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Scorpion4ik [409]
2 years ago
15

Convert (a) 50 oF, (b) 80 oF, (c) 95 oF to Celsius

Physics
1 answer:
storchak [24]2 years ago
5 0
I really need these points thx a lot
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The period T of a simple pendulum depends
Alenkinab [10]

Answer:

too old

Explanation:

7 0
2 years ago
A sinusoidal wave of angular frequency 1,203 rad/s and amplitude 3.1 mm is sent along a cord with linear density 3.9 g/m and ten
kobusy [5.1K]

Answer:

18.7842493212 W

Explanation:

T = Tension = 1871 N

\mu = Linear density = 3.9 g/m

y = Amplitude = 3.1 mm

\omega = Angular frequency = 1203 rad/s

Average rate of energy transfer is given by

P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W

7 0
3 years ago
If the resultant force acting on a 2.0 kg object is equal to (3.0î + 4.0ĵ) N, what is the change in kinetic energy as the object
12345 [234]

Answer:

ΔK = 24 joules.

Explanation:

ΔK = Work done on the object

Work is equal to the dot product of force supplied and the displacement of the object.

W = F * Δs

Δs can be found by subtracting the vectors (7.0, -8.0) and (11.0, -5.0), which is written as Δs = (11.0 - 7.0, -5.0 - -8.0) which equals (4.0, 3.0).

This gives us

W = < 3, 4 > * < 4, 3 > = (3*4)+(4*3) = 24 J

3 0
1 year ago
Lillle is running. She increases her initial speed of 30 km/h to 40 km/h so she
Alex777 [14]

Answer

200km {h}^{ - 2}

Explanation

Acceleration =  \frac{final \:  \:  \: velocity - initial \:  \: velocity }{time}  \\  =  \frac{(40 - 30)km {h}^{ - 1} }{0.05h}  \\  =  \frac{10}{0.05}  \\  = 200km {h}^{ - 2}

Hope this helps you.

Let me know if you have any other questions :-):-)

6 0
3 years ago
Read 2 more answers
Heat is added to a 2kg piece of ice at a rate of 793kW. How long will it take for ice to melt if it was initially 0?
Ede4ka [16]

Answer:

0.84 s

Explanation:

Step 1

Given information:

Mass of the ice (m) = 2.0 kg

Heat transfer rate (Q/T) = 793.0 kW

Latent heat of fusion of ice (Lf) = 334 kJ/kg

\frac{Q}{T}  =  \frac{mLf}{T}

Substituting the corresponding values we have:

793.0 kW=  \frac{2.0kg(334 kJ/kg)}{T} \\  T  =  \frac{2.0kg(334 kJ/kg)}{793.0kW}  =  \frac{668kJ}{793kW}   \\  = 0.84s

8 0
2 years ago
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