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Advocard [28]
3 years ago
8

An airplane is traveling at a speed of 200 m/s relative to the Earth. A person walks toward the front of the plane at a speed of

1 m/s. How fast is that person traveling in reference to the other people on the plane?
Physics
1 answer:
pochemuha3 years ago
6 0

Answer:

1 m/s

Explanation:

This is because, everybody in the plane including the person walking is moving at the speed of the plane which is 200 m/s. Now, when the person starts to walk with a speed of 1 m/s, his relative velocity to the other passengers is 1 m/s since he is already moving at the speed of the plane before starting to walk.

His velocity relative to the ground V' = velocity of plane relative to ground, V + velocity of person relative to plane, v

V = 200 m/s and v = 1 m/s

So, V' = 200 m/s + 1 m/s = 201 m/s

Since the velocity of the other passengers is the velocity of the plane, V = 200 m/s. The velocity of the person traveling in reference to the other people on the plane is V" = V' - V = 201 m/s - 200 m/s = 1 m/s

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As a rain storm passes through a region, there is an associated drop in atmospheric pressure. If the height of a mercury baromet
Pani-rosa [81]

Answer:

new atmospheric pressure is 0.9838 × 10^{5}  Pa

Explanation:

given data

height = 21.6 mm = 0.0216 m

Normal atmospheric pressure = 1.013 ✕ 10^5 Pa

density of mercury = 13.6 g/cm³

to find out

atmospheric pressure

solution

we find first height of mercury when normal pressure that is

pressure p = ρ×g×h

put here value

1.013 × 10^{5}  = 13.6 × 10³ × 9.81 × h

h =  0.759 m

so change in height Δh = 0.759 - 0.0216

new height H = 0.7374 m

so new pressure = ρ×g×H

put here value

new pressure = 13.6 × 10³ × 9.81 × 0.7374

atmospheric pressure = 98380.9584

so new atmospheric pressure is 0.9838 × 10^{5}  Pa

6 0
3 years ago
330 g of water at 55°c are poured into an 855 g aluminum container with an initial temperature of 10°
Olenka [21]
The final temperature of the system is 32.5°
we know,  H = mcT 
where, H = Heat content of the body 
m = Mass,
c = Specific heat
T = Change in temperature
According to to the Principle of Calorimetry 

The net heat remains constant i.e. 
⇒ the heat given by water = heat accepted by the aluminum container.
⇒ 330 x 1 x (45 - T) = 855 x \frac{900}{4200} x (T - 10) 

⇒ 14,850 - 330T = 183.21T  - 1832 

⇒ - 513.21 T = - 16682

or T = 32.5°
3 0
3 years ago
A 123-turn circular coil of radius 2.41 cm is immersed in a uniform magnetic field that is perpendicular to the plane of the coi
klasskru [66]

Answer:

67.44 V

Explanation:

Number of turns N =123

Radius = 2.41 cm =0.0241 m

The magnetic field strength increases from 50.9 T to 96.3 T so change in magnetic field dB = 96.3-50.9=45.4 T

Time interval dt = 0.151 sec

We know that the induced emf e=-NA\frac{dB}{dt}

Area A=\pi r^2=3.14\times 0.0241^2=1.8237\times 10^{-3}m^2

Putting all these values in emf equation e=-123\times 1.8237\times 10^{-3}\times \frac{45.4}{0.151}=-67.44\ V here negative sign indicates that it opposes the cause due to which it is produced

8 0
3 years ago
Equate these two equations for acceleration and express, for constant mass, FΔt.
Mariana [72]
Acceleration is the rate at which velocity changes. 
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Simply equate the two.
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<span>dV / dt = F/m </span>
<span>F*dt = dV * m </span>
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6 0
2 years ago
Consider a large truck carrying a heavy load, such as steel beams. A significant hazard for the driver is that the load may slid
Tanzania [10]

A) 18.4 m

B)

a) mass of the load

b) mass of the truck

Explanation:

A)

In order for the oad not to slide, its acceleration must be the same as the acceleration of the truck.

Since there is only one force acting on the load (the force of static friction), the acceleration of the load will be equal to the force of friction divided by the mass of the load (Newton's second law of motion):

a=\frac{F_f}{m}=\frac{-\mu mg}{m}=-\mu g (1)

where

m is the mass of the load

\mu=0.400 is the coefficient of static friction

g=9.8 m/s^2 is the acceleration due to gravity

The acceleration of the truck (and the load) is also related to the stopping distance of the truck by the suvat equation:

v^2-u^2=2as (2)

where

v = 0 is the final velocity of the car

u = 12.0 m/s is the initial velocity

a is the acceleration

s is the stopping distance

Since the acceleration must be the same, we can substitute (1) into (2), and solving for s we find:

v^2-u^2=-2\mu g s\\s=\frac{v^2-u^2}{-2\mu g}=\frac{0^2-12.0^2}{-2(0.400)(9.8)}=18.4 m

B)

From part A, we see that the data that we have not used in the calculation are:

- The mass of the load

- The mass of the truck

Therefore, the two pieces of data unnecessary for the solution are

a) mass of the load

b) mass of the truck

8 0
3 years ago
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