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lara31 [8.8K]
3 years ago
8

4: in the cabin of a jetliner that cruises at 600 km/h, a pillow drops from an overhead rack. does a passenger walking down the

aisle have to worry about the pillow slamming into her and knocking her over? explain
Physics
2 answers:
Lelu [443]3 years ago
6 0
I think it does matter because when the pillow is in contact to the cabin it is also at a velocity of 600 km/h, but as soon as it leaves the contact of the cabin, inertia tries to slow it down and the passenger being at 600 km/h has a risk of getting hit by it.
Nataly_w [17]3 years ago
5 0

Answer:

To answer this question, we need to recur to the law of inertia.

The law of inertia states<em> "if a body is at rest, or moving at a constant speed in a straight line, it will remain that state until a force act on the body".</em>

<em></em>

So, in this case, the jet is traveling at a speed of 600 km/h, which means that all bodies inside the jet are also traveling at this speed, because they are in it.

Also, if we apply the law of inertia here, imagine that somebody falls from the jet, that person will travel at a speed near to 600km/h, if we discard the wind, following a constant trajectory.

The same phenomenon happens with the pillow, when it falls from an overhead rack, if the passanger is passing under the trajectory of the pillow, it will slam her, because due to intertia, the pillow will continue its movement until something stop it, which can be the person's body.

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Suppose astronomers find an earthlike planet that is twice the size of Earth (that is, its radius is twice the radius of Earth).
JulijaS [17]

Answer:

4 times the mass of Earth

Explanation:

M_1 = Mass of Earth

M_2 = Mass of the other planet

r = Radius of Earth

2r = Radius of the other planet

m = Mass of object

The force of gravity on an object on Earth is

F=\frac{GM_1m}{r^2}

The force of gravity on an object on the other planet is

F=\frac{GM_2m}{(2r)^2}

As the forces are equal

\frac{GM_1m}{r^2}=\frac{GM_2m}{(2r)^2}\\\Rightarrow M_1=\frac{M_2}{4}\\\Rightarrow M_2=4M_1

So, the other planet would have 4 times the mass of Earth

6 0
3 years ago
Lucas is carrying his rock collection into class. He carries 30 pounds of rare rocks 15 feet into Classroom 6. It takes him 3 mi
weqwewe [10]

you would multiply 30 by 15. because its the weight times the distance.

3 0
3 years ago
Read 2 more answers
The change of motion of a body is proportional to the applied force, and it takes place in the direction of a straight line in w
7nadin3 [17]

Answer:

Newton's Second Law of Motion  

Explanation:

According to Newton's second law of motion, the change in velocity of a body is directly proportional to the force applied on it. Velocity is a vector quantity. It measures the magnitude of the speed as well as its direction.

F = m a

where, F is the applied force, m is the mass and a is the acceleration.

It can also be expressed as:

F = \frac{dp}{dt}

where, p = mv ( momentum)

4 0
3 years ago
Read 2 more answers
A beam of light converging to the point of 10 cm is incident on the lens. find the position of the point image if the lens has a
Verizon [17]

Answer:

beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.

To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm

Solution:

As per the given criteria,

the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)

(a) lens is a convex lens with

focal length, f=20cm

object distance, u=12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

20

1

+

12

1

⟹

v

1

=

60

3+5

⟹v=7.5cm

Hence the image formed is real, at 7.5cm from the lens on its right side.

(b) lens is a concave lens with

focal length, f=−16cm

object distance, 12cm

applying the lens formula, we get

f

1

=

v

1

−

u

1

⟹

v

1

=

f

1

+

u

1

⟹

v

1

=

−16

1

+

12

1

⟹

v

1

=

48

−3+4

⟹v=48m

Hence the image formed is real, at 48 cm from the lens on the right side.

6 0
2 years ago
A mass of 0.450 kg rotates at constant speed with a period of 1.45 s at a radius R of 0.140 m in the apparatus used in this labo
Mamont248 [21]

Answer:

1.603 s

Explanation:

Given that

Initial mass, = 0.45 kg

Initial period, = 1.45 s

Initial radius, = 0.14 m

Final mass, = 0.55 kg

Final period, = ?

Final radios, = 0.14 m

Since we are finding the rotation period of two masses of same radius, we can assume that the outward force is the same in both cases. This means that

m₁r₁ω₁² = m₂r₂ω2²

Where, ω = 2π/T, on substituting, we have

0.45 * 0.14 * (2π / 1.45)² = 0.550 * 0.14 * (2π / T₂)²

0.45 / 1.45² = 0.550 / T₂²

T₂² = 0.550 * 1.45² / 0.45

T₂² = 2.56972

T₂ = √2.56972

T₂ = 1.603 sec

7 0
3 years ago
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