All categories

2 years ago

What force is acting on the rainwater in the model?

Physics

2 answers:

Vikki [24]2 years ago

8 0

The force is gravitational because when something is falling is call gravitational

iren [92.7K]2 years ago

6 0

It’s showing gravity

You might be interested in

The magnetic field of an electromagnetic wave in a vacuum is Bz =(2.4μT)sin((1.05×107)x−ωt), where x is in m and t is in s. You

tatiyna

Answer:

Explanation:

Given

$B_z=(2.4\mu T)\sin (1.05\times 10^7x-\omega t)$

Em wave is in the form of

$B=B_0\sin (kx-\omega t)$

where $\omega =frequency\ of\ oscillation$

$k=wave\ constant$

$B_0=Maximum\ value\ of\ Magnetic\ Field$

Wave constant for EM wave k is

$k=1.05\times 10^7 m^{-1}$

Wavelength of wave $\lambda =\frac{2\pi }{k}$

$\lambda =\frac{2\pi }{1.05\times 10^7}$

$\lambda =5.98\times 10^{-7} m$

7 0

2 years ago

A baseball player exerts a force of 100 N on a ball for a distance of 0.5 mas he throws it. If the ball has a mass of 0.15 kg, w

Aloiza [94]

Answer:

25.82 m/s

Explanation:

We are given;

Force exerted by baseball player; F = 100 N

Distance covered by ball; d = 0.5 m

Mass of ball; m = 0.15 kg

Now, to get the velocity at which the ball leaves his hand, we will equate the work done to the kinetic energy.

We should note that work done is a measure of the energy exerted by the baseball player.

Thus;

F × d = ½mv²

100 × 0.5 = ½ × 0.15 × v²

v² = (2 × 100 × 0.5)/0.15

v² = 666.67

v = √666.67

v = 25.82 m/s

4 0

1 year ago

Your latest invention is a car alarm that produces sound at a particularly annoying frequency of 3600 Hz . To do this, the car a

Alex17521 [72]

Answer:

The capacitance and the inductance can choose for a car-alarm circuit are

C = 215.27 μF

L = 9.078 μH

Explanation:

$V =12.0 V$ , $E = 1.55*10^2 J$ , $f = 3600 Hz$

To determine the capacitance can use the equation

$U_c= \frac{1}{2}*C*V^2$

Solve to C'

$C = \frac{U_c*2}{V^2}=\frac{1.55x10^2J*2}{12.0^2V}$

$C=215.27 uF$

To find the inductance can use the frequency of the circuit

$f = \frac{1}{2\pi* \sqrt{C*L} }$

Solve to L'

$L = \frac{1}{4\pi^2*f^2*C}=\frac{1}{4\pi^2*3600^2*215.27 uF}}$

$L = 9.078 uH$

6 0

1 year ago

Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is

ad-work [718]

Answer:

$T_{out}=27.0000077$ ºC

Explanation:

First, let's write the energy balance over the duct:

$H_{out}=H_{in}+Q$

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

$m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q$

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

$m*Cp(T_{out}-T_{in})=Q$

So, let's isolate $T_{out}$ :

$T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}$

The Cp of the air at 27ºC is 1007 $\frac{J}{kgK}$ (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are $T_{out}$ and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

$P=2*H+2*A=2*0.004m+2*0.016m=0.04m$