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Black_prince [1.1K]

3 years ago

13

The operator of a space station observes a space vehicle approaching at a constant speed v. The operator sends a light signal at

speed c toward the space vehicle. What is the speed of the light signal as viewed from the space vehicle

1 answer:

GenaCL600 [577]3 years ago

5 0

Answer:

The speed of the light signal as viewed from the observer is c.

Explanation:

Recall the basic postulate of the theory of relativity that the speed of light is the same in ALL inertial frames. Based on this, the speed of light is independent of the motion of the observer.

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Suppose a scientific team is trying to recreate the energy-producing reactions tht occur in the sun.what would they need for the

Rasek [7]

Abundant hydrogen high temperature high pressure is that they need

7 0

3 years ago

Read 2 more answers

Spymaster Paul, flying a constant 215km/h horizontally in a low-flying helicopter, want to drop secret documents into his contac

aleksandrvk [35]

Answer:

$\theta = 49.56 degree$

Explanation:

Relative horizontal velocity of plane with respect to the car is given as

$v_r = v_p - v_c$

so we have

$v_p = 215 km/h$

$v_c = 155 km/h$

now we have

$v_r = 215 - 155$

$v_r = 60 km/h$

$v_r = 16.67 m/s$

Now time taken by the object to drop the vertical height is given as

$y = \frac{1}{2}gt^2$

$78 = \frac{1}{2}(9.81)t^2$

$t = 3.98 s$

so the distance of the car must be

$d = v_r\times t$

$d = 16.67 \times 3.98$

$d = 66.47 m$

Angle of the car with horizontal is given as

$tan\theta = \frac{y}{x}$

$tan\theta = \frac{78}{66.47}$

$\theta = tan^{-1}(1.17)$

$\theta = 49.56 degree$

5 0

3 years ago

A 12,500 N alien UFO is hovering about the surface of Earth. At time , its position can be given as () = ((0.24 m/s^3)^3 + 25 m)

stepladder [879]

a) $F=(3675i-4543k)N$

b) 5843 N

Explanation:

a)

The position of the UFO at time t is given by the vector:

$r(t)=(0.24t^3+25)i+(4.2t)j+(-0.43t^3+0.8t^2)k$

Therefore it has 3 components:

$r_x=0.24t^3+25\\r_y=4.2t\\r_z=-0.43t^3+0.8t^2$

We start by finding the velocity of the UFO, which is given by the derivative of the position:

$v_x=r'_x=\frac{d}{dt}(0.24t^3+25)=3\cdot 0.24t^2=0.72t^2\\v_y=r'_y=\frac{d}{dt}(4.2t)=4.2\\v_x=r'_z=\frac{d}{dt}(-0.43t^3+0.8t^2)=-1.29t^2+1.6t$

And then, by differentiating again, we find the acceleration:

$a_x=v'_x=\frac{d}{dt}(0.72t^2)=1.44t\\a_y=v'_y=\frac{d}{dt}(4.2)=0\\a_z=v'_z=\frac{d}{dt}(-1.29t^2+1.6t)=-2.58t+1.6$

The weight of the UFO is W = 12,500 N, so its mass is:

$m=\frac{W}{g}=\frac{12500}{9.8}=1276 kg$

Therefore, the components of the force on the UFO are given by Newton's second law:

$F=ma$

So, Substituting t = 2 s, we find:

$F_x=ma_x=(1276)(1.44t)=(1276)(1.44)(2)=3675 N\\F_y=ma_y=0\\F_z=ma_z=(1276)(-2.58t+1.6)=(1276)(-2.58(2)+1.6)=-4543 N$

So the net force on the UFO at t = 2 s is

$F=(3675i-4543k)N$

b)

The magnitude of a 3-dimensional vector is given by

$|v|=\sqrt{v_x^2+v_y^2+v_z^2}$

where

$v_x,v_y,v_z$ are the three components of the vector

In this problem, the three components of the net force are:

$F_x=3675 N\\F_y=0\\F_z=-4543 N$

Therefore, substituting into the equation, we find the magnitude of the net force:

$|F|=\sqrt{3675^2+0^2+(-4543)^2}=5843 N$

7 0

3 years ago

An Amtrak going 250m/s comes to a stop in 12s. What is the<br> acceleration?

astraxan [27]

Answer:

$a=\frac{v-u}{t} \\a = \frac{0-250}{12} = -20.83 m/s$

Explanation:

you mean deceleration right ? because the acceleration is 250m/s

7 0

3 years ago

An intravenous (IV) system is supplying saline solution to a patient at the rate of 0.06 cm3/s through a needle of radius 0.2 mm

horsena [70]

Answer:

Pressure applied to the needle is 7528 Pa

Explanation:

As we know by poiseuille's law of flow of liquid through a cylindrical pipe

the rate of flow through the pipe is given as

$Q = \frac{\Delta P \pi r^4}{8\eta L}$

now we know that

$Q = 0.06 \times 10^{-6} m^3/s$

radius = 0.2 mm

Length = 6.32 cm

$\eta = 1\times 10^{-3} Pa s$

now we have

$6 \times 10^{-8} = \frac{\Delta P \pi (0.2 \times 10^{-3})^4}{8(1 \times 10^{-3})6.32 \times 10^{-2}}$

$3.03 \times 10^{-11} = \Delta P 5.02 \times 10^{-15}$

$\Delta P = 6028 Pa$

now we have

$P - 1500 = 6028 Pa$

$P = 7528 Pa$

8 0

3 years ago