Answer:
Circuit 1 and Circuit 3
Explanation:
If there is 1 Circuit and it runs out of energy, there's another one to produce the power.
I'm guessing that you mean like this:
-- The ruler is held with zero at the bottom, and the centimeter markings
increase as you go up the ruler.
-- You place your fingers with the ruler and the zero mark between them.
-- The number where you catch the ruler is the distance it has fallen.
Then, all we have to find is the time it takes for the ruler to fall 11.3 cm .
Here's the formula for the distance an object falls from rest
in a certain time:
Distance = (1/2) (gravity) (time)²
On Earth, the acceleration of gravity is 9.8 m/s².
So we can write ...
11.2 cm = (1/2) (9.8 m/s²) (time)²
or
0.112 meter = (4.9 m/s²) (time)²
Divide each side
by 4.9 m/s² : (0.112 m) / (4.9 m/s²) = time²
(0.112 / 4.9) sec² = time²
Square root
each side: time = √(0.112/4.9 sec²)
= √ 0.5488 sec²
= 0.74 second (rounded)
Answer:
The ball has an acceleration of -380 m/s², this means the ball slows down
An acceleration of -380 m/s² is the equivalent of 38.736 g's
Explanation:
Step 1: Data given
Velocity of the baseball at time t=0 = 38 m/s
At time t, the ball stops. This means v = 0
time before stops = 0.1s
Step 2: Calculate the acceleration
v= v0+at
with v= the velocity of the ball at time t = 0. v= 0
with v0 = the velocity of the ball at time t=0. v0 = 38 m/s
with a= the acceleration in m/s²
with t = time in seconds
0 = 38 + a*0.1
a = -380 m/s²
The ball has an acceleration of -380 m/s², this means the ball slows down
An acceleration of -380 m/s² is the equivalent of 38.736 g's
Answer:
The answer is A ) High pressure
I hope this helps you :)
please let me know if I am wrong
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.
