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Romashka-Z-Leto [24]
2 years ago
9

An ocean liner is cruising at 10 meters/second and is about to approach a stationary ferryboat. A parcel is released from the oc

ean liner from a height of 5.5 meters above the ferryboat’s deck. Calculate the distance at which the ferryman should position the boat from the point horizontally below the point of release so that the parcel lands inside the boat.
Physics
1 answer:
Afina-wow [57]2 years ago
4 0
The parcel will undergo projectile motion, which means that it will have motion in both the horizontal and vertical direction.

First, we determine how long the parcel will fall using:

s = ut + 1/2 at²

where s will be the height, u is the initial vertical velocity of the parcel (0), t is the time of fall and a is the acceleration due to gravity. 

5.5 = (0)(t) + 1/2 (9.81)(t)²
t = 1.06 seconds

Now, we may use this time to determine the horizontal distance covered by the parcel by using:
distance = velocity * time

The horizontal velocity of the parcel will be equal to the horizontal velocity of the cruise liner.

Distance = 10 * 1.06
Distance = 10.6 meters

The boat should be 10.6 meters away horizontally from the point of release.
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A boat moves through the sea.
sergiy2304 [10]

Answer:

dont you have to times it

Explanation:

4 0
2 years ago
Assume the motions and currents mentioned are along the x axis and fields are in the y direction. (a) does an electric field exe
matrenka [14]
<span> (a) does an electric field exert a force on a stationary charged object? 
Yes. The force exerted by an electric field of intensity E on an object with charge q is
</span>F=qE
<span>As we can see, it doesn't depend on the speed of the object, so this force acts also when the object is stationary.

</span><span>(b) does a magnetic field do so?
No. In fact, the magnetic force exerted by a magnetic field of intensity B on an object with  charge q and speed v is
</span>F=qvB \sin \theta
where \theta is the angle between the direction of v and B.
As we can see, the value of the force F depends on the value of the speed v: if the object is stationary, then v=0, and so the force is zero as well.

<span>(c) does an electric field exert a force on a moving charged object? 
Yes, The intensity of the electric force is still
</span>F=qE
<span>as stated in point (a), and since it does not depend on the speed of the charge, the electric force is still present.

</span><span>(d) does a magnetic field do so?
</span>Yes. As we said in point b, the magnetic force is
F=qvB \sin \theta
And now the object is moving with a certain speed v, so the magnetic force F this time is different from zero.

<span>(e) does an electric field exert a force on a straight current-carrying wire?
Yes. A current in a wire consists of many charges traveling through the wire, and since the electric field always exerts a force on a charge, then the electric field exerts a force on the charges traveling through the wire.

</span><span>(f) does a magnetic field do so? 
Yes. The current in the wire consists of charges that are moving with a certain speed v, and we said that a magnetic field always exerts a force on a moving charge, so the magnetic field is exerting a magnetic force on the charges that are traveling through the wire.

</span><span>(g) does an electric field exert a force on a beam of moving electrons?
Yes. Electrons have an electric charge, and we said that the force exerted by an electric field is
</span>F=qE
<span>So, an electric field always exerts a force on an electric charge, therefore on an electron beam as well.

</span><span>(h) does a magnetic field do so?
Yes, because the electrons in the beam are moving with a certain speed v, so the magnetic force
</span>F=qvB \sin \theta
<span>is different from zero because v is different from zero.</span>
6 0
1 year ago
Suppose you are myopic (nearsighted). You can clearly focus on objects that are as far away as 52.5 cm away. You can clearly foc
Lilit [14]

Answer:

Explanation:

Image of distant object will be made at far point or at 52.5 so

object distance u = infinity

image distance v = - 52.5 cm

focal length required = f

Lens formula

1 / v - 1 / u = 1 / f

1 /  - 52.5 - 0 = 1 / f

f =  -52.5 cm

= -.525 m

Power P = 1 / f = -  1 / .525

= -  1.90

now , for eye with glass we shall find new near point .

v = ?

u = - 17.2 cm

f = -  52.5 cm

1 / v - 1 / u = 1 / f

  1 / v + 1 / 17.2 = -  1 / 52.5

1 / v  = - 1 / 17.2 -    1 / 52.5

= - .05813 -  .019

= - .07713

u = - 12.96 cm

so new near point will be 12.96 cm

5 0
1 year ago
A baseball catcher puts on an exhibition by catching a 0.15-kg ball dropped from a helicopter at a height of 101 m. What is the
yaroslaw [1]

Answer:

The speed of the ball 1.0 m above the ground is 44 m/s (Answer A).

Explanation:

Hi there!

To solve this problem, let´s use the law of conservation of energy. Since there is no air resistance, the only energies that we should consider is the gravitational potential energy and the kinetic energy. Because of the conservation of energy, the loss of potential energy of the ball must be compensated by a gain in kinetic energy.

In this case, the potential energy is being converted into kinetic energy as the ball falls (this is only true when there are no dissipative forces, like air resistance, acting on the ball). Then, the loss of potential energy (PE) is equal to the increase in kinetic energy (KE):

We can express this mathematically as follows:

-ΔPE = ΔKE

-(final PE - initial PE) = final KE - initial KE

The equation of potential energy is the following:

PE = m · g · h

Where:

PE = potential energy.

m = mass of the ball.

g = acceleration due to gravity.

h = height.

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the ball.

v = velocity.

Then:

-(final PE - initial PE) = final KE - initial KE          

-(m · g · hf - m · g · hi) = 1/2 · m · v² - 0     (initial KE = 0 because the ball starts from rest)  (hf = final height, hi = initial height)

- m · g (hf - hi) = 1/2 · m · v²

2g (hi - hf) = v²

√(2g (hi - hf)) = v

Replacing with the given data:

√(2 · 9.8 m/s²(101 m - 1.0 m)) = v

v = 44 m/s

The speed of the ball 1.0 m above the ground is 44 m/s.

3 0
2 years ago
A stranded soldier shoots a signal flare into the air to attract the attention of a nearby plane. The flare has an initial verti
Ipatiy [6.2K]

Answer:

Explanation:

h = ut - 16 t² = ut - 1/2 x32 t² = ut - 1/2 g t² , g = acceleration  = - 32 ft / s²

1) v² = u² - 2 g h , v = 0 so

h = u² / 2g = 1500² / 2 x 32 = 35156.25 ft

2) v = u - gt

t = u / g = 1500 / 32 = 46.875 s

3) It will hit the ground after 2 x 46.875 = 93.75 s

4 ) time to reach 30000 ft height  t is given by

h = ut - 16 t²

30000 = 1500t - 16t²

16t²-1500t + 30000 = 0

t = 28.92 s  and 64.82 s

Time required to travel 50000 by plane

= 50000/880 = 56.82 . There is no match of timing so plane will not hit it.

8 0
2 years ago
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