# A 10 kg block moving at 10 m/s in a direction 45 degrees above the horizontal. When it has fallen to a point that is 10 m below the initial point measured vertically (without air friction), the block's kinetic energy is closest to

The block's kinetic energy is closest to 1500 Joules.

### Kinetic energy :

The energy is always conserved.

So that, the total kinetic energy will be sum of initial potential energy and kinetic energy during falling.

Given that, mass(m)=10kg, v=10m/s, h=10m,g=10m/s^2

K.E=(1/2)mv^2 + mgh

K.E=(1/2)*10*100 + (10*10*10)

K.E=500 + 1000=1500Joule

The  block's kinetic energy is closest to 1500 Joules.

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Kinetic energy = 1500 J

Explanation:

The computation of the block's kinetic energy is shown below:

As we know that

Conservation of energy is

PE_i + KE_i = PE_f + KE_f

where,

Initial Potential energy = PE_i = m gh = 10kg× 10m/s^2 × 10m = 1000 J

Initial Kinetic energy = KE_i = (0.5) m V^2 = (0.5) (10 kg) (10 m/s)^2 = 500 J

Final potential energy = PE_f = mgh = 0

As h = 0 which is at reference line

So

PE_i + KE_i = PE_f + KE_f

Now put these valeus to the above formulas

1000 J + 500 J = 0 + KE_f

After solving this

Kinetic energy = 1500 J

## Related Questions

John and Linda are arguing about the definition of density. John says the density of an object is proportional to itsmass. Linda says the object's mass is proportional to its density and to its volume. Which one, if either, is correct?A. They are both wrong
B. John is correct, but Linda is wrong
C. John is wrong, but Linda is correct
D. They are both correct.
E. John must be wrong, because Linda always wins these arguments.

They are both correct.

Explanation:

The density of an object is defined as the ratio of its mass to its volume. This implies that the density of the object is both proportional to the mass and also to the volume of the object. John only mentioned mass which is correct. Linda mentioned the second variable on which density depends which is the volume of the object.

Hence considering the both statements objectively, one can say that they are both correct.

Two particles are traveling through space. At time t the first particle is at the point (−1 + t, 4 − t, −1 + 2t) and the second particle is at (−7 + 2t, −6 + 2t, −1 + t). (a) (5 Points) Do the paths of the two particles cross? If so, where?

Yes, the paths of the two particles cross.

Location of path intersection = ( 1 , 2 , 3)

Explanation:

In order to find the point of intersection, we need to set both locations equal to one another. It should be noted however, that the time for each particle can vary as we are finding the point where the paths meet, not the point where the particles meet themselves.

So, we can name the time of the first particle ,  and the time of the second particle .

Setting the locations equal, we get the following equations to solve for and :

Equation 1

Equation 2

Equation 3

Solving these three equations simultaneously we get:

2 seconds

4 seconds

Since, we have an answer for when the trajectories cross, we know for a fact that they indeed do cross.

The point of crossing can be found by using the value of or in the location matrices. Doing this for the first particle we get:

Location of path intersection = ( -1 + 2 , 4 - 2 , -1 + 2(2) )

Location of path intersection = ( 1 , 2 , 3)

While David was riding his bike around the circular cul-de-sac by his house, he wondered if the constant circular motion was having any effect on his tires. What would be the best way for David to investigate this?A.
Measure the circumference of the tire before and after riding.
B.
Measure the total distance traveled on his bike and divide this by how long it took him.
C.
Measure the wear on his treads before and after riding a certain number of laps.
D.
Time how long it takes him to ride 5 laps around his cul-de-sac.

C.

Measure the wear on his treads before and after riding a certain number of laps.

Measure the wear on his treads before and after riding a certain number of laps.

Explanation:

By riding in a circular motion the inside of the tire will be in contact with the road more than the outside of the tire. Thus, to see if the constant circular motion had any effect on his tires David should measure the tread depth on both the inside and the outside of the tires before the experiment and measure the inside and the outside of the tires (at the same location on the tires) after the experiment. Then he can compare the tread loss on the inside of the tire to the tread loss on the outside of the tire.

A shot-putter exerts an unbalanced force of 128 N on a shot giving it an acceleration of 19m/s2. What is the mass of the shot?

128 is the ans cuz N is also lnown as mass

Explanation:

128

A large convex lens stands on the floor. The lens is 180 cm tall, so the principal axis is 90 cm above the floor. A student holds a flashlight 120 cm off the ground, shining straight ahead (parallel to the floor) and passing through the lens. The light is bent and intersects the principal axis 60 cm behind the lens. Then the student moves the flashlight 30 cm higher (now 150 cm off the ground), also shining straight ahead through the lens. How far away from the lens will the light intersect the principal axis now?A. 30 cm
B. 60 cm
C. 75 cm
D. 90 cm

B. 60 cm

All parallel light rays are bent through the focal point of a convex lens, so the rays from the flashlight 150 cm above the floor must go through the same point on the principal axis as the rays from the flashlight 120 cm above the floor. The location of the focal point does not change when the position of the object is moved either vertically or horizontally.

A mass of 0.14 kg is attached to a spring and set into oscillation on a horizontal frictionless surface. The simple harmonic motion of the mass is described by x(t) = (0.28 m)cos[(8 rad/s)t]. Determine the following. (a) amplitude of oscillation for the oscillating mass .

The amplitude of oscillation for the oscillating mass is 0.28 m.

Explanation:

Given that,

Mass = 0.14 kg

Equation of simple harmonic motion

....(I)

We need to calculate the amplitude

Using general equation of simple harmonic equation

Compare the equation (I) from general equation

The amplitude is 0.28 m.

Hence, The amplitude of oscillation for the oscillating mass is 0.28 m.