Answer:
The answer to the question is
The object would fall 57.625 m in the first 5 seconds
Explanation:
To solve the question, we note that
the height of fall = 490 ft = 149.352 m
Time to touch the ground = 7 seconds
We are required to find out how far the object falls in the first 5 seconds
We apply the relation
S = u·t + 0.5×g·t ² = We then have
149.352 = U×7+0.5*9.81*49 From where u = -13 m/s
Therefore to find how far it falls in the first 5 seconds, we have
-13*5 + 0.5*9.81*25 = 57.625 m
Answer:
9.01amp
Explanation:
Power = V^2/R
Given that v = 11volts, P = 99watts
99 = 11^2/R
11×11 = 99R
121= 99R
R = 121/99
R= 1.22ohms
From ohms Law; V = IR
11volts = I × 1.22ohms
I = 11/1.23
I = 9.01 amp
Answer:3.51
Explanation:
Given
Coefficient of Friction 
Consider a small element at an angle \theta having an angle of 
Normal Force

Friction 

and 







Answer:
v₃ = 3.33 [m/s]
Explanation:
This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.
In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.

where:
m₁ = mass of the car = 1000 [kg]
v₁ = velocity of the car = 10 [m/s]
m₂ = mass of the truck = 2000 [kg]
v₂ = velocity of the truck = 0 (stationary)
v₃ = velocity of the two vehicles after the collision [m/s].
Now replacing:
![(1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s]](https://tex.z-dn.net/?f=%281000%2A10%29%2B%282000%2A0%29%3D%281000%2B2000%29%2Av_%7B3%7D%5C%5Cv_%7B3%7D%3D3.33%5Bm%2Fs%5D)