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3 years ago

Give your answer with solution(Easy Question)

Physics

1 answer:

Furkat [3]3 years ago

5 0

Volume of any rectangular prism is (length) x (width) x (height).

The box in the picture is definitely a rectangular prism, so its volume is

(length) x (width) x (height) .

That's (4.0 cm) x (6.0 cm) x (2.0 cm) = 48 cm³ .

So the volume of the box is 48 cm³ fer sher, but that's NOT the preferred answer to this question. If you take your time and read all the fine print carefully, you'll see that this sneaky question wants the answer in units of meter³ . In order to produce the answer that it's fishing for, YOU have to be sneaky too. You have to convert units, either before you start doing the calculating, or else convert the units of your result before you choose the answer from the list.

Before you calculate: Convert the given dimensions to (meters). Then

Volume = (0.04 m) x (0.06 m) x (0.02 m)

Volume = 0.000048 m³

Volume = 4.8 x 10⁻⁵ m³

After calculating but before picking the answer:

Volume = (4.0 cm) x (6.0 cm) x (2.0 cm)

Volume = (48 cm³)

Convert:

Volume = (48 cm³) x (1 meter / 100 cm)³

Volume = (48 cm³) x (1 meter³ / 10⁶ cm³)

Volume = (48 cm³ meter³ / 10⁶ cm³)

Volume = 48 x 10⁻⁶ m³

Volume = 4.8 x 10⁻⁵ m³

This is exactly the same number as 48 cm³, but this is the way they want to see it, and 48 cm³ isn't. So you have to select choice - D .

You might be interested in

This sphere contains all of the frozen water on Earth

erica [24]

Cryosphere is the sphere that contains all of the frozen water on Earth. The correct option among all the options that are given in the question is the fourth option or the last option. Cryosphere is actually those parts of the surface of the earth where the water is in a frozen state, like in seas and oceans.

7 0

3 years ago

If you measure the amount of work accomplished in a particular time interval, u have measured-

serious [3.7K]

Answer:

The power

Explanation:

We know that the work definition is given by the following expression:

W = F * d

where:

F = force [Newtons] [N]

d = distance [meters] [m]

W = work [Joules]

And the expression that defines the work done by unit of time is called - Power, therefore:

P = W/t

where:

P = power [watts] [w]

W = work [Joules] [J]

t = time [seconds] [s]

6 0

3 years ago

WILL GIVE BRAINLIEST TO THE CORRECT ANSWER

LenKa [72]

Answer:

See the answers below.

Explanation:

We can solve both problems using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F =m*a

where:

F = force [N] (units of newtons)

m = mass = 1000 [kg]

a = acceleration = 3 [m/s²]

$F = 1000*3\\F=3000[N]$

And the weight of any body can be calculated by means of the mass product by gravitational acceleration.

$W=m*g\\W=1000*9.81\\W=9810 [N]$

4 0

3 years ago

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent

azamat

The wavelengths of the constituent travelling waves CANNOT be 400 cm.

The given parameters:

Length of the string, L = 100 cm

The wavelengths of the constituent travelling waves is calculated as follows;

$L = \frac{n \lambda}{2} \\\\n\lambda = 2L\\\\\lambda = \frac{2L}{n}$

for first mode: n = 1

$\lambda = \frac{2\times 100 \ cm}{1} \\\\\lambda = 200 \ cm$

for second mode: n = 2

$\lambda = \frac{2L}{2} = L = 100 \ cm$

For the third mode: n = 3

$\lambda = \frac{2L}{3} \\\\\lambda = \frac{2 \times 100}{3} = 67 \ cm$

For fourth mode: n = 4

$\lambda = \frac{2L}{4} \\\\\lambda = \frac{2 \times 100}{4} = 50 \ cm$

Thus, we can conclude that, the wavelengths of the constituent travelling waves CANNOT be 400 cm.

The complete question is below:

A string of length 100 cm is held fixed at both ends and vibrates in a standing wave pattern. The wavelengths of the constituent travelling waves CANNOT be:

A. 400 cm

B. 200 cm

C. 100 cm

D. 67 cm

E. 50 cm

Learn more about wavelengths of travelling waves here: brainly.com/question/19249186

5 0

2 years ago

A man is standing on a weighing machine on a ship which is bobbing up and down with simple harmonic motion of period T=15.0s.Ass

STALIN [3.7K]

Well, first of all, one who is sufficiently educated to deal with solving
this exercise is also sufficiently well informed to know that a weighing
machine, or "scale", should not be calibrated in units of "kg" ... a unit
of mass, not force. We know that the man's mass doesn't change,
and the spectre of a readout in kg that is oscillating is totally bogus.

If the mass of the man standing on the weighing machine is 60kg, then
on level, dry land on Earth, or on the deck of a ship in calm seas on Earth,
the weighing machine will display his weight as 588 newtons or as
132.3 pounds. That's also the reading as the deck of the ship executes
simple harmonic motion, at the points where the vertical acceleration is zero.

If the deck of the ship is bobbing vertically in simple harmonic motion with
amplitude of M and period of 15 sec, then its vertical position is

   y(t) = y₀ + M sin(2π t/15) .

The vertical speed of the deck is   y'(t) = M (2π/15) cos(2π t/15)

and its vertical acceleration is y''(t) = - (2πM/15) (2π/15) sin(2π t/15)

      = - (4 π² M / 15²) sin(2π t/15)

      = - 0.1755 M sin(2π t/15) .

There's the important number ... the 0.1755 M.
That's the peak acceleration.
From here, the problem is a piece-o-cake.

The net vertical force on the intrepid sailor ... the guy standing on the
bathroom scale out on the deck of the ship that's "bobbing" on the
high seas ... is (the force of gravity) + (the force causing him to 'bob'
harmonically with peak acceleration of 0.1755 x amplitude).

At the instant of peak acceleration, the weighing machine thinks that
the load upon it is a mass of 65kg, when in reality it's only 60kg.
The weight of 60kg = 588 newtons.
The weight of 65kg = 637 newtons.
The scale has to push on him with an extra (637 - 588) = 49 newtons
in order to accelerate him faster than gravity.

Now I'm going to wave my hands in the air a bit:

Apparent weight = (apparent mass) x (real acceleration of gravity)

(Apparent mass) = (65/60) = 1.08333 x real mass.

Apparent 'gravity' = 1.08333 x real acceleration of gravity.

The increase ... the 0.08333 ... is the 'extra' acceleration that's due to
the bobbing of the deck.

0.08333 G = 0.1755 M

The 'M' is what we need to find.

Divide each side by 0.1755 : M = (0.08333 / 0.1755) G

'G' = 9.0 m/s²
   M = (0.08333 / 0.1755) (9.8) = 4.65 meters .

That result fills me with an overwhelming sense of no-confidence.
But I'm in my office, supposedly working, so I must leave it to others
to analyze my work and point out its many flaws.
In any case, my conscience is clear ... I do feel that I've put in a good
5-points-worth of work on this problem, even if the answer is wrong .

8 0

3 years ago