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2 years ago

A 1150 kg car is on a 8.70° hill. using x-y axis tilted down the plane, what is the x-component of the weight?

1 answer:

6 0

I assume the x-y axis are tilted such that the x-axis is parallel to the surface of the hill while the y-axis is perpendicular to it.

In this case, the x-component of the weight is given by:
$W_x =mg \sin \theta$
where
m is the mass of the car
g is the acceleration of gravity
$\theta$ is the angle of the hill

Substituting numbers into the formula, we find
$W_x=(1150 kg)(9.81 m/s^2)(\sin 8.70^{\circ})=1706 N$

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A physics professor wants to perform a lecture demonstration of Young's double-slit experiment for her class using the 633-nm li

babunello [35]

Answer:

0.00001266 m

Explanation:

D = Distance from source to screen

m = Order

d = Slit separation

The distance from a point on the screen to the center line

$y=\frac{m\lambda D}{d}$

At m = 0

$y_0=0$

$y_1-y_0=35\ cm\\\Rightarrow y_1=35\ cm$

At m = 1

$y_1=\frac{1\times 633\times 10^{-9}\times 7}{d}\\\Rightarrow d=\frac{1\times 633\times 10^{-9}\times 7}{0.35}\\\Rightarrow d=0.00001266\ m$

The slit separation is 0.00001266 m

3 0

2 years ago

Hot combustion gases enter the nozzle of a turbojet engine at 260 kpa, 747oc, and 80 m/s. the gases exit at a pressure of 85 kpa

Aleksandr-060686 [28]

Hot combustion gases are accelerated in a 92% efficient adiabatic nozzle from low velocity to a specified velocity. The exit velocity and the exit temp are to be determined.

Given:

T1 = 1020 K à h1 = 1068.89 kJ/kg, Pr1 = 123.4

P1 = 260 kPa

T1 = 747 degrees Celsius

V1 = 80 m/s ->nN = 92% -> P2 = 85 kPa

Solution:

From the isentropic relation,

Pr2<span> = (P2 / P1)PR1 = (85 kPa / 260 kPa) (123.4) = 40.34 = h2s = 783.92 kJ/kg</span>

There is only one inlet and one exit, and thus, m1 = m2 = m3. We take the nozzle as the system, which is a control volume since mass crosses the boundary.

h2a = 1068.89 kJ/kg – (((728.2 m/s)2 – (80 m/s)2) / 2) (1 kJ/kg / 1000 m2/s2) = 806.95 kJ/kg\

From the air table, we read T2a = 786.3 K

5 0

2 years ago

There are devices to put in a light socket that control the current through a lightbulb, thereby increasing its lifetime. Which

Dmitrij [34]

Answer: B

Explanation:

Limiting the maximum current through the bulb. This will help in preserving or improving the bulb's lifetime and also this won't have an effect on the brightness of the bulb as brightness is affected by the average value. Although brightness is a factor of current, reducing the maximum current won't have any bearing on the average current the bulb is getting.

4 0

2 years ago

Steel blocks A and B, which have equal masses, are at TA = 300 oC and T8 = 400 oC. Block C, with mc - 2mA, is at TC = 350 oC. Bl

shepuryov [24]

Answer:

b) TA = TB = TC

Explanation:

When put in contact each other, and isolated, both blocks will exchange heat till they reach to thermal equilibrium.
During this process, the body at a higher temperature, will loss heat, tat it will be gained by the other body.
The equilibrium condition will be reached when the following equation be met:

$\Delta Q = c_{st}* m_{A} * (T_{fin} - T_{0A} ) = c_{st}* m_{B} * (T_{0B} - T_{fin} )$

Replacing by the values of T₀A = 300º C, and T₀B = 400ºC, and simplifying common terms as mA = mB, we can solve for Tfin, as follows:

$(400 \ºC - T_{fin}) = (T_{fin} - 300 \ºC) \\ \\ 2* T_{fin} = 700\ºC\\ \\ T_{fin} = 350\ºC$

So, when both blocks reach to equilibrium, they will be at a common final temperature, 350ºC.
When put in contact with block C, at the same temperature, at that instant, the three blocks will have the same common temperature of 350 ºC.
So, option b) is the right one.

8 0

2 years ago

A 0.0427 kg racquet-ball is moving

Gwar [14]

Answer:

Mass of the box = 0.9433 kg

Explanation:

Mass of racket-ball $(m_1)$ = 0.00427 kg