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3 years ago

Which of the following will cause water to change to ice!

Physics

2 answers:

Contact [7]3 years ago

7 0

Explanation:

D removing heat from the water

olganol [36]3 years ago

3 0

Answer:

Explanation:

removing heat from water

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Write the relation between horsepower and Watt?

Vadim26 [7]

Explanation:

Horsepower is related to watt as: 1 horsepower = 746 W = 750 W approximately. 1 horsepower (HP) = 750 watt (approx.) The power is one watt when one joule of work is done in one second.

8 0

2 years ago

Read 2 more answers

What is most likely the amount of energy available at a trophic level of primary consumers if the amount of energy available to

docker41 [41]

Answer:

200 kilocalories

Explanation:

6 0

3 years ago

1. Determine the magnitude of two equal but opposite charges if they attract one another with a force of 0.7N when at distance o

adell [148]

Answer:

q = 2.65 10⁻⁶ C

Explanation:

For this exercise we use Coulomb's law

F = $k \frac{q_1q_2}{r^2}$

In this case they indicate that the load is of equal magnitude

q₁ = q₂ = q

the force is attractive because the signs of the charges are opposite

F = $k \ \frac{q^2}{r^2}$

q = $\sqrt{\frac{F \ r^2}{k} }$

we calculate

q = $\sqrt{\frac{0.7 \ 0.3^2 }{9 \ 10^9} }$

q = $\sqrt{7 \ 10^{-12} }$ Ra 7 10-12

q = 2.65 10⁻⁶ C

7 0

3 years ago

1 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supp

Tanzania [10]

Answer with explanation:

We are given that

Mass of ball, $m_1=$ 75 g= $\frac{75}{1000}=0075kg$

1 kg=1000 g

Height, $h_1=1.6 m$

$h_2=0.6 m$

Horizontal velocity, $v_x=2 m/s$

Mass of plate $m_2=400 g=\frac{400}{1000}=0.4 kg$

a.Initial velocity of plate, $u_2=0$

Velocity before impact= $u_1=\sqrt{2gh_1}=\sqrt{2\times 9.8\times 1.6}=5.6m/s$

Where $g=9.8 m/s^2$

Velocity after impact, $v_1=\sqrt{2gh_2}=\sqrt{2\times 9.8\times 0.6}=3.4m/s$

According to law of conservation of momentum

$m_1u_1+m_2u_1=-m_1v_1+m_2v_2$

Substitute the values

$0.075\times 5.6+0=-0.075\times 3.4+0.4v_2$

$0.4v_2=0.075\times 5.6+0.075\times 3.4$

$v_2=\frac{0.075\times 5.6+0.075\times 3.4}{0.4}=1.69 m/s$

Velocity of plate=1.69 m/s

b.Initial energy= $\frac{1}{2}m_1v^2_x+m_1gh_1=\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 1.6=1.326 J$

Final energy= $\frac{1}{2}m_1v^2_x+m_1gh_2+\frac{1}{2}m_2v^2_2$

Final energy= $\frac{1}{2}(0.075)(2^2)+0.075\times 9.8\times 0.6+\frac{1}{2}(0.4)(1.69)^2=1.162 J$

Energy lost due to compact=Initial energy-final energy=1.326-1.162=0.164 J

6 0

3 years ago

What type of circuit is illustrated?

Reil [10]

Answer:

I beleive its B

Explanation:

If not then A but I'm positive its B

6 0

3 years ago