Answer:
The pressure exerted by the brick on the table is 18,933.3 N/m².
Explanation:
Given;
height of the brick, h = 0.1 m
density of the brick, ρ = 19,300 kg/m³
acceleration due to gravity, g = 9.81 m/s²
The pressure exerted by the brick on the table is calculated as;
P = ρgh
P = (19,300)(9.81)(0.1)
P = 18,933.3 N/m²
Therefore, the pressure exerted by the brick on the table is 18,933.3 N/m².
We have that the maximum height reached by the basketball from its release point is

From the question we are told
- A basketball is tossed upwards with a speed of 5.0 m/s. We can ignore air resistance.
- What is the maximum height reached by the basketball from its release point?
Generally the Newtons equation for Motion is mathematically given as


Therefore
The maximum height reached by the basketball from its release point is

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Answer:2.4m
Explanation:
Velocity=360m/s
Frequency=150Hz
Wavelength=velocity ➗ frequency
Wavelength=360 ➗ 150
Wavelength=2.4m
A guessing answer the best answer but you have had to subtract the answer by the equation that it was giving u
Answer:
x = 11.23 m
Explanation:
For this interesting exercise, we must use angular kinematics, linear kinematics and the relationship between angular and linear quantities.
Let's reduce to SI system units
θ = 155 rev (2pi rad / rev) = 310π rad
α = 2.00rev / s2 (2pi rad / 1 rev) = 4π rad / s²
Let's look for the angular velocity at the time the piece is released, with starting from rest the initial angular velocity is zero (wo = 0)
w² = w₀² + 2 α θ
w =√ 2 α θ
w = √(2 4pi 310pi)
w = 156.45 rad / s
The relationship between angular and linear velocity
v = w r
v = 156.45 0.175
v = 27.38 m / s
In this part we have the linear speed and the height that it travels to reach the floor, so with the projectile launch equations we can find the time it takes to arrive
y =
t - ½ g t²
As it leaves the highest point its speed is horizontal
y = 0 - ½ g t²
t = √ (-2y / g)
t = √ (-2 (-0.820) /9.8)
t = 0.41 s
With this time we calculate the horizontal distance, because the constant horizontal speed
x = vox t
x = 27.38 0.41
x = 11.23 m