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Gwar [14]
3 years ago
10

When the volcano Krakatoa erupted in 1883, it was heard 5000 km away. Which statement about the sound from the volcano is not co

rrect?
a. If such a loud sound were to be made today, an astronaut orbiting in space (a vacuum) at a height of 400 km could hear it.
b. People further from the volcano heard the sound later than people nearer to the volcano.
c. The amplitude of the sound waves would have been smaller further from the volcano.
d. The sound was very loud because a lot of energy was transferred to vibrations of the air.
Physics
1 answer:
lina2011 [118]3 years ago
7 0
Answer a) is incorrect as sound does not travel in a vacuum.
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Kevin is refinishing his rusty wheelbarrow. He moves his sandpaper back and forth 45 times over a rusty area, each time moving a
dmitriy555 [2]
W = _|....F*dx*cos(a)........With F=force, x=distance over which force acts on object,
.......0.............................and a=angle between force and direction of travel.

Since the force is constant in this case we don't need the equation to be an integral expression, and since the force in question - the force of friction - is always precisely opposite the direction of travel (which makes (a) equal to 180 deg, and cos(a) equal to -1) the equation can be rewritted like so:

W = F*x*(-1) ............ or ............. W = -F*x

The force of friction is given by the equation: Ffriction = Fnormal*(coeff of friction)

Also, note that the total work is the sum of all 45 passes by the sandpaper. So our final equation, when Ffriction is substituted, is:

W = (-45)(Fnormal)(coeff of friction)(distance)
W = (-45)...(1.8N).........(0.92).........(0.15m)
W = ................-11.178 Joules
5 0
3 years ago
A cup of coffee is sitting on a table in a train that is moving with a constant velocity. The coefficient of static friction bet
Vikki [24]

Answer:

a = 2.94 m/s²

Explanation:

In order for the cup not to slip, the unbalanced force on cup must be equal to the frictional force:

Unbalanced Force = Frictional Force

ma = μR = μW

ma = μmg

a = μg

where,

a = maximum acceleration for the cup not to slip = ?

μ = coefficient of static friction = 0.3

g = acceleration due to gravity = 9.8 m/s²

Therefore,

a = (0.3)(9.8 m/s²)

<u>a = 2.94 m/s²</u>

3 0
3 years ago
you're reading from the journal of a European explorer from the early 1600s. In one passage, the explorer describes itting on th
Zepler [3.9K]

Answer: horse latitudes

Explanation:

8 0
3 years ago
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A block has a volume of 0.09 m3 and a density of 4,000 kg/m3. What's the force of gravity acting on the block in water?
12345 [234]

                                       Density = (mass) / (volume)

                                4,000 kg/m³ = (mass) / (0.09 m³)

Multiply each side
by  0.09 m³ :           (4,000 kg/m³) x (0.09 m³) = mass

                                 mass = 360 kg .

Force of gravity = (mass) x (acceleration of gravity)

                           = (360 kg) x (9.8 m/s²)

                           = (360 x 9.8)  kg-m/s²

                           =   3,528 newtons . 

That's the force of gravity on this block, and it doesn't matter
what else is around it.  It could be in a box on the shelf or at
the bottom of a swimming pool . . . it's weight is 3,528 newtons
(about 793.7 pounds).

Now, it won't seem that heavy when it's in the water, because
there's another force acting on it in the upward direction, against
gravity.  That's the buoyant force due to the displaced water.

The block is displacing 0.09 m³ of water.  Water has 1,000 kg of
mass in a m³, so the block displaces 90 kg of water.  The weight
of that water is  (90) x (9.8) = 882 newtons (about 198.4 pounds),
and that force tries to hold the block up, against gravity.

So while it's in the water, the block seems to weigh

       (3,528  -  882) = 2,646 newtons  (about 595.2 pounds) .

But again ... it's not correct to call that the "force of gravity acting
on the block in water".  The force of gravity doesn't change, but
there's another force, working against gravity, in the water.
5 0
3 years ago
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An athlete whirls a 7.00 kg hammer 1.8 m from the axis of rotation in a horizontal circle, as
iogann1982 [59]

Answer:

A-500 N

Explanation:

The computation of the tension in the chain is shown below

As we know that

F = ma

where

F denotes force

m denotes mass = 7

And, a denotes acceleration

Now for the acceleration we have to do the following calculations

The speed (v) of the hammer is

v = Angular speed × radius

where,

Angular seed = 2 × π ÷ Time Period

So, v = 2 × π × r ÷ P

v = 2 × 3.14 × 1.8 ÷ 1

= 11.304 m/s

Now

a = v^2 ÷ r

= 70.98912 m/s^2

Now the tension is  

T = F = m × a

= 7 × 70.98912

= 496.92384 N

= 500 N

5 0
3 years ago
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