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nata0808 [166]
4 years ago
9

A brick falls from a wall to the ground 3.42 m below. How much time will it take to

Physics
1 answer:
ddd [48]4 years ago
6 0

Answer:

The time taken by the brick to hit the ground, t = 0.84 s

Explanation:

Given that,

A brick falls from a height, h = 3.42 m

The initial velocity of the brick is zero.

Since the brick is under free-falling. The time equation of a free-falling body when the displacement is given is

                                     t = \sqrt{2h/g}

where,

                                 h - height from surface in meters

                                 g - acceleration due to gravity

on substituting the values in the above equation,

                                 t =  \sqrt{2X3.14/9.8}

                                   = 0.84 s

Hence, time taken by the brick to hit the ground is t = 0.84 s

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A uniform rod is 2. 0 m long and has mass 15 kg. What is most nearly the rod's mass moment of inertia?
trapecia [35]

The rod's mass moment of inertia is 5kgm².

<h3>Moment of Inertia:</h3>

The "sum of the product of mass" of each particle with the "square of its distance from the axis of rotation" is the formula for the moment of inertia.

The Parallel axis Theorem can be used to compute the moment of inertia about the end of the rod directly or to derive it from the center of mass expression. I = kg m². We can use the equation for I of a cylinder around its end if the thickness is not insignificant.

If we look at the rod we can assume that it is uniform. Therefore the linear density will remain constant and we have;

or = M / L = dm / dl

dm = (M / L) dl

I =  \int\limits^M_0 {r^2} \, dm

I = \int\limits^\frac{L}{2} _\frac{-L}{2}  {I^2 (M/L)} \, dl

Here the variable of the integration is the length (dl). The limits have changed from M to the required fraction of L.

I = \int\limits^\frac{L}{2} _\frac{-L}{2}  {I^2 (M/L)} \, dl

I = \frac{M}3L}[(\frac{L^3}{2^3}   - \frac{-L^3}{2^3} )]\\\\I = \frac{1}{12}ML^2

Mass of the rod = 15 kg

Length of the rod = 2.0 m

Moment of Inertia, I = \frac{1}{12}15 (2)^2

                               = 5 kgm²

Therefore, the moment of inertia is 5kgm².

Learn more about moment of inertia here:

brainly.com/question/14119750

#SPJ4

4 0
2 years ago
wide tube that extends down from the bag of solution, which hangs from a pole so that the fluid level is 90.0 cm above the needl
Bingel [31]

Answer:

The average gauge pressure inside the vein is 110270.58 Pa

Explanation:

This question can be solved using the Bernoulli's Equation. First, in order to determine the outlet pressure of the needle, we need to find the total pressure exerted by the atmosphere and the fluid.

P_f: fluid's\ pressure\\P_f= \rho g h=1025\frac{kg}{m^3} \times 9.8 \frac{m}{s^2} \times 0.9 m=9040.5 Pa \\P_T: total\ pressure\\P_T=P_{atm}+P_f\\P_T=101325 Pa + 9040.5 Pa=110275.5 Pa\\

Then, we have to find the fluid's outlet velocity with the transversal area of the needle, as follows:

S: transversal\ area \\S= \pi r^2=\pi (0.200 \times 10^{-3})^2=5.65 \times 10^{-7} m^2\\v=\frac{F}{S}=\frac{5.55 \times 10^{-8} \frac{m^3}{s}}{5.65 \times 10^{-7} m^2}=0.98\times 10^{-1} \frac{m}{s}

As we have all the information, we can complete the Bernoulli's expression and solve to find the outlet pressure as follows:

P_T-P_{out}=\frac{1}{2} \rho v^2\\P_{out}=P_T-\frac{1}{2} \rho v^2=110275.5 Pa-\frac{1}{2} 1025\frac{kg}{m^3} (0.98\times 10^{-1} \frac{m}{s})^2=110275.5 Pa-4.92 Pa =110270.58 Pa

6 0
3 years ago
If a 5.0 kg box is pulled simultaneously by a 10.0 N force and a 5.0 N force, then its acceleration must be Group of answer choi
vivado [14]

Answer:

We cannot tell from the information given

Explanation:

Given;

mass of the box, m = 5 kg

first force, F₁ = 10 N

second force, F₂ = 5 N

(I) Assuming the two forces are acting horizontally in opposite direction, the resultant force on the box is calculated as;

∑Fx = 10 N - 5 N

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Apply Newton's second law of motion;

∑Fx = ma

a = ∑Fx/m

a = 5 / 5

a = 1 m/s² in the direction of the 10 N force.

(II) Also, if the two forces are acting in the same direction, the resultant force is calculated as;

∑Fx = 10 N + 5 N

∑Fx = 15 N

a = 15 / 5

a = 3 m/s²

Therefore, the information given is not enough to determine the acceleration of the box.

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