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prohojiy [21]
3 years ago
14

When 108 g of water at a temperature of 22.5?

Physics
1 answer:
FrozenT [24]3 years ago
3 0
<span>ΔT for the first sample is the total samples final temp, minus the first sample's initial temp (47.9-22.5), so 25.4oC. Calculating q for the first sample as 108g x 4.18 J/g C x 25.4oC = 11466.58 Joules Figuring that since the first sample gained heat, the second sample must have provided the heat, so doing the calculation for the second sample, I used q=mCΔT 11466.58 Joules = 65.1g x 4.18 J / g C x ΔT 11466.58/(65.1gx4.18)=ΔT ΔT=42.14oC So, since second sample lost heat, it's initial temperature was 90.04oC (47.9oC final temperature of mixture + 42.14oC ΔT of second sample).</span>
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Answer:

Explanation:

It can be increased by: increasing the rate of rotation. Increasing the strength of the magnetic field. Increasing the number of turns on the coil.

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4 0
3 years ago
What is the acceleration of a 349 kg object that moved with a force of 750 N?
zavuch27 [327]

Answer:

<h3>The answer is 2.15 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

a =  \frac{f}{m}  \\

where

f is the force

m is the mass

From the question we have

a =  \frac{750}{349}  \\  = 2.14899713...

We have the final answer as

<h3>2.15 m/s²</h3>

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4 0
3 years ago
How large must the coefficient of static friction be between the tires and road, if a car rounds a level curve of radius 85 m at
tatuchka [14]

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0.66

Explanation:

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4 0
3 years ago
What is the main difference between magma and lava
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4 0
3 years ago
On the Apollo 14 mission to the moon, astronaut Alan Shepard hit a golf ball with a golf club improvised from a tool. The free-f
aliya0001 [1]

Answer:

15.3 s and 332 m

Explanation:

With the launch of projectiles expressions we can solve this problem, with the acceleration of the moon

    gm = 1/6 ge

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We calculate the range

    R = Vo² sin 2θ  / g

    R = 25² sin (2 30) / 1.63

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We will calculate the time of flight,

   Y = Voy t – ½ g t2  

   Voy = Vo sin θ

When the ball reaches the end point has the same initial  height Y=0

0 = Vo sin  t – ½  g t2

0 = 25 sin (30)  t – ½ 1.63 t2

0= 12.5 t –  0.815 t2

We solve the equation

0= t ( 12.5 -0.815 t)

 t=0 s

t= 15.3 s

The value of zero corresponds to the departure point and the flight time is 15.3 s

Let's calculate the reach on earth

R2 = 25² sin (2 30) / 9.8

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R/R2 = 6

Therefore the ball travels a distance six times greater on the moon than on Earth

5 0
3 years ago
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