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Studentka2010 [4]
3 years ago
5

To a man running east at the rate of 3m/s vain appears to fall vertically with a speed of 4m/s. Find the actual speed and direct

ion of rain...​
Physics
1 answer:
maria [59]3 years ago
5 0

Answer:

The actual speed of the rain is 5 m/s and its direction is -53.13°

Explanation:

The actual speed of the rain V = speed of man, v + speed of rain relative to man, v'.

V = v + v'

We add these vectorially.

Since the man's speed is 3 m/s east, in the positive x - direction, we have v = 3i  and the rain's speed is falling vertically at 4 m/s, in the negative y- direction, we have v' = -4j

So, V = v + v'

V = 3i + (-4j)

V = 3i - 4j

So, the magnitude of V which is its speed is V = √(3² + (-4)²) = √(9 + 16) = √25 = 5 m/s

The direction of V, Ф = tan⁻¹(vertical component/horizontal component) = tan⁻¹(-4/3) = tan⁻¹(-4/3) = tan⁻¹(-1.333) = -53.13°

So, the actual speed of the rain is 5 m/s and its direction is -53.13°

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A 1.25-kg ball begins rolling from rest with constant angular acceleration down a hill. If it takes 3.60 s for it to make the fi
miv72 [106K]

Answer:

The time taken is  \Delta t  = 1.5 \ s

Explanation:

From the question we are told that

   The mass of the ball is  m =  1.25 \ kg

    The time taken to make the first complete revolution is  t= 3.60 s

    The displacement of the first complete revolution is  \theta  =  1 rev  =  2 \pi \  radian

Generally the displacement for one  complete revolution is mathematically represented as

       \theta =  w_i t  +  \frac{1}{2} *  \alpha  * t^2

Now given that the stone started from rest w_i  = 0 \ rad / s

     2 \pi =0   +  0.5*  \alpha  *(3.60)^2

     \alpha   =  0.9698 \  s

Now the displacement for two  complete revolution is

         \theta_2  =  2 *  2\pi

         \theta_2  = 4\pi

Generally the displacement for two complete revolution is mathematically represented as  

     4 \pi =   0  +  0.5 * 0.9698 * t^2

=>   t^2  =  25.9187

=>   t=  5.1 \ s

So

 The  time taken to complete the next oscillation is mathematically evaluated as

     \Delta t  =  t_2  - t

substituting values

      \Delta t  = 5.1 -  3.60

     \Delta t  = 1.5 \ s

           

 

7 0
2 years ago
Which of the following is a result of gravitational forces in the Solar System?
postnew [5]
The correct answer is d
3 0
3 years ago
Astronomers can now report that active star formation was going on at a time when the universe was only 20% as old as it is toda
disa [49]

Answer:

First, as you may know, the light travels at a given velocity.

In vaccum, this velocity is c = 3x10^8 m/s.

And we know that:

distance = velocity*time

Now, if some object (like a star ) is really far away, the light that comes from that star may take years to reach the Earth.

This means that the images that the astronomers see today, actually happened years and years ago (So the night sky is like a picture of the "past" of the universe)

Also, for example, if an astronomer sees some particular thing, he can apply a model (a "simplification" of some phenomena that is used to simplify it an explain it) and with the model, the scientist can infer the information of the given thing some time before it was seen.

8 0
3 years ago
Freight car A with a gross weight of 200,000 lbs is moving along the horizontal track in a switching yard at 4 mi/hr. Freight ca
zhenek [66]

Answer: a) 4.7 mi/hr.  b) 86,500 lbs. mi²/Hr²

Explanation:

As in any collision, under the assumption that no external forces exist during the very small collision time, momentum must be conserved.

If the collision is fully inelastic, both masses continue coupled each other as a single mass, with a single speed.

So, we can write the following:

p₁ = p₂ ⇒m₁.v₁ + m₂.v₂ = (m₁ + m₂). vf

Replacing by the values, and solving for vf, we get:

vf = (200,000 lbs. 4 mi/hr + 100,000 lbs. 6 mi/hr) / 300,000 lbs = 4.7 mi/hr

If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

K₁ = 1/2 (m₁.v₁² + m₂.v₂²) = 3,400,000 lbs. mi² / hr²

After the collision, total kinetic energy is as follows:

K₂ = 1/2 ((m₁ + m₂) vf²) = 3,313,500 lbs. mi²/hr²

So we have an Energy loss, equal to the difference between initial kinetic energy and final kinetic energy, as follows:

DE = K₁ - K₂ = 86,500 lbs. mi² / hr²

This loss is due to the impact, and is represented by the work done by friction forces (internal) during the impact.

8 0
2 years ago
How much heat is released when 432 g of water cools down from 71'c to 18'c?
maria [59]
The heat released by the water when it cools down by a temperature difference \Delta T is
Q=mC_s \Delta T
where
m=432 g is the mass of the water
C_s = 4.18 J/g^{\circ}C is the specific heat capacity of water
\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ} is the decrease of temperature of the water

Plugging the numbers into the equation, we find
Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J
and this is the amount of heat released by the water.
7 0
3 years ago
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