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3 years ago

Which are the upthrust forces?

Physics

2 answers:

anygoal [31]3 years ago

6 0

Explanation:

An object that is partly or completely submerged experiences a greater pressure on its bottom surface than on its top.

Bond [772]3 years ago

4 0

Answer:

upthrust force is an upward force exerted by a fluid that opposes the weight of a partially or fully immersed object

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A 357 kg box is pulled 8.00 m up a 30° frictionless, inclined plane by an external force of 4925 N that acts parallel to the pla

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3 years ago

A net force, the magnitude of which is 3800 N, accelerates a 1260-kg vehicle for 10.0 s. The vehicle travels 50.0 m during this

Novay_Z [31]

Answer:

SEE EXPLANATION

Explanation:

$p = \frac{fd}{t} \\ where \: \\p = power \\ f = force \\ d = distance \\ and \: t = time \\ \\ p = \frac{3800 \times 50}{10} \\ p = \frac{190000}{10} \\ p = 19000w$

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3 years ago

If a ball is thrown straight up into the air with an initial velocity of 65 ft/s, its height in feet after t seconds is given by

fgiga [73]

Answer:

a) $v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) $v=65-32(2)=1ft/s$

Explanation:

From the exercise we got the ball's equation of position:

$y=65t-16t^{2}$

a) To find the average velocity at the given time we need to use the following formula:

$v=\frac{y_{2}-y_{1} }{t_{2}-t_{1} }$

Being said that, we need to find the ball's position at t=2, t=2.5, t=2.1, t=2.01, t=2.001

$y_{t=2}=65(2)-16(2)^{2} =66ft$

$y_{t=2.5}=65(2.5)-16(2.5)^{2} =62.5ft$

$v_{1}=\frac{(62.5-66)ft}{(2.5-2)s}=-7ft/s$

$y_{t=2.1}=65(2.1)-16(2.1)^{2} =65.94ft$

$v_{2}=\frac{(65.94-66)ft}{(2.1-2)s}=-0.6ft/s$

$y_{t=2.01}=65(2.01)-16(2.01)^{2} =66.0084ft$

$v_{3}=\frac{(66.0084-66)ft}{(2.01-2)s}=0.84ft/s$

$y_{t=2.001}=65(2.001)-16(2.001)^{2} =66.001ft$

$v_{4}=\frac{(66.001-66)ft}{(2.001-2)s}=1ft/s$

b) To find the instantaneous velocity we need to derivate the equation

$v=\frac{df}{dt}=65-32t$

$v=65-32(2)=1ft/s$

7 0

3 years ago