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lana [24]
3 years ago
9

Please help me to solve this and give me the summary of answer

Physics
1 answer:
Neko [114]3 years ago
7 0
Using the principle of floatation.

u = w............(a)

Upthrust of fluid is equal to the weight of the object.

Let the volume of the wood be V.

The upthrust u, is related to the volume submerged in water, and that is 1/5 of it volume, that is (1/5)V = 0.2V

Formula for upthrust, u = vdg

where v = volume of fluid displaced
d = density of fluid
g = acceleration due to gravity

weight, w = mg
where m = mass
g = acceleration due to gravity

From (a)

                     u = w

                 vdg =  mg      Cancel out g

                   vd  =  m
 
The v  is equal to 0.2V, which is the submerged volume. Notice that the small letter v is volume of fluid displaced, and capital V is the volume of the solid.

d is density of fluid which is water in this case, 1000 kg/m³

         0.2V * 1000 =  m

           200V =  m

Hence the mass of the object is  200V  kg.

But Density of solid =  Mass of solid / Volume of solid

                                 =    200V / V

                                 =    200 kg/m³

Density of solid = 200 kg/m³     
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A "590-W" electric heater is designed to operate from 120-V lines.
lukranit [14]

Answer:

a) 24.4 Ω

b) 4.92 A

c) 495.9 W

d)

c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.

Explanation:

b)

The formula for power is:

P = IV

where,

P = Power of heater = 590 W

V = Voltage it takes = 120 V

I = Current Drawn = ?

Therefore,

590 W = (I)(120 V)

I = 590 W/120 V

<u>I = 4.92 A</u>

<u></u>

a)

From Ohm's Law:

V = IR

R = V/I

Therefore,

R = 120 V/4.92 A

<u>R = 24.4 Ω</u>

<u></u>

c)

For constant resistance and 110 V the power becomes:

P = V²/R

Therefore,

P = (110 V)²/24.4 Ω

<u>P = 495.9 W</u>

<u></u>

d)

If the resistance decreases, it will increase the current according to Ohm's Law. As a result of increase in current the power shall increase according to formula (P = VI). Therefore, correct option is:

<u>c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.</u>

7 0
2 years ago
This problem is based on the whole idea of pressure but I’m having trouble on when the area circle formula is included.
Mice21 [21]

Answer:

6.23x10^6Pa

Explanation:

Data obtained from the question include:

F (force) = 490N

r (radius) = 0.005m

A (area of the circlular heel) =?

P (pressure) =.?

First, we'll begin by calculating the area of the circlular heel. This is illustrated below:

Area of circle = πr^2

Area = 22/7 x (0.00)^2

Area = 7.86x10^-5m^2

Pressure is simply force per unit area. It represented mathematically as

Pressure = Force /Area

Pressure = 490/7.86x10^-5

Pressure = 6.23x10^6N/m2

Recall: 1N/m2 = 1Pa

Therefore, 6.23x10^6N/m2 = 6.23x10^6Pa

Therefore, the woman exert a pressure of 6.23x10^6Pa on the floor

8 0
3 years ago
A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with
Julli [10]

Explanation:

Given that,

Capacitor C=1.66\ \mu F

Resistor R=80.0\ \Omega

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

f_{c}=\dfrac{1}{2\pi R C}

Where, R = resistor

C = capacitor

Put the value into the formula

f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}

f_{c}=1198.45\ Hz

(B). We need to calculate the V_{R} when f = \dfrac{1}{2f_{c}}

Using formula of  V_{R}

V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})

Put the value into the formula

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=2.280\ Volt

(C). We need to calculate the V_{R} when f = f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=3.606\ Volt

(D). We need to calculate the V_{R} when f = 2f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=4.561\ Volt

Hence, This is the required solution.

8 0
3 years ago
B. Find the velocity of a rider on the ride.
Stells [14]

D

Explanation:

Mutiply and alagrab to subterc

5 0
3 years ago
Why might the term red hot be misleading? (relating to stars)
Allushta [10]

red hot is hot, but other colours are even hotter. stars may be hotter than red hot.

there is also something calle the red shift.

3 0
3 years ago
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