a hippopotamus produces a pressure of 250000 pa when it is standing on all four feet if the weight of the hippo is 40000 N what

Answer:

The mass will be 4.437 kg

Explanation:

The force constant k is given as 7 N/m

The time period of oscillation T is 5 sec

Thus, angular frequency

It is known that angular frequency is computed via

Squaring both sides gives us

The mass equals 4.437 kg

Response:

Explanation:

Let T denote the tension.

By employing Newton's second law to analyze the bucket's downward motion, we have:

mg - T = ma

A torque, TR, acts on the drum, inducing an angular acceleration α in it. If I refers to the moment of inertia of the drum, then:

TR = Iα

Rearranging gives: TR = Ia/R

This leads to T =  Ia/R²

Substituting this expression for T back into the previous equation yields:

mg - T = ma

mg - Ia/R² = ma

Consequently, we find that mg =  Ia/R² + ma

Therefore, a (I/R² + m) = mg

This results in: a = mg / (I/R² + m)

Next, we aim to express T as:

mg - T = ma

which simplifies to mg - ma  = T

Rearranging gives mg - m²g / (I/R² + m) = T

Thus, we arrive at: mg - mg / (1 + I / m R²) = T

For part (b), T =  Ia/R²

and for part (c), the moment of inertia of a hollow cylinder calculates to:

I = 1/2  M (R² - (R² / 4))

This simplifies to 3/4 x 1/2 MR², yielding 3/8 MR²

Thus, I / R² = 3/8 M

When we substitute, we find a = mg / (3/8 M + m)

and subsequently T =  Ia/R²

= 3/8 MR² × mg / (3/8 M + m) × 1/R²

Results in:

1) The projectile's motion follows
,
In order to determine the velocity, we must compute the derivative of h(t): Next, we will compute the speed at t=2 s and t=4 s: The negative value of the second speed suggests that the projectile has already attained its highest point and is now descending. 2) The maximum height of the projectile occurs when its speed equals zero: Thus, we have And solving yields


3) To determine the maximum height, we substitute the time at which the projectile reaches this peak into h(t), specifically t=2.30 s: 4) The time at which the projectile lands is when the height reaches zero; h(t)=0, which leads to This results in a second-degree equation, producing two answers: the negative root can be disregarded as it lacks physical significance; the second root is

, which indicates the landing time of the projectile. 5) The moment the projectile impacts the ground corresponds to the velocity at time t=4.68 s:

, carrying a negative sign to denote a downward direction.

The formula to apply is expressed as:

ρ = nA/VcNₐ
where
ρ signifies density
n represents the number of atoms within a unit cell (for FCC, n=4)
A indicates the atomic weight
Vc stands for the cubic cell volume equal to a³, with a being the side length (for FCC, a = 4r/√2, where r is the radius)\
Nₐ denotes Avogadro's number, which is 6.022×10²³ atoms/mol

Calculating the radius: r = 0.1387 nm*(10⁻⁹ m/nm)*(100 cm/1m) = 1.387×10⁻⁸ cm
Then find a: a = 4(1.387×10⁻⁸ cm)/√2 = 3.923×10⁻⁸ cm
Then calculate V: V = a³ = (3.923×10⁻⁸ cm)³ = 6.0376×10⁻²³ cm³

Now compute density:
ρ = [(4 atoms)(195.08 g/mol)]/[(6.0376×10⁻²³ cm³)(6.022×10²³ atoms/mol)]Finally, we get ρ = 21.46 g/cm³

The terminal velocity of the object is 12.58 m/s. Explanation: Terminal velocity occurs when the drag force equals the weight of the object. The gravitational force is calculated as mg = 80 * 9.8 = 784 N. The drag force can then be equated to solve for the terminal velocity, resulting in v = 12.58 m/s or v = -15.58 m/s (which is unfeasible). Thus, the terminal velocity of the falling object is 12.58 m/s.