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Romashka-Z-Leto [24]
3 years ago
6

An aluminum bar has a mass of 9 kg in the air. Calculate its volume. Now imagine that you submerge the aluminum bar in water han

ging it from a rope. Calculate the apparent weight it would have in the water.
Physics
1 answer:
yan [13]3 years ago
7 0

Answer:

55.7 N

Explanation:

The density of aluminum is 2710 kg/m³.  So its volume is:

V = (9 kg) / (2710 kg/m³)

V = 0.00332 m³

The apparent weight is the actual weight minus the buoyant force.

N = mg − B

N = mg − ρVg

N = g (m − ρV)

N = (9.8 m/s²) (9 kg − (1000 kg/m³) (0.0332 m³))

N = 55.7 N

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The diagram below shows a closed system of two tanks that each contain water.
lidiya [134]
Lmalemwlsnlenekenfndelenekf
4 0
3 years ago
Will Mark Brainliest if Correct PLZ!!!!! A bullet is shot at some angle above the horizontal at an initial velocity of 87m/s on
qaws [65]

Answer:

≅50°

Explanation:

We have a bullet flying through the air with only gravity pulling it down, so let's use one of our kinematic equations:

Δx=V₀t+at²/2

And since we're using Δx, V₀ should really be the initial velocity in the x-direction. So:

Δx=(V₀cosθ)t+at²/2

Now luckily we are given everything we need to solve (or you found the info before posting here):

  • Δx=760 m
  • V₀=87 m/s
  • t=13.6 s
  • a=g=-9.8 m/s²; however, at 760 m, the acceleration of the bullet is 0 because it has already hit the ground at this point!

With that we can plug the values in to get:

760=(87)(cos\theta )(13.6)+\frac{(0)(13.6^{2}) }{2}

760=(1183.2)(cos\theta)

cos\theta=\frac{760}{1183.2}

\theta=cos^{-1}(\frac{760}{1183.2})\approx50^{o}

3 0
3 years ago
A train travels 70 kilometers in 1 hours, and then 63 kilometers in 2 hours. What is its average speed?
lozanna [386]
The answer is:  " 44 \frac{1}{3} km " ; 
                              or; write as:  " 44.333 km " .
___________________________________________________________
Explanation:
___________________________________________________________
(70 km + 63 km) ÷ (2 + 1 ) =  133 km ÷ 3 =  " 44 \frac{1}{3} km "  ; 
                                                                            or; write as:  " 44.333 km " .
___________________________________________________________
3 0
4 years ago
A meter stick balances horizontally on a knife-edge at the 51 cm mark. With two nickels stacked over the 6.0 cm mark, the stick
Oliga [24]

Answer:

65g

Explanation:

Two main conditions for equilibrium are:

I. The resultant force must be equal to zero. That is, sum of the forces acting in one direction about a point must be equal to the sum of the forces acting in the opposite direction about the same point.

II. The resultant moment must be equal to zero. That is, sum of the moments in one direction about a point must be equal to the sum of the moments in another direction about the same point.

For the above question,

the 51cm mark is the point where the resultant weight of the meter stick lies,

the pivot or point is the 45cm mark where the stick balanced when 2 nickels ( total mass (5.0g x 2) 10g were placed at the 6cm mark.

Using the conversion factor:

1000g(1kg) = 10N, we can convert mass to weight, calculate the weight of the meter stick then reconvert to mass.

That is,

mass of 2 nickels = 10g = 10/1000 = 0.01N.

Moment = Force x distance from line of force to pivot of rotation

Applying the principle of equilibrium,

Moment of left side = Moment of right side

0.01 x (45-6) = W x (51-45)

Where W = weight of the meter stick

W x 6 = 0.01 x 39

W x 6 = 0.39

W = 0.39/6

W= 0.065N

Therefore, mass of meter stick = 0.065 x 1000 = 65g.

4 0
3 years ago
How does the Sun's gravity influence tides?
nevsk [136]

Answer: <em>Around each new moon and full moon, the sun, Earth, and moon arrange themselves more or less along a line in space. Then the pull on the tides increases, because the gravity of the sun reinforces the moon's gravity. ... This is the spring tide: the highest (and lowest) tide. Spring tides are not named for the season.</em>

8 0
3 years ago
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