Forces are needed to lift, turn, move, open, close, push, pull, and so on. When you throw a ball, you are using force to make the ball move through the air. More than one force can act on an object at the same time.
For the same reason that you can skate around a curve at constant speed but not with constant velocity.
The DIRECTION you're going is part of your velocity, but it's not part of your speed.
If the DIRECTION changes, that's a change of velocity.
The object doesn't have to change speed to have a different velocity. A change of direction is enough to do it.
And any change of velocity is called acceleration.
Answer:
Explanation:
1) acceleration is the change in velocity of a body with respect to time.
acceleration = velocity/time
Given
velocity = 139m/s
time = 20secs
acceleration = 139/20
acceleration = 6.95m/s²
Hence its average acceleration during the first 20 seconds of the launch is 6.95m/s²
2) Speed is the rate of change of distance with respect to time.
average speed = distance/time
Time = distance/speed
Time = 100/7823
Time = 0.013s
3) Using the equation of motion v² = u²+2as
v² = 0²+2(15)(60)
v² = 30*60
v² =1800
v = √1800
v = 42.43m/s
Answer:
A) M
Explanation:
The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:
Box with mass M

Box with mass 2M

Box with mass 3M

On the third equation, acceleration can be modelled in terms of F'':

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.



Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:




Afterwards, F' as function of the external force can be obtained by direct substitution:

The net forces of each block are now calculated:
Box with mass M


Box with mass 2M


Box with mass 3M

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.
Answer:
4,800?
First I subtracted 8 from 24 because there are 24 hours in a day ad she spends 8 of those hours sleeping. I then took the answer which was 16, and multiplied that by 60 to find out how many minutes were in 16 hours and I got 960. I then multiplied 960 by 5 because she winks 5 times each minute. This left me with the answer 4,800.