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zysi [14]
3 years ago
13

It takes 4.186 J of energy to raise the temperature of lg of water by 1°C. How fast would a 2 g

Physics
1 answer:
mezya [45]3 years ago
5 0

Answer:

1. D - 3.6 x 10^4 J

2. D - 0.04

3. A - 30.3 m/s

4. C - 3.67 x 10^8 J

5. A - 38.6 m

6. C - 7.07 m/s

7. B - 4.21 g

8. B - -6.56 J

9. A - 65 m/s

10. B - 116 Days

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Identify one type of electromagnetic energy that has a shorter wave length than green light
11111nata11111 [884]
When you look at the acronym ROY G. BIV, you get the wavelengths from longest to shortest.  Since Green is in the middle of ROY G. BIV, anything to the right of it would have a shorter wavelength.  You can choose in the visible range Blue, Indigo, or Violet.
5 0
4 years ago
Consider a car engine running at constantspeed. That is, the crankshaft of the en-gine rotates at constant angular velocity whil
Irina-Kira [14]

Answer:

The value is  |v| = 7.39 \  m/s

Explanation:

From the question we are told that

    The angular speed is  w = 2030 \ rpm = \frac{2030 * 2 * \pi }{ 60} =  212.61 \ rad/s

    The distance between the minimum and maximum external position is  d = 6.95 \ cm  = 0.0695 \ m

Generally the amplitude of the crank shaft is mathematically represented as

         A =  \frac{d}{2}      

=>     A =  \frac{0.0695}{2}    

=>     A =  0.03475 \  m

Generally the maximum speed of the piston is mathematically represented as

        |v| = A * w  

=>    |v| = 0.03475 * 212.61

=>    |v| = 7.39 \  m/s

     

7 0
3 years ago
If a loaded truck that accelerates at 1 meter per second squared loses its load and has three-fourths of the origional mass, wha
Nina [5.8K]
I believe it'd accelerate at 1.25 m/s^2 instead of 1, as it lost 1/4 of its mass (.25), so now it is .25 of 1 faster.

8 0
3 years ago
49. In what direction does the force of friction act?
DedPeter [7]
I believe it always acts in the direction opposite to that of the object in motion, so if the force was applied to the left then friction acts in the right

hope that helps
5 0
3 years ago
Two projectiles A and B are fired simultaneously from a level, horizontal surface. The projectiles are initially 62.2 m apart. P
Svetach [21]

Let the point where A is launched act as the origin, so that the horizontal positions at time <em>t</em> of the respective projectiles are

• A : <em>x</em> = (19.5 m/s) cos(30°) <em>t</em>

• B : <em>x</em> = 62.2 m + (19.5 m/s) cos(60°) <em>t</em>

These positions are the same at the moment one projectile is directly above the other, which happens for time <em>t</em> such that

(19.5 m/s) cos(30°) <em>t</em> = 62.2 m + (19.5 m/s) cos(60°) <em>t</em>

Solve for <em>t</em> :

(19.5 m/s) (cos(30°) - cos(60°)) <em>t</em> = 62.2 m

<em>t</em> = (62.2 m) / ((19.5 m/s) (cos(30°) - cos(60°))

<em>t</em> ≈ 8.71 s

6 0
3 years ago
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