The speed of the runner is 300 m /38 seconds. You can simplify this answer to be about 7.9 m/s
Answer:
19.2m/s
Explanation:
Assuming that 2.4m/s^2 was the acceleration and not a typo, we can use the equation v=at, where v=velocity, a=acceleration, and t=time,
plug in known varibles,
v=2.4*8
v=19.2m/s
Answer:
27.1 m/s
Explanation:
Given that at a race car driving event, a staff member notices that the skid marks left by the race car are 9.06 m long. The very experienced staff member knows that the deceleration of a car when skidding is -40.52 m/s2.
Using third equation of motion,
V^2 = U^2 + 2aS
Since the car is decelerating, the final velocity V = 0
Substitute all the parameter into the equation above,
0 = U^2 - 2 * 40.52 * 9.06
U^2 = 734.22
U = 
U = 27.096
U = 27.1 m/s approximately
Therefore, the staff member can estimate for the original speed of the race car to be 27.1 m/s if it came to a stop during the skid
Gravitational force is given by, 
Where, m and M are the masses of the objects, R is the distance between them and G gravitational constant.
Gravitational force of the star on planet 1, 
Gravitational force of the star on planet 2, 
Ratio, 

Therefore, the gravitational force of the star on the planet 1 is three times that on planet 2.
Answer:
P = 4000 [Pa]
Explanation:
Pressure is defined as the relationship between Force and the area where the body rests.
The support area is equal to:
![A=50*20=1000[cm^{2} ]](https://tex.z-dn.net/?f=A%3D50%2A20%3D1000%5Bcm%5E%7B2%7D%20%5D)
But we must convert from square centimeters to square meters.
![1000[cm^{2}]*\frac{1^{2}m^{2} }{100^{2}m^{2} }=0.1[m^{2} ]](https://tex.z-dn.net/?f=1000%5Bcm%5E%7B2%7D%5D%2A%5Cfrac%7B1%5E%7B2%7Dm%5E%7B2%7D%20%20%7D%7B100%5E%7B2%7Dm%5E%7B2%7D%20%20%7D%3D0.1%5Bm%5E%7B2%7D%20%5D)
And the pressure is:
![P=\frac{F}{A} \\P=400/0.1\\P=4000[N/m^{2} ]or 4000[Pa]](https://tex.z-dn.net/?f=P%3D%5Cfrac%7BF%7D%7BA%7D%20%5C%5CP%3D400%2F0.1%5C%5CP%3D4000%5BN%2Fm%5E%7B2%7D%20%5Dor%204000%5BPa%5D)