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3 years ago

The temperature of a room is 77°F what would be the temperature in Celsius scale

Physics

1 answer:

anygoal [31]3 years ago

5 0

Formulas change from F to degree C : C = ( F - 32 )/1.8

so we have (77-32)/1.8 = 25 oC

ok done. Thank to me :>

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HELP PLS!!!! Light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is th

maw [93]

You've already told us the speed in ft/s . It's right there in the question. You said that light travels about 982,080,000 ft/s.

We don't know how accurate that number is, but for purposes of THIS question, that's the number we're going with.

In scientific notation, it's written . . . 9.8208 x 10⁸ ft/s .

We don't know where you were going with the number of seconds in a year. But to answer the question that you eventually asked, it turned out that we don't even need it.

6 0

3 years ago

Two identical waves are moving in the same direction with the same speed. If the amplitude of the combination of the two waves i

umka2103 [35]

Answer:

The phase difference between these two waves is 141.1⁰

Explanation:

The displacement of the wave is given as;

$Y = y_xSin(Kx - \omega t)+y_xSin(Kx- \omega t + \phi)\\\\Y = 2y_xCos(\frac{1}{2} \phi)Sine(Kx- \omega t + \frac{1}{2} \phi)$

Amplitude, A = 2yₓCos(¹/₂Φ)

Since the amplitude of the combination is 1.5 times that of one of the original amplitudes = yₓ = 1.5 × A = 1.5A

A = 2(1.5A)Cos(¹/₂Φ)

A = 3ACos(¹/₂Φ)

¹/₃ = Cos(¹/₂Φ)

(¹/₂Φ) = Cos ⁻(0.3333)

(¹/₂Φ) = 70.55°

Φ = 141.1°

The phase difference between these two waves is 141.1⁰

4 0

3 years ago

3. What conclusion can you make about the electric field strength between two parallel plates? Explain your answer referencing P

KIM [24]

Answer:

From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases

Explanation:

I cannot find any attached photo, but we can proceed anyways theoretically.

The electric field strength (E) at any point in an electric field is the force experienced by a unit positive charge (Q) at that point

i.e

$E=\frac{F}{Q}$

But the force F

$F= \frac{kQ1Q2}{r^2}$

But the electric field intensity due to a point charge Q at a distance r meters away is given by

$E= \frac{\frac{kQ1Q2}{r^2}}{Q} \\\\\E= \frac{Q1}{4\pi er^2 }$

From the relation above we can conclude that the as the distance between the two plate increases the electric field strength decreases

6 0

3 years ago

A girl on a spinning amusements park is 12m from the center of the ride and has a centripetal acceleration of 17 m/s^2. What is

Tanya [424]

Answer: 14.28 m/s

Explanation:

Assuming the girl is spinning with uniform circular motion, her centripetal acceleration $a_{c}$ is given by the following equation:

$a_{c}=\frac{V^{2}}{r}$ (1)

Where:

$a_{c}=17 m/s^{2}$ is the centripetal acceleration

$V$ is the tangential speed

$r=12 m$ is the radius of the circle

Isolating $V$ from (1):

$V=\sqrt{a_{c}r}$ (2)

$V=\sqrt{(17 m/s^{2})(12 m)}$

Finally:

$V=14.28 m/s$ This is the girl's tangential speed

3 0

3 years ago

A large blue marble of mass 3.5 g is moving to the right with a velocity of 15 cm/s. The large marble hits a small red marble of

Aneli [31]

Explanation:

Let's solve the problem by using conservation of momentum:

m1 u1 + m2 u2 = m1 v1 + m2 v2

where:

m1 = 3.5 g is the mass of the blue marble

m2= 1.2 g is the mass of the red marble

u1 = 15 cm/s is the initial velocity of the blue marble

u2 = 3.5 cm/s is the initial velocity of the red marble

v1 = 5.5 cm/s is the final velocity of the blue marble

We can find the final velocity of the red marble by re-arranging the equation and solving for v2 :

v2 = 1/m²(m1 u1 + m2 u2 - m1 v1)=

=1/1.2(3.5×15+1.2×3.5−3.5×5.5 )=31cm/s

7 0

4 years ago

The temperature of a room is 77°F what would be the temperature in Celsius scale​

The temperature of a room is 77°F what would be the temperature in Celsius scale