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Ugo [173]
3 years ago
15

A rectangular dam is 101 ft long and 54 ft high. If the water is 35 ft deep, find the force of the water on the dam (the density

of water is 62.4 lb/ft3).
Physics
1 answer:
blsea [12.9K]3 years ago
6 0

To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.

P = \rho h

Here,

h = Depth average

\rho = Density

Moreover,

\text{Density of water}= \rho = 62.4lb/ft^3

Replacing,

P = (62.4lb/ft^3)(\frac{35}{2}ft)

P = 1092 lb/ft^2

Finally the force

\text{Force} = \text{Pressure}\times \text{Area of dam with water acting on it}

F = 1092lb/ft^2(101ft*52ft)

F = 5.735*10^6lbf

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A car accelerates uniformly from rest to speed 6.6 m/s in 6.5 s .Find the distance the car travel during this time .​
kirill [66]

Answer:

<em>The distance the car traveled is 21.45 m</em>

Explanation:

<u>Motion With Constant Acceleration </u>

It occurs when an object changes its velocity at the same rate thus the acceleration is constant.

The relation between the initial and final speeds is:

v_f=v_o+at\qquad\qquad [1]

Where:

a   = acceleration

vo = initial speed

vf  = final speed

t    = time

The distance traveled by the object is given by:

\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]

Solving [1] for a:

\displaystyle a=\frac{v_f-v_o}{t}

Substituting the given data vo=0, vf=6.6 m/s, t=6.5 s:

\displaystyle a=\frac{6.6-0}{6.5}

a = 1.015\ m/s^2

The distance is now calculated with [2]:

\displaystyle x=0*6.5+\frac{1.015*6.5^2}{2}

x = 21.45 m

The distance the car traveled is 21.45 m

6 0
3 years ago
(3.16_Q2) Which weights would you use on a single thread to create a 6.86 N force? Question 2 options: Weight IDs A, B, C, D Wei
Tomtit [17]

1. E,F

2. D,E,F

3. B,C,E,G

4. A,B,C,D

I did the test! :)

5 0
3 years ago
A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is
lesantik [10]

Answer:

35870474.30504 m

Explanation:

r = Distance from the surface

T = Time period = 24 h

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

m = Mass of the Earth =  5.98 × 10²⁴ kg

Radius of Earth = 6.38\times 10^6\ m

The gravitational force will balance the centripetal force

\dfrac{GMm}{R^2}=m\dfrac{v^2}{R}\\\Rightarrow v=\sqrt{\dfrac{GM}{R}}

T=\dfrac{2\pi r}{v}\\\Rightarrow T=\dfrac{2\pi r}{\sqrt{\dfrac{GM}{r}}}

From Kepler's law we have relation

T^2=\dfrac{4\pi^2r^3}{GM}\\\Rightarrow r^3=\dfrac{T^2GM}{4\pi^2}\\\Rightarrow r=\left(\dfrac{(24\times 3600)^2\times 6.67\times 10^{-11}\times 5.98\times 10^{24}}{4\pi^2}\right)^{\dfrac{1}{3}}\\\Rightarrow r=42250474.30504\ m

Distance from the center of the Earth would be

42250474.30504-6.38\times 10^6=\mathbf{35870474.30504\ m}

8 0
3 years ago
Often called velocity this is the velocity of an object at a particular moment in time
Yuliya22 [10]
Instantaneous speed?
8 0
3 years ago
Even when shut down after a period of normal use, a large commercial nuclear reactor transfers thermal energy at the rate of 150
Phoenix [80]

Answer:

The temperature of the core raises by 2.8^{o}C every second.

Explanation:

Since the average specific heat of the reactor core is 0.3349 kJ/kgC

It means that we require 0.3349 kJ of heat to raise the temperature of 1 kg of core material by 1 degree Celsius

Thus reactor core whose mass is 1.60\times 10^{5}kg will require

0.3349\times 1.60\times 10^{5}kJ\\\\=0.53584\times 10^{5}kJ

energy to raise it's temperature by 1 degree Celsius in 1 second

Hence by the concept of proportionately we can infer 150 MW of power will increase the temperature by

\frac{150\times 10^{6}}{0.53584\times 10^{8}}=2.8^{o}C/s

5 0
3 years ago
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