-processing theory is the idea that our minds have conscious and unconscious tracks. A

Answer:

conscious and unconscious

Explanation:

3. A block of mass m1=1.5 kg on an inclined plane of an angle of 12° is connected by a cord over a mass-less, frictionless pulley to a second block of a mass m2= 1.2 kg hanging vertically. If the hanging block falls 0.92min 1.23 s, what is the coefficient of friction between m1 and the inclined plane?

Please show work

Answers

Answer:

$\mu=0.377$

Explanation:

we need to start by drawing the free body diagram for each of the masses in the system. Please see attached image for reference.

We have identified in green the forces on the blocks due to acceleration of gravity ( $w_1$ and $w_2$ ) which equal the product of the block's mass times "g".

On the second block ( $m_2$ ), there are just two forces acting: the block's weight ( $m_2\,*\,g$ ) and the tension (T) of the string. We know that this block is being accelerated since it has fallen 0.92 m in 1.23 seconds. We can find its acceleration with this information, and then use it to find the value of the string's tension (T). We would need both these values to set the systems of equations for block 1 in order to find the requested coefficient of friction.

To find the acceleration of block 2 (which by the way is the same acceleration that block 1 has since the string doesn't stretch) we use kinematics of an accelerated object, making use of the info on distance it fell (0.92 m) in the given time (1.23 s):

$x_f-x_i=v_i\,t-(1)/(2) a\,t^2$ and assume there was no initial velocity imparted to the block:

$x_f-x_i=v_i\,t-(1)/(2) a\,t^2\n-0.92\,m=0\,-(1)/(2) a\,(1.23)^2\na=(0.92\,*\,2)/(1.23^2) \na=1.216 \,(m)/(s^2)$

Now we use Newton's second law in block 2, stating that the net force in the block equals the block's mass times its acceleration:

$F_(net)=m_2\,a\nw_2-T=m_2\,a\nm_2\,g-T=m_2\,a\nm_2\,g-m_2\,a=T\nm_2\,(g-a)=T\n1.2\,(9.8-1.216)\,N=T\nT=10.3008\,N$

We can round this tension (T) value to 10.3 N to make our calculations easier.

Now, with the info obtained with block 2 (a - 1.216 $(m)/(s^2)$ , and T = 10.3 N), we can set Newton's second law equations for block 1.

To make our study easier, we study forces in a coordinate system with the x-axis parallel to the inclined plane, and the y-axis perpendicular to it. This way, the motion in the y axis is driven by the y-component of mass' 1 weight (weight1 times cos(12) -represented with a thin grey trace in the image) and the normal force (n picture in blue in the image) exerted by the plane on the block. We know there is no acceleration or movement of the block in this direction (the normal and the x-component of the weight cancel each other out), so we can determine the value of the normal force (n):

$n-m_1\,g\,cos(12^o)=0\nn=m_1\,g\,cos(12^o)\nn=1.5\,*\,9.8\,cos(12^o)\nn=14.38\,N$

Now we can set the more complex Newton's second law for the net force acting on the x-axis for this block. Pointing towards the pulley (direction of the resultant acceleration $a$ ), we have the string's tension (T). Pointing in the opposite direction we have two forces: the force of friction (f ) with the plane, and the x-axis component of the block's weight (weight1 times sin(12)):

$F_(net)=m_1\,a\nT-f-w_1\,sin(12)=m_1\,a\nT-w_1\,sin(12)-m_1\,a=f\nf=[10.3-1.5\,*\,9.8\,sin(12)-1.5\,*1.216]\,N\nf=5.42\,N$

And now, we recall that the force of friction equals the product of the coefficient of friction (our unknown $\mu$ ) times the magnitude of the normal force (14.38 N):

$f=\mu\,n\n5.42\,N=\mu\,*\,14.38\,N\n\mu=(5.42)/(14.38)\n\mu=0.377$

with no units.

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