When did galileo discover projectile motion?

How much energy must be absorbed by water with a mass of 0.5 kg in order to raise the temperature from 30°C to 65°C? Note: Water

kvasek [131]

Answer:

73325J

Explanation:

Given parameters:

Mass of water = 0.5kg

Initial temperature = 30°C

Final temperature = 65°C

Specific heat capacity = 4190J/kg°C

Unknown:

Amount of energy absorbed = ?

Solution:

The amount of energy absorbed can be derived using the expression below;

H = m c Δt

H is the amount of energy

m is the mass

c is the specific heat

Δt is the change in temperature

H = 0.5 x 4190 x (65 - 30 )

H = 73325J

6 0

3 years ago

Can someone help me w this ?

lukranit [14]

Answer:

i can't sorry

Explanation:

I didn't really pay attention in that class

7 0

2 years ago

If the coefficient of friction is 0.3900 and the cylinder has a radius of 2.700 m, what is the minimum angular speed of the cyli

Aleks04 [339]

Answer:

w=3.05 rad/s or 29.88rpm

Explanation:

k = coefficient of friction = 0.3900

R = radius of the cylinder = 2.7m

V = linear speed of rotation of the cylinder

w = angular speed = V/R or to rewrite V = w*R

N = normal force to cylinder

N= $=\frac{m(V)^{2}}{R}=m*(w)^2*R$

$Friction force\\Ff = k*N\\Ff= k*m*w^2*R$

$Gravitational force \\Fg = m*g$

These must be balanced (the net force on the people will be 0) so set them equal to each other.

$Fg = Ff$

$m*g = k*m*w^2*R$

$g=k*w^{2}*R$

$w^2 =\frac{g}{k*R}$

$w=\sqrt{\frac{g}{k*R}} \\w =\sqrt{\frac{9.8\frac{m}{s^{2}}}{0.3900*2.7m}}\\ w=\sqrt{9.306}=3.05 \frac{rad}{s}$

There are 2*pi radians in 1 revolution so:

$RPM=\frac{w}{2\pi }*60\\RPM=\frac{3.05\frac{rad}{s}}{2\pi}*60\\RPM= 0.498*60\\RPM=29.88$

So you need about 30 RPM to keep people from falling out the bottom

7 0

4 years ago

2. Use the diagram below to answer this question. As the ball moves from point A

Thepotemich [5.8K]

Answer:

at point A the ball possess pontetial energy , point B kinetic energy then point C pontetial energy

5 0

3 years ago

Air enters a turbine operating at steady state with a pressure of 75 Ibf/in.^2, a temperature of 800º R and velocity of 400 ft/s

Arturiano [62]

Answer:

(a) W/m = 49.334 Btu/lb

(b) $\frac{E_{d} }{m}$ = 22.12 Btu/lb

Explanation:

For the given problem, it can be assumed that the system is operating at steady state and the effects of potential energy can be neglected.

(a) Using the thermodynamic table for air.

At the temperature ( $T_{1}$ )of 800 ºR and pressure ( $P_{1}$ ) of 75 Ibf/in.^2, we can deduce that:

Specific enthalpy ( $h_{1}$ ) = 191.81 BTu/lb

Specific entropy ( $s_{1}$ ) = 0.6956 Btu/(lb.ºR)

At the temperature ( $T_{2}$ )of 600 ºR and pressure ( $P_{2}$ ) of 15 Ibf/in.^2, we can deduce that:

Specific enthalpy ( $h_{2}$ ) = 143.47 BTu/lb

Specific entropy ( $s_{2}$ ) = 0.6261 Btu/(lb.ºR)

The work done can be calculated using energy rate equation:

$\frac{W}{m} = \frac{Q}{m} + (h_{1} - h_{2}) + \frac{V_{1}^{2} - V_{2}^{2}}{2}$

Q/m = heat transfer = -2 Btu/lb

$V_{1}$ = 400 ft/s

$V_{2}$ = 100 ft/s

$\frac{W}{m} = -2 + (191.81 - 143.47) + \frac{400^{2} - 100^{2}}{2}*[tex]\frac{1}{2*32.2*778}$ [/tex] = -2 + 48.34 + 29.938 = 49.334 Btu/lb

(b) To calculate the exergy destruction, we will use the equation for exergy rate:

$\frac{E_{d} }{m} = [1-\frac{T_{o} }{T_{b} }](\frac{Q}{m}) - \frac{W}{m} + [(h_{1} - h_{2}) -T_{o}(s_{1} - s_{2}) + \frac{V^{2} _{1} - V_{2} ^{2}}{2}]$

The equation above is further simplified to:

$\frac{Ed}{m} = T_{o}[(s_{2} -s_{1}) - Rln\frac{P_{2} }{P_{1} } - \frac{Q/m}{T_{b} }]$

Using a reference temperature (To) = 500 °R

Average surface temperature (Tb = 620°R

$\frac{Ed}{m} = 500*[(0.6261 -0.6956) - (1.986/28.97)ln\frac{15 }{75 } - \frac{-2}{620}}]$

$\frac{E_{d} }{m}$ = 500*[-0.0695 +0.068688*1.609 +0.003225] = 22.12 Btu/lb

5 0

4 years ago