Explanation:
A centripetal force (from Latin centrum, "center" and petere, "to seek") is a force that makes a body follow a curved path. (not sure but hope this helps )
Answer:
according to the method of their formation
Explanation:
Answer:
11250 N
Explanation:
From the question given above, the following data were obtained:
Normal force (R) = 15000 N
Coefficient of static friction (μ) = 0.75
Frictional force (F) =?
Friction and normal force are related by the following equation:
F = μR
Where:
F is the frictional force.
μ is the coefficient of static friction.
R is the normal force.
With the above formula, we can calculate the frictional force acting on the car as follow:
Normal force (R) = 15000 N
Coefficient of static friction (μ) = 0.75
Frictional force (F) =?
F = μR
F = 0.75 × 15000
F = 11250 N
Therefore, the frictional force acting on the car is 11250 N
The troposphere<span> is the first layer above the surface and contains half of the Earth's atmosphere. </span>Weather<span> occurs in this layer. </span><span> Many jet aircrafts fly in the </span>stratosphere<span> because it is very stable. Also, the ozone layer absorbs harmful rays from the Sun.</span><span> Meteors or rock fragments burn up in the </span>mesosphere.<span>The </span>thermosphere<span> is a layer with auroras. It is also where the space shuttle orbits. </span><span>The atmosphere merges into space in the extremely thin </span>exosphere<span>. This is the upper limit of our atmosphere.
I'm not sure if this helps you but I would say most of the changes happen in the troposphere because that is where the weather occurs. Sorry if this isn't the answer you were looking for.
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Answer:
Vi = 8.28 m/s
Explanation:
This problem is related to the projectile motion.
As we know there are two components of motion associated with this, the horizontal component and vertical component.
The horizontal distance covered by the ball is
Vx*t = x
Vx*t = 5.3
Vx = 5.3/t eq. 1
Also we know that
Vx = Vicos(60)
Vx = Vi*0.5 eq. 2
equate eq. 1 and eq. 2
5.3/t = Vi*0.5
5.3/0.5 = Vi*t
Vi*t = 10.6 eq. 3
The vertical distance is
Vy = y1 + Vyi*t - 0.5gt²
also we know that
Vyi = Visin(60)
Vyi = Vi*0.866
It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance
3 = 1.9 + Vi*0.866*t - 0.5gt²
3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
1.1 = 0.866(Vi*t) - 4.9t²
0.866(Vi*t) = 4.9t² + 1.1
substitute Vi*t = 10.6 in above equation
0.866(10.6) = 4.9t² + 1.1
9.18 = 4.9t² + 1.1
4.9t² = 8.08
t² = 8.08/4.9
t² = 1.648
t = 1.28 sec
Finally, initial speed can be found by substituting the value of t into eq. 3
Vi*t = 10.6
Vi = 10.6/t
Vi = 10.6/1.28
Vi = 8.28 m/s