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2 years ago

A school is creating a small parking lot next to the school. They will need

Physics

1 answer:

bagirrra123 [75]2 years ago

7 0

Answer:

2200.9lb

Explanation:

This is a conversion problem.

We have been given that:

Mass of rock the school needed = 998140g

Unknown:

Pound of rocks the park needed = ?

To solve this problem, we have to convert from:

grams to kilograms and then to pounds

1000g of the rock will weigh 1kg

So; 998140g of the rock will weight 998.14kg

Therefore:

1kg of a substance weighs 2.205lb

998.14g will weight2.205 x 998.14 = 2200.9lb

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When four people with a combined mass of 310 kg sit down in a 2000-kg car, they find that their weight compresses the springs an

luda_lava [24]

Answer:

Explanation:

F=kx

x=F/k

F=2000 kg

x=100 cm=9*10^-3

effective spring constant=k=F/x

k=2000/9*10^-3=2.2*10^-5

now frequency

f=1/2π√k/m

f=1/2*3.14√2.2*10^-5/310

f=1/6.28√7.097*10^-8

f=1/6.28*2.7*10^-4

f=0.16*2.7*10^-4

f=4.32*10^-5

4 0

3 years ago

What are the two sources of energy for the earth system

spin [16.1K]

Solar energy and Geothermal energy are your answers :)

8 0

3 years ago

Read 2 more answers

The current theory of the structure of the

Mariana [72]

Answers:

a) $2.82(10)^{21} kg$

b) $1410 J$

c) $36.62 m/s$

Explanation:

<h3>a) Mass of the continent</h3>

Density $\rho$ is defined as a relation between mass $m$ and volume $V$ :

$\rho=\frac{m}{V}$ (1)

Where:

$\rho=2720 kg/m^{3}$ is the average density of the continent

$m$ is the mass of the continent

$V$ is the volume of the continent, which can be estimated is we assume it as a a slab of rock 5300 km on a side and 37 km deep:

$V=(length)(width)(depth)=(5300 km)(5300 km)(37 km)=1,030,330,000 km^{3} \frac{(1000 m)^{3}}{1 km^{3}}=1.03933(10)^{18} m^{3}$

Finding the mass:

$m=\rho V$ (2)

$m=(2720 kg/m^{3})(1.03933(10)^{18} m^{3})$ (3)

$m=2.82(10)^{21} kg$ (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy $K$ is given by the following equation:

$K=\frac{1}{2}mv^{2}$ (5)

Where:

$m=2.82(10)^{21} kg$ is the mass of the continent

$v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s$ is the velocity of the continent

$K=\frac{1}{2}(2.82(10)^{21} kg)(1(10)^{-9} m/s)^{2}$ (6)

$K=1410 J$ (7) This is the kinetic energy of the continent

<h3>c) Speed of the jogger</h3>

If we have a jogger with mass $m=77 kg$ and the same kinetic energy as that of the continent $1413 J$ , we can find its velocity by isolating $v$ from (5):

$v=\sqrt{\frac{2 K}{m}}$ (6)

$v=\sqrt{\frac{2 (1413 J)}{77 kg}}$

Finally:

$v=36.62 m/s$ This is the speed of the jogger

5 0

3 years ago

HELP ME HOW ARE BLACK HOLES FROMED

V125BC [204]

Answer:

Stellar black holes form when the center of a very massive star collapses in upon itself.

6 0

3 years ago

Read 2 more answers

At a certain location, wind is blowing steadily at 9 m/s. Determine the mechanical energy of air per unit mass and the power gen

Misha Larkins [42]

Answer:

The specific mechanical energy of the air in the specific location is 40.5 J/kg.
The power generation potential of the wind turbine at such place is of 2290 kW
The actual electric power generation is 687 kW

Explanation:

The mechanical energy of the air per unit mass is the specific kinetic energy of the air that is calculated using: $\frac{1}{2} V^2$ where V is the velocity of the air.
The specific kinetic energy would be: $\frac{1}{2}(9\frac{m}{s})^2=40.5\frac{m^2}{s^2}=40.5\frac{m^2 }{s^2}\frac{kg}{kg}=40.5\frac{N*m }{kg}=40.5\frac{J}{kg}$ .
The power generation of the wind turbine would be obtained from the product of the mechanical energy of the air times the mass flow that moves the turbine.
To calculate mass flow it is required first to calculate the volumetric flow. To calculate the volumetric flow the next expression would be: $\frac{V\pi D_{blade}^2}{4} =\frac{9\frac{m}{s}\pi(80m)^2}{4} =45238.9\frac{m^3}{s}$
Then the mass flow is obtain from the volumetric flow times the density of the air: $m_{flow}=1.25\frac{kg}{m^3}45238.9\frac{m^3}{s}=56548.7\frac{kg}{s}$
Then, the Power generation potential is: $40.5\frac{J}{kg} 56548.7\frac{kg}{s} =2290221W=2290.2kW$
The actual electric power generation is calculated using the definition of efficiency: $\eta=\frac{E_P}{E_I}}$ , where η is the efficiency, $E_P$ is the energy actually produced and, $E_I$ is the energy input. Then solving for the energy produced: $E_P=\eta*E_I=0.30*2290kW=687kW$

6 0

3 years ago