Answer:8.1 m
Explanation:
Given
ball is launched from height of 3 m
initial velocity
considering the ball is thrown vertically upward
Using
where,
u=initial velocity
v=final Velocity
a=acceleration
s=distance
At maximum height final velocity will be zero
Therefore maximum height w.r.t ground is
The lighter one travel with the speed of 0.574 m/sec. Force is defined as the product of mass and acceleration. Its unit is Newton.
<h3>What is force?</h3>
Force is defined as the push or pull applied to the body. Sometimes it is used to change the shape, size, and direction of the body.
The two skaters push off against each other on frictionless ice, then torque act by the one skater on the other is equal.
P₁₂=P₂₁
F₁₂V₁ =F₂₁V₂
625 ×V₁= 725 N× 1.5 m/s
V₁=0.574 m/sec
Hence, the lighter one travel with the speed of 0.574 m/sec.
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Answer:
1)
2)
Explanation:
<u>Projectile Motion</u>
When an object is launched near the Earth's surface forming an angle with the horizontal plane, it describes a well-known path called a parabola. The only force acting (neglecting the effects of the wind) is the gravity, which acts on the vertical axis.
The heigh of an object can be computed as
Where is the initial height above the ground level, is the vertical component of the initial velocity and t is the time
The y-component of the speed is
1) We'll find the vertical component of the initial speed since we have not enough data to compute the magnitude of
The object will reach the maximum height when . It allows us to compute the time to reach that point
Solving for
Thus, the maximum heigh is
We know this value is 8 meters
Solving for
Replacing the known values
2) We know at t=1.505 sec the ball is above Julie's head, we can compute
Answer:
Explanation:
Let's start by calculating the angular velocity of the Moon. We know that the period is:
So now we can calculate its angular velocity:
The centripetal acceleration is given by
where
is the radius of the orbit
Substituting,
Answer:
Explanation:
a )
Depth of hole from surface of water d = .50 m - .03 m = .47 m
velocity of efflux v = √ 2gd
v = √ (2 x 9.8 x .47 )
v = 3.03 m /s
b )
Volume flow rate = π R² v where R is radius of hole at the bottom .
= 3.14 x ( .005 ) ² x 3.03 m/s
= 2.378 x 10⁻⁴ m³ /s
c )
Volume of water collected in 60 s
= 2.378 x 10⁻⁴ x 60
= 1.4268 x 10⁻² m³
If height attained in collecting container be h
π R² h = 1.4268 x 10⁻² m³ where R is radius of container
3.14 x ( .1 )² x h = 1.4268 x 10⁻²
h = .4544 m .
Pressure at the bottom of container = hρ g
where h is height of water , ρ is density of water
Pressure = .4544 x 1000 x 9.8 N /m²
= 4453.12 N /m²