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3 years ago

You have a 16-kg suitcase. what work did you do when you have slowly lifted it 0.80 m upward?

1 answer:

8 0

The formula is:
Work = Force · Displacement
F = m · g
F = 16 kg · 9.8 m/s² = 156.8 N
and we know that:
d = 0.8 m
W = 156.8 N · 0.8 m = 125.44 J
Answer:
W = 125.44 J.

You might be interested in

¿A qué velocidad debe circular un auto de carreras para recorrer 87 km en 20 min? (pasar a metros por segundo m/s)

Alexxx [7]

Answer:

Velocidad en m / s = 72,25 m / s

Explanation:

Dado

Distancia a recorrer por el coche de carreras = 87 Km

1 km = 1000 m

Por lo tanto, 87 km = 87000 m

Tiempo necesario para viajar 87 km / 87000 metros = 20 minutos o 20 * 60 = 1200 segundos

Velocidad en m / s = 87000/1200

Velocidad en m / s = 72,25 m / s

7 0

3 years ago

Diffusion and osmosis are forms of passive transport.<br><br> True<br> False

pishuonlain [190]

Answer:

True. Diffusion and osmosis are forms of passive transport.

Explanation:

In diffusion, particles move from an area of higher concentration to one of lower concentration until equilibrium is reached.

In osmosis, a semipermeable membrane is present, so only the solvent molecules are free to move to equalize concentration.

8 0

3 years ago

Read 2 more answers

In the Sun, fusion reactions create helium nuclei. To form each helium

seraphim [82]

D is the answer. Bc the pointy

6 0

3 years ago

Read 2 more answers

What is the mass of an object that weighs 500 newtons on earth?

Eva8 [605]

The mass would be 51 kg. So b is the right answer.

6 0

3 years ago

A speedboat is approaching a dock at 25 m/s (56 mph). When the dock is 150 m away, the driver begins to slow down. a) What accel

trasher [3.6K]

Answer:

a) -2.038 m/s²

b) 40.33 mph

c) 312.5 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-25^2}{2\times 150}\\\Rightarrow a=-2.083\ m/s^2$

Acceleration of the boat is -2.083 m/s² if the boat will stop at 150 m.

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -1\times 150+25^2}\\\Rightarrow v=18.03\ m/s$

Speed of the boat by when it will hit the dock is 18.03 m/s

Converting to mph

$1\ mile=1609.34\ m$

$1\ h=3600\ seconds$

$18.03\times \frac{3600}{1609.34}=40.33\ mph$

Speed of the boat by when it will hit the dock is 40.33 mph

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-25^2}{2\times -1}\\\Rightarrow s=312.5\ m$

The distance at which the boat will have to start decelerating is 312.5 m

5 0

3 years ago