if you have to apply 40 n of force on a crowbar to lift a 400 n rock what is the actual machanical andvatage of the crow bar

Ratling [72]

question 2 answer is ALL OF THE ABOVE

question 3 answer is WARM AND MOIST.

8 0

3 years ago

Water enters a baseboard radiator at 180 °F and at a flow rate of 2.0 gpm. Assuming the radiator releases heat into the room at

beks73 [17]

Answer:

Temperature of water leaving the radiator = 160°F

Explanation:

Heat released = (ṁcΔT)

Heat released = 20000 btu/hr = 5861.42 W

ṁ = mass flowrate = density × volumetric flow rate

Volumetric flowrate = 2 gallons/min = 0.000126 m³/s; density of water = 1000 kg/m³

ṁ = 1000 × 0.000126 = 0.126 kg/s

c = specific heat capacity for water = 4200 J/kg.K

H = ṁcΔT = 5861.42

ΔT = 5861.42/(0.126 × 4200) = 11.08 K = 11.08°C

And in change in temperature terms,

10°C= 18°F

11.08°C = 11.08 × 18/10 = 20°F

ΔT = T₁ - T₂

20 = 180 - T₂

T₂ = 160°F

8 0

3 years ago

Shanti is riding on a train that is moving at a speed of 90 km/h. He is carrying a power cord for his phone that is 1.2 m long.

faltersainse [42]

Answer:equal to 1.2 m

Explanation: i took the test

3 0

2 years ago

Read 2 more answers

About 65 million years ago an asteroid struck Earth in the area of the Yucatán Peninsula and wiped out the dinosaurs and many ot

9966 [12]

Answer:

$2.44156\times 10^{13}\ m^3$

29010.53917 m

Explanation:

$\rho$ = Density of asteroid = 2 g/cm³

V = Volume

d = Diameter = 10 km

r = Radius = $\dfrac{d}{2}=\dfrac{10}{2}=5\ km$

v = Velocity = 11 km/s

$H_v$ = Heat vaporization of water = $2.26\times 10^6\ J/kg$

$\Delta T$ = Change in temperature = 100-20

Mass is given by

$m=\rho V\\\Rightarrow m=\rho\dfrac{4}{3}\pi r^3\\\Rightarrow m=2000\dfrac{4}{3}\times \pi\times 5000^3\\\Rightarrow m=1.0472\times 10^{15}\ kg$

The kinetic energy is

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1.0472\times 10^{15}\times 11000^2\\\Rightarrow K=6.33556\times 10^{22}\ J$

Heat is given by

$Q=mc\Delta T+mH_v\\\Rightarrow 6.33556\times 10^{22}=m\times (4186\times (100-20)+2.26\times 10^6)\\\Rightarrow m=\dfrac{ 6.33556\times 10^{22}}{4186\times (100-20)+2.26\times 10^6}\\\Rightarrow m=2.44156\times 10^{16}\ kg$

Mass of water is $2.44156\times 10^{16}\ kg$

Volume is $\dfrac{2.44156\times 10^{16}}{10^3}=2.44156\times 10^{13}\ m^3$

Amount of water is $2.44156\times 10^{13}\ m^3$

If it were a cube

$h=V^{\dfrac{1}{3}}\\\Rightarrow h=(2.44156\times 10^{13})^{\dfrac{1}{3}}\\\Rightarrow h=29010.53917\ m$

The height of the water would be 29010.53917 m

4 0

2 years ago

The flow rate in a firehose is 0.524 m3/s. It is able to shoot water to the top of a building 40.4 m tall, but not higher. You r

Marrrta [24]

Answer:

A fire hose must be able to shoot water to the top of a building 35.0 m tall ... Water enters this hose at a steady rate of 0.500 m3/s and shoots out of a round nozzle. ... I know that Flow rate=0.500 m3/s=A*V. I know the pressure needed to ... The first equation has no potential while the second has no kinetic.

Explanation:

3 0

3 years ago