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mash [69]
2 years ago
5

The 500 pages of a book have a mass of 2.50 kg. What is the mass of each page A in kg B in mg?

Physics
1 answer:
Furkat [3]2 years ago
3 0

Answer:

A.

Mass per page = mass of book / number of pages

Mass per page = 2.5kg / 500 pages = 0.005kg

B.

Mass per page = mass of book / number of pages

Total mass = 2.5kg = 2,500grams = 2,500,000mili grams

Mass per page = 2500000mg / 500 pages = 5000mg

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To a cyclist riding due west with a speed of 4 m/s a wind appears to
rusak2 [61]

Answer:

The actual velocity of the wind blowing from south to west is, V = 3.08 m/s

Explanation:

Given data,

The velocity of the cyclist towards west, v = 4 m/s

The wind appears to blow from the south, U = 2 m/s

Let Ф be the angle formed by the direction of the wind with the west axis,

Therefore, the actual velocity of the wind is,

                                 V = (v - U) secФ

                                     = 2 sec Ф

Le the value of Ф = 30°, (The student may add the necessary value of Ф)

                                  V =  2(1.54)

                                   V = 3.08 m/s

Hence, the actual velocity of the wind blowing from south to west is, V = 3.08 m/s

7 0
2 years ago
13. An aircraft heads North at 320 km/h rel:
AURORKA [14]

The velocity of the aircraft relative to the ground is 240 km/h North

Explanation:

We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.

Mathematically:

v' = v + v_a

where

v' is the velocity of the aircraft relative to the ground

v is the velocity of the aircraft relative to the air

v_a is the velocity of the air relative to the ground.

Taking north as positive direction, we have:

v = +320 km/h

v_a = -80 km/h (since the air is moving from North)

Therefore, we find

v'=+320 + (-80) = +240 km/h (north)

Learn more about vector addition:

brainly.com/question/4945130

brainly.com/question/5892298

#LearnwithBrainly

7 0
2 years ago
A 30-m-long rocket train car is traveling from Los Angeles to New York at 0.5c when a light at the center of the car flashes. Wh
Novosadov [1.4K]

Given that,

Distance =30 m

speed = 0.5c

(A). We need to find the bell and siren simultaneous events for a passenger seated in the car

According to given data

The distance travelled by the light to reach either side of the rocket  train car is same.

So, The two events are simultaneous and the bell and siren are the simultaneous events for a passenger seated in the car.

(B). We need to calculate time interval between the events

Using formula of time dilation

\Delta t=\dfrac{\Delta t'}{\sqrt{1-\dfrac{v^2}{c^2}}}.....(I)

Where, \delta t' = proper time

\delta t = time interval between the events

The time interval between the events measured in a reference frame

The proper time in this case is

\Delta t'=\Delta t_{1}-\dfrac{v\Delta x}{c^2}

For the second interval,

Put the value of \Delta t' in the equation (I)

\Delta t_{2}=\dfrac{\Delta t_{1}-\dfrac{v\Delta x}{c^2}}{\sqrt{1-\dfrac{v^2}{c^2}}}

Put the value in the equation

\Delta t_{2} = \dfrac{0-\dfrac{0.5c\times30}{c^2}}{\sqrt{1-\dfrac{0.5^2c^2}{c^2}}}

\Delta t_{2}=\dfrac{-15}{3\times10^{8}\sqrt{1-0.25}}

\Delta t_{2}=-5.77\times10^{8}\ s

Negative sign shows the siren rings before the bell ring.

Hence, (A). Yes, the bell and siren are simultaneous events.

(B). The siren sounds before the bell rings.

8 0
2 years ago
Charge q1 is distance s from the negative plate of a parallel-plate capacitor. Charge q2=q1/3 is distance 2s from the negative p
Svetlanka [38]

Answer:

The ratio (U₁/U₂) = 6

Explanation:

U, the potential energy is given as

U = kqQ/r

k = Coulomb's constant

q = charge we're concerned about

Q = charge of the negative plate of the capacitor

r = distance of q from the negative plate of the capacitor.

For charge q₁

U₁ = kq₁Q/s

U₂ = kq₂Q/2s

But q₂ = q₁/3

U₂ becomes U₂ = kq₁Q/6s

U₁ = kq₁Q/s

U₂ = kq₁Q/6s

(U₁/U₂) = 6

5 0
2 years ago
A 10-kg dog is running with a speed of 5.0 m/s. what is the minimum work required to stop the dog in 2.40 s?
Tasya [4]
<h3><u>Answer;</u></h3>

<em>Work = 125 joules </em>

<h3><u>Explanation and solution</u>;</h3>
  • Work is the product of force and the distance covered. Therefore, Work = force × distance.
  • Work is measured in joules.
  • Work is also a change in energy, such that work is done when energy changes, so when kinetic energy, or potential energy changes the there is work being done.

Thus; kinetic energy = work done

Kinetic energy = 1/2mv²

                       = 1/2 × 10× 5²

                       = 5 × 25

                       = 125 joules

Hence, work done is 125 joules.

3 0
2 years ago
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