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____ [38]
3 years ago
12

A steel block has a volume of 0.08 m³ and a density of 7,840 kg/m³. What is the force of gravity acting on the block (the weight

) in water?
Physics
2 answers:
Blababa [14]3 years ago
3 0
d=\frac{m}{V}\\
\\
0.08 = \frac{m}{7,840}\rightarrow m=0.08 * 7,840=627.2 \ kg

\vec{F_g}=m.\vec{g}=627.2 *9.8=6,146.56 \ N
Anni [7]3 years ago
3 0
Density = 7840 kg per m³
Volume = 0.08 m³
Mass = 0.08 of 7840 kg = 627.2 kg

Now I'm going to give you the correct answer to your question, and then,
after that, I'm also going to give you the answer you expect.

The correct answer is:  (Ryan2 the Expert in Answer #1 is correct) . . . . .

The force of gravity on a mass doesn't depend on what it's sunk in, surrounded by,
hanging from, or resting on.  As long as it's on or near the Earth's surface, the force
of gravity acting on it is the same, no matter what else is around.

Force of gravity (weight) = (mass) x (gravity) = 627.2 kg x 9.8 m/s² = 6146.56 newtons
=============================================

The apparent weight is less, because when it's immersed in a fluid, there's
a buoyant force acting on it, which cancels part of the force of gravity.

The buoyant force = weight of the displaced water.

Assume the density of water is 1 kg per liter = 1,000 kg per m³.

Then the weight of 0.08 m³ of water is (80 kg x 9.8 m/s²) = 784 newtons.

This is the upward buoyant force on the steel when it's in water,
and it makes the steel seem 784 newtons lighter in water.  So the
apparent weight of the steel in water is

6146.56 minus 784 = 5362.56 newtons.

That's why you can pick up your big brother in the swimming pool.
The force of gravity on him doesn't change, but the buoyant force
in the water balances part of the force of gravity, and makes him
seem to weigh less.

If he could blow himself up like a balloon, and displace enough water
to weigh just as much as he really does, then the buoyant force would
cancel his weight completely, and he would totally float !

That's how steel ships float in water.  They're shaped in a shape that can
displace a lot of water.  The whole secret is in the shape.
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2) The center of the earth is even hotter! It's possible for people to melt, so you would most likely melt if you were trying to get to the center of the earth.

Hope this helps! :D

4 0
3 years ago
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7 0
3 years ago
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A long jumper can jump a distance of 7.4 m when he takes off at an angle of 45° with respect to the horizontal. Assuming he can
GenaCL600 [577]

Answer:

0.02 m

Explanation:

R₁ = initial distance jumped by jumper = 7.4 m

R₂ = final distance jumped by jumper = ?

θ₁ = initial angle of jump = 45°

θ₂ = final angle of jump = 42.9°

v = speed at which jumper jumps at all time

initial distance jumped is given as

R_{1}=\frac{v^{2}Sin2\theta _{1} }{g}

final distance jumped is given as

R_{2}=\frac{v^{2}Sin2\theta _{2} }{g}

Dividing final distance by initial distance

\frac{R_{2}}{R_{1}}=\frac{Sin2\theta _{1}}{Sin2\theta _{2}}

\frac{R_{2}}{7.4}=\frac{Sin2(42.9)}{Sin2(45))}

R_{2} =7.38

distance lost is given as

d = R_{1} - R_{2}

d = 7.4 - 7.38

d = 0.02 m

8 0
3 years ago
A ski gondola is connected to the top of a hill by a steel cable oflength 620 m and diameter 1.5 cm. As thegondola comes to the
klemol [59]

Answer:

Explanation:

Given

Length of cable L=620 m

Diameter of cable d=1.5 cm

time taken to return to original position T=14 s

time taken to cover distance L

t=\frac{T}{2}=7 s

velocity

v=\frac{L}{t}=\frac{620}{7}=88.57 m/s

(b)Relation between velocity of wave Tension is

v=\sqrt{\frac{T}{\mu }} , where \mu =mass per unit Length

T=v^2\cdot \mu

T=(88.57)^2\cdot \frac{m}{L}

T=(88.57)^2\cdot \frac{\rho AL}{L}

where \rho =density\ of\ steel =7850 kg/m^3

A=area\ of\ cross-section=\frac{\pi }{4}d^2=1.76\times 10^{-4} cm^2

T=(88.57)^2\cdot 7850\times 1.76\times 10^{-4}

T=10,883 N

5 0
3 years ago
A self-driving car traveling along a straight section of road starts from rest, accelerating at 2.00 m/s2 until it reaches a spe
Rasek [7]

Answer:

56.5\ \text{s}

21.13\ \text{m/s}

Explanation:

v = Final velocity

u = Initial velocity

a = Acceleration

t = Time

s = Displacement

Here the kinematic equations of motion are used

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{25-0}{2}\\\Rightarrow t=12.5\ \text{s}

Time the car is at constant velocity is 39 s

Time the car is decelerating is 5 s

Total time the car is in motion is 12.5+39+5=56.5\ \text{s}

Distance traveled

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{25^2-0}{2\times 2}\\\Rightarrow s=156.25\ \text{m}

s=vt\\\Rightarrow s=25\times 39\\\Rightarrow s=975\ \text{m}

v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{0-25}{5}\\\Rightarrow a=-5\ \text{m/s}^2

s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0-25^2}{2\times -5}\\\Rightarrow s=62.5\ \text{m}

The total displacement of the car is 156.25+975+62.5=1193.75\ \text{m}

Average velocity is given by

\dfrac{\text{Total displacement}}{\text{Total time}}=\dfrac{1193.75}{56.5}=21.13\ \text{m/s}

The average velocity of the car is 21.13\ \text{m/s}.

6 0
3 years ago
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