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belka [17]
3 years ago
15

How much heat is required to raise the temperature of 45.5 g of water from its melting point to its boiling point?

Physics
1 answer:
KATRIN_1 [288]3 years ago
3 0
First, melt the ice, H2O(s): 

<span>45.5gH2O x 333J/1gH2O = 15,152 J </span>

<span>Next, raise the temperature of the water, H2O(l): </span>

<span>45.5gH2O x 100degC x 4.184J/1gH2O-1degC = 19,037 J </span>

<span>Finally, addem all up: </span>

<span>34,200 J = 34.2 kJ to three significant figures</span>
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I need help please..its too complicated
Charra [1.4K]

Question: What was his initial velocity?

Answer: 3.62 m/s

3 0
2 years ago
What is the power output of a pump which can raise 60kg of water height to height of 10m every minute
sattari [20]

Answer:

<h3>Power = Work Done/time</h3>

=> Power = 60×10×10/60

=> Power = 6000/60

=> Power = 100 Watt

Hence the power output of a pump is 100 Watts.

8 0
2 years ago
Can someone please help with this. "A ball is thrown vertically upward with a speed of 26.6 m/s. How high does it rise? The acce
lubasha [3.4K]

v^2=u^2+2as

0=26.6^2-2x9.8xs

-26.6^2/-2x9.8=s

calculator

8 0
3 years ago
From the top of a tall building, a gun is fired. The bullet leaves the gun at a speed of 340 m/s, parallel to the ground. As the
Ivahew [28]

Answer:

The launching point is at a distance D = 962.2m and H = 39.2m

Explanation:

It would have been easier with the drawing. This problem is a projectile launching exercise, as they give us data after the window passes and the wall collides, let's calculate with this data the speeds at the point of contact with the window.

X axis

           x = Vox t

           t = x / vox

           t = 7.1 / 340

           t = 2.09 10-2 s

In this same time the height of the window fell

           Y = Voy t - ½ g t²

Let's calculate the initial vertical speed, this speed is in the window

           Voy = (Y + ½ g t²) / t

           Voy = [0.6 + ½ 9.8 (2.09 10⁻²)²] /2.09 10⁻² = 0.579 / 0.0209

            Voy = 27.7 m / s

We already have the speed at the point of contact with the window. Now let's calculate the distance (D) and height (H) to the launch point, for this we calculate the time it takes to get from the launch point to the window; at this point the vertical speed is Vy2 = 27.7 m / s

             Vy = Voy - gt₂

             Vy = 0 -g t₂

             t₂ = Vy / g

             t₂ = 27.7 / 9.8

             t₂ = 2.83 s

This is the time it also takes to travel the horizontal and vertical distance

            X = Vox t₂

            D = 340 2.83

            D = 962.2 m

           

            Y = Voy₂– ½ g t₂²

            Y = 0 - ½ g t2

            H = Y = - ½ 9.8 2.83 2

            H = 39.2 m

The launching point is at a distance D = 962.2m and H = 39.2m

6 0
3 years ago
A bar magnet is dropped toward a conducting ring lying on the floor. As the magnet falls toward the ring, does it move as a free
REY [17]

Explanation:

According to the Faraday-Lenz law, a conductive ring generates an induced current due to the change in the magnetic flux caused by the motion of the bar magnet. This induced current generates a magnetic field opposite to the magnetic field of the bar, generating an upward force that opposes the weight of the bar magnet, Therefore, it does not move as a freely  falling object.

3 0
3 years ago
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