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2 years ago

A 1,811 kg car goes over the top of a hill of radius 20 m. What is the maximum speed the car can

1 answer:

8 0

If I had an answer to the question, I would tell you but I dont. I’m just doing this because I have to for free answers. And it has to be 20 characters long.

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What did scientists create using scientific measurements?

Alexxx [7]

Answer:

lines?

Explanation:

3 0

3 years ago

1 Verify the identity. Show your work. cot θ ∙ sec θ = csc θ

Fittoniya [83]

To verify the identity, we can make use of the basic trigonometric identities:
cot θ = cos θ / sin θ
sec θ = 1 / cos θ
csc θ = 1 / sin θ

Using these identities:
cot θ ∙ sec θ = (cos θ / sin θ ) ( 1 / cos θ)

We can cancel out cos θ, leaving us with
cot θ ∙ sec θ = 1 / sin θ
cot θ ∙ sec θ = = csc θ

6 0

3 years ago

In a hydroelectric power plant, water flows from an elevation of 400 ft to a turbine, where electric power is generated. For an

VashaNatasha [74]

Answer:

$V=3.475ft^3/s=3.48ft^3/s$

Explanation:

We have here values from SI and English Units. I will convert the units to English Units.

We hace for the power P,

$P= 100kW \rightarrow 100kW*(\frac{737.56ft.lbf/s}{1kW})$

$P= 73.7*10^{3}ft.lbf/s$

we have other values such $h=400ft$ and $\gamma= 62.42lb/ft^3$ (specific weight of the water), and 0.85 for \eta

We need to figure the flow rate of the water (V) out, that is,

$V=\frac{P}{\gamma h \eta_0}$

Where $\eta_0$ is the turbine efficiency, at which is,

$\eta_0 = \frac{P}{\gamma Vh}$

Replacing,

$V=\frac{73.7*10^{3}}{62.42*400*0.85}$

$V=3.475ft^3/s$

With this value (the target of this question) we can also calculate the mass flow rate of the waters,

through the density and the flow rate,

$m=\rho V\\m= 3.475*1.94 \\m=6.7415 slugs/s$

converting the slugs to lbm, 1slug = 32.174lbm, we have that the mass flow rate of the water is,

$m= 217lbm/s$

6 0

3 years ago

An object is launched from the top of a building which is 100 m tall (relative to the ground) at a speed of 22 m/s at an angle o

klio [65]

Answer:

$3.58\:\mathrm{s}$

Explanation:

We can use the kinematics equation $\Delta y=v_it+\frac{1}{2}at^2$ to solve this problem. To find the initial vertical velocity, find the vertical component of the object's initial velocity using basic trigonometry for right triangles:

$\sin28^{\circ}=\frac{y}{22},\\y=22\sin28^{\circ}=10.3283743813\:\mathrm{m/s}$

Now we can substitute values in our kinematics equation:

$\Delta y=-100$
$a=-9.8\:\mathrm{m/s^2}$ (acceleration due to gravity)
$v_i=-10.3283743813\:\mathrm{m/s}$
Solving for $t$

$-100=-10.3283743813t+\frac{1}{2}\cdot -9.8\cdot t^2,\\\\-4.9t^2-10.3283743813t+100=0,\\\\\boxed{t=3.5849312673637455}, t=-5.692762773751501\:\text{(Extraneous)}$

6 0

3 years ago

Efficiency of a machine is always less than 100% why?

kifflom [539]

The reason the efficiency is always less than 100% is because there is always some sort of friction in the machine.

Answer: because of friction

7 0

2 years ago