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PolarNik [594]
3 years ago
10

A ball is launched horizontally from a height of 172 m above the ground. its initial horizontal velocity is 16m/s. how long does

it take to reach the ground? how far away does it land?​
please answer quickly!!!
Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
8 0

Answer:

172*16 times it and that is your answer

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Find the kinetic energy of a 0.1-kilogram toy truck moving at the speed of 1.1 meters per second.
Viktor [21]

Answer:

0.061 J

Explanation:

The kinetic energy of an object is given by

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

where

m is the mass of the object

v is its speed

For the toy truck in the problem, we have

m = 0.1 kg is its mass

v = 1.1 m/s is its speed

Putting the numbers into the equation, we find

K=\frac{1}{2}(0.1 kg)(1.1 m/s)^2=0.061 J

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A spaceship moving with an initial velocity of 58.0 meters/second experiences a uniform acceleration and attains a final velocit
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The ship's average speed during the 12 seconds is

                         (1/2) (58 + 153)

                     =  (1/2) (211 m/s)

                     =        105.5 m/s .

Traveling for 12 seconds at an average speed of  105.5 m/s ,
the ship covers

               (12 sec) x (105.5 m/s)

           =        1,266 meters  during the acceleration
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4 years ago
Based on the Law of Conservation of Energy, which of the below is true?(1 point)
Veseljchak [2.6K]

Answer:

C

Explanation:

why because if something is conserved, it is constant, and does not change with time. A moving body may change its position, acceleration, and velocity with time, but it's energy is constant. The conversation of energy law states that: In any closed system (isolated system) the total energy of the system remain constant.

Mathematically it is written as

pe + ke = pe + ke

8 0
2 years ago
How is social status directly related to health care?
Inessa [10]

Its A according to Ed Genuity

7 0
4 years ago
Read 2 more answers
Kinematics
leonid [27]

Answer:

a)

a = 2 [m/s^2]

b)

a = 1.6 [m/s^2]

c)

xt = 2100 [m]

Explanation:

In order to solve this problem we must use kinematics equations. But first we must identify what kind of movement is being studied.

a)

When the car moves from rest to 40 [m/s] by 20 [s], it has a uniformly accelerated movement, in this way we can calculate the acceleration by means of the following equation:

v_{f} = v_{i}+(a*t)

where:

Vf = final velocity = 40 [m/s]

Vi = initial velocity = 0 (starting from rest)

a = acceleration [m/s^2]

t = time = 20 [s]

40 = 0 + (a*20)

a = 2 [m/s^2]

The distance can be calculates as follows:

v_{f} ^{2} =  v_{i} ^{2}+(2*a*x)

where:

x1 = distance [m]

40^2 = 0 + (2*2*x1)

x1 = 400 [m]

Now the car maintains its speed of 40 [m/s] for 30 seconds, we must calculate the distance x2 by means of the following equation, it is important to emphasize that this movement is at a constant speed.

v = x2/t2

where:

x2 = distance [m]

t2 = 30 [s]

x2 = 40*30

x2 = 1200 [m]

b)

Immediately after a change of speed occurs, such that the previous final speed becomes the initial speed, the new Final speed corresponds to zero, since the car stops completely.

v_{f} = v_{i}-a*t

Note: the negative sign of the equation means that the car is stopping, i.e. slowing down.

0 = 40 - (a *25)

a = 40/25

a = 1.6 [m/s^2]

The distance can be calculates as follows:

v_{f} ^{2}  = v_{i} ^{2} -2*a*x3\\

0 = (40^2) - (2*1.6*x3)

x3 = 500 [m]

c)

Now we sum all the distances calculated:

xt = x1 + x2 + x3

xt = 400 + 1200 + 500

xt = 2100 [m]

8 0
3 years ago
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