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Dennis_Churaev [7]

4 years ago

10

ALWAYS use significant figure rules. Remember that these rules apply to all numbers that are measurements. In working this probl

em, assume the value of "g" to be 9.8 m/s2 with two (2) significant digits unless otherwise stated.
Calculate the total energy of 2.0 kg object moving horizontally at 10 m/s 50 meters above the surface.

a.10. J
b.100 J
c.200 J
d.1080 J

2 answers:

Anni [7]4 years ago

8 0

Answer:

i think it is b sorry if im worng

iris [78.8K]4 years ago

8 0

Answer : The correct option is, (d) 1080 J

Explanation :

Total energy = Kinetic energy + Potential energy

$\text{Total energy}=\frac{1}{2}mv^2+mgh$

where,

m = mass of an object = 2.0 kg

v = speed of an object = 10 m/s

h = height of an object = 50 m

g = acceleration due to gravity = $9.8m/s^2$

Now put all the given values in the above expression, we get:

$\text{Total energy}=\frac{1}{2}\times (2.0kg)\times (10m/s)^2+(2.0kg)\times (9.8m/s^2)\times (50m)$

Total energy = 1080 J

Thus, the total energy will be, 1080 J

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3 0

3 years ago

Air as an ideal gas enters a diffuser operating at steady state at 5 bar, 280 K with a velocity of 510 m/s. The exit velocity is

Nataly [62]

Answer:

Explanation:

Calculating the exit temperature for K = 1.4

The value of $c_p$ is determined via the expression:

$c_p = \frac{KR}{K_1}$

where ;

R = universal gas constant = $\frac{8.314 \ J}{28.97 \ kg.K}$

k = constant = 1.4

$c_p = \frac{1.4(\frac{8.314}{28.97} )}{1.4 -1}$

$c_p= 1.004 \ kJ/kg.K$

The derived expression from mass and energy rate balances reduce for the isothermal process of ideal gas is :

$0=(h_1-h_2)+\frac{(v_1^2-v_2^2)}{2}$ ------ equation(1)

we can rewrite the above equation as :

$0 = c_p(T_1-T_2)+ \frac{(v_1^2-v_2^2)}{2}$

$T_2 =T_1+ \frac{(v_1^2-v_2^2)}{2 c_p}$

where:

$T_1 = 280 K \\ \\ v_1 = 510 m/s \\ \\ v_2 = 120 m/s \\ \\c_p = 1.0004 \ kJ/kg.K$

$T_2= 280+\frac{((510)^2-(120)^2)}{2(1.004)} *\frac{1}{10^3}$

$T_2 = 402.36 \ K$

Thus, the exit temperature = 402.36 K

The exit pressure is determined by using the relation: $\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{k}{k-1}$

$P_2=P_1(\frac{T_2}{T_1})^\frac{k}{k-1}$

$P_2 = 5 (\frac{402.36}{280} )^\frac{1.4}{1.4-1}$

$P_2 = 17.79 \ bar$

Therefore, the exit pressure is 17.79 bar

7 0

3 years ago

Suppose that a public address system emits sound uniformly in all directions and that there are no reflections . The intensity a

Marina86 [1]

Answer:

$I_{2}=2.39*10^{-5} W/m^{2}$

Explanation:

The definition of the intensity in terms of power is given by:

$I=\frac{P}{A}$

Where:

P is the power
A is the area

If the sound emits uniformly in all directions and that there are no reflections, we can assume the geometry of the wave sound is spherical.

Let's recall the area of a sphere is $A = 4\pi R^{2}$

To the first location we have:

$I_{1}=3*10^{-4} W/m^{2}=\frac{P}{4\pi 22^{2}}$

and to the second location we have:

$I_{2}=\frac{P}{4\pi 78^{2}}$

Now, we can divide each intensity to find the second intensity.

$\frac{I_{2}}{I_{1}}=\frac{\frac{P}{4\pi 78^{2}}}{\frac{P}{4\pi 22^{2}}}$

$I_{2}=I_{1}* \frac{22^{2}}{78^{2}}$

$I_{2}=3*10^{-4}\frac{22^{2}}{78^{2}}$

$I_{2}=2.39*10^{-5} W/m^{2}$

I hope it helps you!

5 0

3 years ago

Read 2 more answers

A 0.5 kg rock is dropped from a height of 1.0 m above the ground. Approximately how much kinetic energy will be stored in the ro

irina1246 [14]

Answer:

2.45 J

Explanation:

The following data were obtained from the question:

Mass (m) = 0.5 kg

Height (h) = 1 m

Kinetic energy (KE) =?

Next, we shall determine the velocity of the rock after it has fallen half way. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Height (h) = 1/2 = 0.5 m

Final velocity (v) =?

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 0.5)

v² = 9.8

Take the square root of both side

v = √9.8

v = 3.13 m/s

Finally, we shall determine the kinetic energy of the rock after it has fallen half way. This can be obtained as follow:

Mass (m) = 0.5 kg

Velocity (v) = 3.13 m/s

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 0.5 × 3.13²

KE = 0.25 × 9.8

KE = 2.45 J

Therefore, the kinetic energy of the rock after it has fallen half way is 2.45 J

8 0

3 years ago

You have a grindstone (a disk) that is 90.0 kg, has a 0.340-m radius, and is turning at 90.0 rpm, and you press a steel axe agai

QveST [7]

Answer:

Explanation:

Given mass of grindstone $m=90\ kg$

radius of stone $r=0.34\ m$

angular speed of disc $\omega =90\ rpm$

Steel Axle applying a force of $F=20\ N$

coefficient of kinetic friction $\mu =0.2$

Frictional Torque applied by steel is given by

$\tau=r\times f_r$

$\tau =r\times \mu F$

where $f_r$ =frictional force

$\tau =r\times \mu \times F$

$\tau =0.34\times 0.2\times 20$

$\tau =1.36\ N-m$

Torque is also given by

$\tau =I\cdot \alpha$

where $\alpha$ =angular acceleration

I=moment of Inertia

$\tau =0.5Mr^2\times \alpha$

$0.5Mr^2\times \alpha =1.36\ N-m$

$0.5\times 90\times 0.34^2\times \alpha =1.36$

$\alpha =0.261\ rad/s^2$

3 0

4 years ago