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3 years ago

On a cold winter day, the flow of heat is from the outside in. A.True B.False

Physics

2 answers:

fredd [130]3 years ago

8 0

Your answer s obviously false

lbvjy [14]3 years ago

7 0

Answer:

False

Explanation:

ooga booga, cold winter day cold outside, heat not outside ooga booga.

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If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

vovikov84 [41]

Complete Question

The speed of a transverse wave on a string of length L and mass m under T is given by the formula

$v=\sqrt{\frac{T}{(m/l)}}$

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

Answer:

$(m/l)=\frac{10}{V^2}$

Explanation:

From the question we are told that

Speed of a transverse wave given by

$v=\sqrt{\frac{T}{(m/l)}}$

Maximum Tension is $T=10.0N$

Generally making $(m/l)$ subject from the equation mathematically we have

$v=\sqrt{\frac{T}{(m/l)}}$

$v^2=\frac{T}{(m/l)}$

$(m/l)=\frac{T}{V^2}$

$(m/l)=\frac{10}{V^2}$

Therefore the Linear mass in terms of Velocity is given by

$(m/l)=\frac{10}{V^2}$

8 0

3 years ago

How does the frequency of infrared electromagnetic waves compare with the frequency of radio and microwaves?

iVinArrow [24]

Answer:

Answer is B.

Because the wavelength of infrared is shorter than microwave radiation

8 0

3 years ago

If he leaves the ramp with a speed of 35.0 m/s and has a speed of 33.0 m/s at the top of his trajectory, determine his maximum h

nadezda [96]

Answer:

H = 6.93 m

Explanation:

given data

velocity v = 35 m/s

horizontal component Vx = 33 m/s

solution

we get here maximum height so first we get vertical component here that is express as

Vy = $\sqrt{v^2- Vx^2}$ .........................1

put here value

Vy = $\sqrt{35^2- 33^2}$

Vy = 11.66 m/s

and

now we get height

H = $\frac{Vy^2}{2g}$ .............................2

put here value

H = $\frac{11.66^2}{2\times 9.8}$

H = 6.93 m

7 0

3 years ago

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

Elena L [17]

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

$a_n = \frac{330 }{260}$

$a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

$a_e = \frac{F_{keel}}{m_b }$

substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

The resultant acceleration is

$a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

$a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

5 0

3 years ago

The owner of a company that manufactures drinking cups decides it would be impressive to build an inground swimming pool that is

garri49 [273]

Answer:

The depth is 5.15 m.

Explanation:

Lets take the depth of the pool = h m

The atmospheric pressure ,P = 101235 N/m²

The area of the top = A m²

The area of the bottom = a m²

Given that A= 1.5 a

The force on the top of the pool = P A

The total pressure on the bottom = P + ρ g h

ρ =Density of the water = 1000 kg/m³

The total pressure at the bottom of the pool = (P + ρ g h) a

The bottom and the top force is same

(P + ρ g h) a = P A

P a +ρ g h a = P A

ρ g h a = P A - P a

$h=\dfrac{P ( A-a)}{\rho g a}$

$h=\dfrac{P ( 1.5 a-a)}{\rho g a}$

$h=\dfrac{P ( 1.5- 1)}{\rho g}$

$h=\dfrac{101235 ( 1.5- 1)}{1000\times 9.81}\ m$

h=5.15 m

The depth is 5.15 m.

7 0

3 years ago