Answer:
Explanation:
First we need to determine the distance covered during deceleration. According to the equation of motion.
S = ut+1/2at²
Given:
u = 20m/s
t = 0.50s
a = -10m/s (deceleration is negative acceleration)
S = 20²+1/2(-10)(0.5)²
S = 400-5(0.5)²
S = 400-5(0.25)
S = 400-1.25
S = 398.75m
If the deer steps onto the road 35m in front of you, the distance between you and the deer when you come to a stop will be 398.75-35 = 363.75m
Answer:
0.5 , 54.5
Explanation:
for acceleration we should derivate the equation 2 times
x=3t³+t²/4
v=9t²+t/2
a=18t+1/2
a(0)=0.5
a(3)=54.5
|Momentum| = (mass) x (speed)
225 kg-m/s =(50kg) x (speed)
Divide each side by (50kg): Speed=(225 kg-m/s) / (50 kg) = 4.5 m/s .
Regarding the velocity, nothing can be said other than the speed, because
we have no information regarding the direction of the object's motion.
Answer:
<h2>602.08 N</h2>
Explanation:
The force supplied by the train can be found by using the formula
w is the workdone
d is the distance
From the question we have
We have the final answer as
<h3>602.08 N</h3>
Hope this helps you
Answer:
The magnitude of the force is 34.59 N.
Explanation:
Given that,
Inside pressure 
Area 
Outside pressure = 1 atm
We need to calculate the magnitude of the force
Using formula of force
Where, 
=inside Pressure
=outside Pressure
A = area
Put the value into the formula
Hence, The magnitude of the force is 34.59 N.
