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vladimir2022 [97]

2 years ago

9

A 75 W lightbulb is being run on a 110 V outlet. Determine the resistance of the lightbulb. Provide a detailed description of th

e process.

1 answer:

Sauron [17]2 years ago

8 0

Power is defined as

P = I*V

where I is the current and V is the voltage

Ohm's law gives us the relation betwen Voltage and current in a resistive component

V = I*R , Then

P = V² / R

We solve for R,

R = (110 V)²/ 75W = 161.33 ohms

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Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45-m-diamete

Margarita [4]

Explanation:

It is given that,

Diameter of the semicircle, d = 45 m

Radius of the semicircle, r = 22.5 m

Speed of greyhound, v = 15 m/s

The greyhound is moving under the action of centripetal acceleration. Its formula is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(15)^2}{22.5}$

$a=10\ m/s^2$

We know that, $g=9.8\ m/s^2$

$a=\dfrac{10\times g}{9.8}$

$a=1.02\ g$

Hence, this is the required solution.

5 0

2 years ago

4.2 mol of monatomic gas A interacts with 3.2 mol of monatomic gas B. Gas A initially has 9500 J of thermal energy, but in the p

Komok [63]

Answer:

14657.32 J

Explanation:

Given Parameters ;

Number of moles mono atomic gas A , n 1 = 4 .2 mol

Number of moles mono atomic gas B , n 2 = 3.2mol

Initial energy of gas A , K A = 9500 J

Thermal energy given by gas A to gas B , Δ K = 600 J

Gas constant R = 8.314 J / molK

Let K B be the initial energy of gas B.

Let T be the equilibrium temperature of the gas after mixing.

Then we can write the energy of gas A after mixing as

(3/2)n1RT = KA - ΔK

⟹ (3/2) x 4.2 x 8.314 x T = 9500 - 600

T = (8900 x 3 )/(2x4.2x8.314) = 382.32 K

Energy of the gas B after mixing can be written as

(3/2)n2RT = KB + ΔK

⟹ (3/2) x 3.2 x 8.314 x 382.32 = KB + 600

⟹ KB = [(3/2) x 3.2 x 8.314 x 382.32] - 600

⟹ KB = 14657.32 J

6 0

3 years ago

What is the net force at the equilibrium point? Derive an equation for the location of the equilibrium point based on the accele

Vitek1552 [10]

To find a general equilibrium point for a spring based on the hook law, it is possible to start from the following premise:

Hook's law is given by:

$F = k\Delta X$

Where,

k= Spring Constant

$\Delta X =$ Change in Length

F = Force

When there is a Mass we have two force acting at the System:

W= mg

Where W is the force product of the weigth. Then the force net can be defined as,

$F_{net} = F+W$

But we have a system in equilibrium, so

$0 = K\Delta X -mg$

We find the equilibrium for any location when

$\Delta X = \frac{mg}{k}$

4 0

2 years ago

The freezer compartment in a conventional refrigerator can be modeled as a rectangular cavity 0.3 m high and 0.25 m wide with a

Mkey [24]

Answer:

Thickness of Styrofoam insulation is 0.02741 m.

Explanation:

Given that,

Height = 0.25 m

Depth = 0.5 m

Power = 400 W

Temperature = 33°C

We need to calculate the area of Styrofoam

Using formula of area

$A=2(lb+bh+hl)$

Put the value into the formula

$A=2(0.3\times0.5+0.25\times0.5+0.5\times0.3)$

$A=0.85\ m^2$

Inner surface temperature of freezer

$T_{i}=-10°C=263\ K$

Outer surface temperature of freezer

$T_{o}=33+273=306\ K$

We need to calculate the thickness of Styrofoam insulation

Using Fourier law,

$q=\dfrac{kA}{L}(T_{o}-T_{i})$

$L=\dfrac{kA}{q}(T_{o}-T_{i})$

Put the value into the formula

$L=\dfrac{0.30\times0.85}{400}(306-263)$

$L=0.02741\ m$

Hence, Thickness of Styrofoam insulation is 0.02741 m.

3 0

2 years ago

A ball is hurled straight up at a speed of 15 m/s, leaving the hand of the thrower 2.00 m above the ground. Compute the times an

SVETLANKA909090 [29]

Answer:

5.37 m/s

0.98 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -9.81\times 10+15^2}\\\Rightarrow v=5.37\ m/s$

Velocity of the ball when it passes an observer sitting at a window is 5.37 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{5.37-15}{-9.81}\\\Rightarrow t=0.98\ s$

Time taken by the ball to pass the observer sitting at a window is 0.98 seconds

3 0

2 years ago