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exis [7]
3 years ago
11

Q2.

Physics
1 answer:
zzz [600]3 years ago
3 0

Answer:

800W

Explanation:

Given the following data;

Voltage = 230 V

Model = VSO Ha

Power = 800 W

Power is calculated by the multiplication of voltage and current flowing through an electric circuit.

Mathematically, power is given by the formula;

Power = current * voltage

The S.I unit for power is Watts and it is a measure of the energy consumption of an electronic device.

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A temperature of 200 degrees F is equivalent to approximately
Ann [662]
(1) Changing Fahrenheit to Celsius:
The formula used to convert from Fahrenheit to Celsius is as follows:
C = <span>(F - 32) * 5/9
</span>We are given that F=200, substitute in the above formula to get the corresponding temperature in Celsius as follows:
C = (200-32) * (5/9) = 93.333334 degrees Celsius

(2) Changing the Fahrenheit to kelvin:
The formula used to convert from Fahrenheit to kelvin is as follows:
K = <span>(F - 32) * 5/9 + 273.15
</span>We are given that F = 200. substitute in the above formula to get the corresponding temperature in kelvin as follows:
K = (200-32)*(5/9) + 273.15 = 366.483334 degrees kelvin


5 0
2 years ago
Read 2 more answers
Any one want to talk i'm so boreed
Sedbober [7]

Answer:

<h3>We can talk. ❤❤</h3>

Explanation:

..........

3 0
2 years ago
Canadian geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at
Wewaii [24]

Answer:

θ=19.877⁰

Explanation:

Given data

Velocity Va=34.0 km/h

Velocity Va=100 km/h

To find

Angle θ

Solution  

We want the bird to fly with velocity Vb=100 km/h with an angle θ relative to the ground so that the bird fly due south relative to the ground.From figure which is attached we got

Sinθ=(Va/Vb)

Sinθ=(34.0/100)

θ=Sin⁻¹(34.0/100)

θ=19.877⁰

5 0
3 years ago
He does whatever a spider can. Spider-Man, who has a mass of 76 [kg], is clinging onto an inclined wall forming an inclination a
Aliun [14]

Answer:570.54 N

Explanation:

Given

mass of man=76 kg

\theta =50^{\circ}

As man is standing over inclined building therefore

its weight has two components i.e. sin and cos component

Force perpendicular to inclined wall

F=mgcos\theta =76\times 9.8\times \sin 50

F=570.54 N

4 0
3 years ago
Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless
drek231 [11]

Answer:

v_f = 15 \frac{m}{s}

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum \vec{L} is

\vec{L}  = \vec{r} \times \vec{p}

where \vec{r} is the position and \vec{p} the linear momentum.

We also know that the torque is

\vec{\tau} = \frac{d\vec{L}}{dt}  = \frac{d}{dt} ( \vec{r} \times \vec{p} )

\vec{\tau} =  \frac{d}{dt}  \vec{r} \times \vec{p} +   \vec{r} \times \frac{d}{dt} \vec{p}

\vec{\tau} =  \vec{v} \times \vec{p} +   \vec{r} \times \vec{F}

but, as the linear momentum is \vec{p} = m \vec{v} this means that is parallel to the velocity, and the first term must equal zero

\vec{v} \times \vec{p}=0

so

\vec{\tau} =   \vec{r} \times \vec{F}

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

\vec{\tau}_{rod} =   0

this means, for the angular momentum measure from the rod:

\frac{d\vec{L}_{rod}}{dt} =   0

that means :

\vec{L}_{rod} = constant

So, the magnitude of initial angular momentum is :

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)

but the angle is 90°, so:

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|

| \vec{L}_{rod_i} | = r_i * m * v_i

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}

| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}

For our final angular momentum we have:

| \vec{L}_{rod_f} | = r_f * m * v_f

and the radius is 0.250 m and the mass is 2.00 kg

| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f

but, as the angular momentum is constant, this must be equal to the initial angular momentum

7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f

v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}

v_f = 15 \frac{m}{s}

8 0
3 years ago
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