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3 years ago

10

An ant does 1 N-m of work in dragging a 0.0020-N grain of sugar. How far does the ant drag the sugar.

1 answer:

algol133 years ago

8 0

The ant would drag the sugar 500m

Good luck!

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State the law of universal gravitation, and use examples to explain how changes in mass and changes in distance affect gravitati

juin [17]

F(g)= Gm1m2/ r^2 If mass is increased, so will the force of gravity because it is in direct relationship with the gravitational force, but if distance is increased, the force of gravity will decrease because it is indirectly related ( since it is on the bottom of the equation)

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How does space promote science education

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3 years ago

Two 84.5 g ice cubes are dropped into 30 g of water in a glass. If the water is initially at a temperature of 50 C and if the ic

Setler [38]

Answer:

final temperature will be 0 degree C

Total amount of ice will be

$m_{ice} = 182 g$

total amount of water

$m_{water} = 17 g$

Explanation:

After thermal equilibrium is achieved we can say that

Heat given by water = heat absorbed by ice cubes

so we will have

Heat given by water to reach 0 degree C

$Q_1 = m_1s_1 \Delta T_1$

$Q_1 = 0.030(4186)(50 - 0)$

$Q_1 = 6279 J$

heat absorbed by ice cubes to reach 0 degree

$Q_2 = m_2 s_2 \Delta T_2$

$Q_2 = (0.169)(2100)(30)$

$Q_2 = 10647 J$

so we will have

$Q_2 > Q_1$

so here we can say that few amount of water will freeze here to balance the heat

$10647 - 6279 = mL$

$m = \frac{10647 - 6279}{335000}$

$m = 13 g$

so final temperature will be 0 degree C

Total amount of ice will be

$m_{ice} = 84.5 + 84.5 + 13$

$m_{ice} = 182 g$

total amount of water

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$m_{water} = 17 g$

4 0

3 years ago

Lydia is 9kg underweight. Caroline has an ideal weight for her height. Viv is 8kg overweight. Who is most likely to have the hig

Crank

Answer: Viv

Explanation:

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3 years ago

Read 2 more answers

Each of the space shuttle's main engines is fed liquid hydrogen bya high-pressure pump. Turbine blades inside the pump rotateat

kvasek [131]

Answer:

a) $a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}$

b) $k = \frac{9.8}{9.74}=1.006$

Explanation:

Part a

For this case we can begin finding the period like this:

$T= \frac{1}{w} =\frac{1}{617 rad/s}=0.00162 s$

Then we know that the centripetal acceleration is given by:

$a= \frac{v^2}{r}$

And the velocity is given by:

$v=\frac{2\pi r}{T}$

If we replace this into the acceleration we got:

$a = \frac{(\frac{2\pi r}{T})^2}{r}= \frac{4 \pi^2 r}{T^2}$

And we can replace the values and we got:

$a= \frac{4 \pi^2 (0.02 m)^2}{0.00162 s}=9.74 \frac{m}{s^2}$

Part b

For this case we want to find a value of k such that:

$a= k 9.8$

Where a = 9.74, so then we can solve for k like this:

$k = \frac{9.8}{9.74}=1.006$

8 0

3 years ago