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3 years ago

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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The graph below represents changes in water. What process is occurring during section “B” of the graph?

IrinaK [193]

The scientific inquiry process

6 0

3 years ago

A 75-g projectile traveling at 600 m /s strikes and becomes embedded in the 40-kg block, which is ini-tially stationary. Compute

levacccp [35]

Explanation:

The given data is as follows.

Mass, m = 75 g

Velocity, v = 600 m/s

As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.

$m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

where, $m_{1}$ = mass of the projectile

$m_{2}$ = mass of block

v = velocity after the impact

Now, putting the given values into the above formula as follows.

$m_{1}v_{1_{i}} = (m_{1} + m_{2})vi$

$75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v$

= $\frac{45}{50.075}$

v = 0.898 m/s

Now, equation for energy is as follows.

E = $\frac{1}{2}mv^{2}$

= $\frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}$

= 13500 J

Now, energy after the impact will be as follows.

E' = $\frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}$

= 20.19 J

Therefore, energy lost will be calculated as follows.

$\Delta E$ = E E'

= (13500 - 20) J

= 13480 J

And, n = $\frac{\Delta E}{E}$

= $\frac{13480}{13500} \times 100$

= 99.85

= 99.9%

Thus, we can conclude that percentage n of the original system energy E is 99.9%.

7 0

3 years ago

Read 2 more answers

saad has mass 80kg when resting on the ground at the equator what will be the centripetal acceleration on saad if the radius of

Sidana [21]

Thank you for posting your question here at brainly. Below is the answer:

m = 80 kg
R = 6.4 x 10^6 m
Earth completes one rotation in 24 hours
Thus, T = 24 x 3600 s
Centripetral acceleration is given by
ac = w^2R
ac = (2 pie/T)^2R
ac = (2 pie/T)^2R = 4 pie^2/(24 x 3600)^2 x 6.4 x 10^6
ac = 0.034 m/s^2

8 0

4 years ago

A small sphere has a harge of 9uC and other small sphere has a charge of 4uC.

Helga [31]

Answer:

Electrical force, F = 90 N

Explanation:

It is given that,

Charge on sphere 1, $q_1=9\ \mu C=9\times 10^{-6}\ C$

Charge on sphere 2, $q_1=4\ \mu C=4\times 10^{-6}\ C$

Distance between two spheres, d = 6 cm = 0.06 m

Let F is the electrical force between them. It is given by the formula of electric force which is directly proportional to the product of charges and inversely proportional to the square of distance between them such that,

$F=k\dfrac{q_1q_2}{d^2}$

$F=9\times 10^9\times \dfrac{9\times 10^{-6}\times 4\times 10^{-6}}{(0.06)^2}$

F = 90 N

So, the electrical force between them is 90 N. Hence, this is the required solution.

7 0

3 years ago

Suppose a 1300 kg car is traveling around a circular curve in a road at a constant

Scrat [10]

Answer:

F = 4212 N

Explanation:

Given that,

Mass of a car, m = 1300 kg

Speed of car on the road is 9 m/s

Radius of curve, r = 25 m

We need to find the magnitude of the unbalanced force that steers the car out of its natural straight- line path. The force is called centripetal force. It can be given by :

$F=\dfrac{mv^2}{r}\\\\F=\dfrac{1300\times 9^2}{25}\\\\F=4212\ N$

So, the force has a magnitude of 4212 N

4 0

3 years ago