Answer:
12.495m/s
Explanation:
Horizontal displacement is the range of the projectile motion.
The range is expressed as;
R = 2U/g
U is the speed at which the rock is thrown (initial speed)
g is the acceleration due to gravity.
Given
R = 255cm = 2.55m
g = 9.8m/s²
Required
Speed U
Substitute the given parameters into the formula as shown;
2.55 = 2U/9.8
Cross multiply
2U = 2.55×9.8
2U = 24.99
U = 24.99/2
U = 12.495m/s
Hence the speed that you thew the rock is 12.495m/s
Answer:0.27
Explanation:
Given
One worker Pushes with force 
other Pulls it with a rope of rope 
mass of crate 
both forces are horizontal and crate slides with a constant speed
Both forces are in the same direction so Friction will oppose the forces and will be equal in magnitude of sum of two forces because crate is moving with constant speed i.e. net force is zero on it

where
is the friction force



where
is the coefficient of static friction



Answer:
M
Explanation:
To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after
As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved
⇒
-
is the moment of interia of earth before impact -
is the angular velocity of earth about an axis passing through the center of earth before impact
is moment of interia of earth and asteroid system
is the angular velocity of earth and asteroid system about the same axis
let 
since 

⇒ if time period is to increase by 25%, which is
times, the angular velocity decreases 25% which is
times
therefore

(moment of inertia of solid sphere)
where M is mass of earth
R is radius of earth

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)
where
is mass of asteroid
⇒ 

=
+ 

⇒

The vertical weight carried by the builder at the rear end is F = 308.1 N
<h3>Calculations and Parameters</h3>
Given that:
The weight is carried up along the plane in rotational equilibrium condition
The torque equilibrium condition can be used to solve
We can note that the torque due to the force of the rear person about the position of the front person = Torque due to the weight of the block about the position of the front person
This would lead to:
F(W*cosθ) = mgsinθ(L/2) + mgcosθ(W/2)
F(1cos20)= 197/2(3.10sin20 + 2 cos 20)
Fcos20= 289.55
F= 308.1N
Read more about vertical weight here:
brainly.com/question/15244771
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