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Svetradugi [14.3K]
2 years ago
5

A radar used to detect the presence of aircraft receives a pulse that has reflected off an object 5 ✕ 10−5 s after it was transm

itted. What is the distance (in m) from the radar station to the reflecting object?
Physics
1 answer:
Sunny_sXe [5.5K]2 years ago
7 0

Answer:

7500 m

Explanation:

The radar emits an electromagnetic wave that travels towards the object and then it is reflected back to the radar.

We can call L the distance between the radar and the object; this means that the electromagnetic wave travels twice this distance, so

d = 2L

In a time of

t=5\cdot 10^{-5}s

Electromagnetic waves travel in a vacuum at the speed of light, which is equal to

c=3.0\cdot 10^8 m/s

Since the electromagnetic wave travels with constant speed, we can use the equation for uniform motion ,so:

d=vt (1)

where

v=c=3.0\cdot 10^8 m/s

t=5\cdot 10^{-5}s

d=2L, where L is the distance between the radar and the object

Re-arranging eq(1) and substituting, we find L:

L=\frac{vt}{2}=\frac{(3.0\cdot 10^8)(5\cdot 10^{-5})}{2}=7500 m

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Answer : Zamir's displacement and Talia's displacement is equal.

Explanation :

Displacement is explained to be the changing position of an object.

Zamir covers total distance 27 m and Talia covers total distance 19 m but  Zamir's initial and final position and Talia's initial and final position is same.

So, we can say that Zamir's displacement and Talia's displacement is equal.

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1 year ago
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An astronaut holds a rock 100m above the surface of Planet X . The rock is then thrown upward with a speed of 15m/s , as shown i
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Answer:5 m/s^{2}

Explanation:

The described situation is is related to vertical motion (and free fall). So, we can use the following equation that models what happens with this rock:

y=y_{o}+V_{o}sin\theta t-\frac{1}{2}gt^{2} (1)

Where:

y=0m is the rock's final height

y_{o}=100 m is the rock's initial height

V_{o}=15 m/s is the rock's initial velocity

\theta=90\° is the angle at which the rock was thrown (directly upwards)

t=10 s is the time

g is the acceleration due gravity in Planet X

Then, isolating g and taking into account sin(90\°)=1:

g=(-\frac{2}{t^{2}})(y-y_{o}-V_{o}t) (2)

g=(-\frac{2}{(10 s)^{2}})(0 m-100 m-(15 m/s)(10 s)) (3)

Finally:

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7 0
1 year ago
Plastic foam is about 0.10 times as dense as water. What weight of bricks could you stack on a 1m x 1m x 0.10m slab of foam, so
goblinko [34]

Answer: Weight = 98.1N

Explanation:

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Given that the Plastic foam is about 0.10 times as dense as water. That is,

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The volume V = 1 ×1×0.1 = 0.1 m^3

Density is the ratio of mass to volume

Density = mass/volume

Let us substitute for density and volume to get mass.

100 = M/0.1

Make M the subject of formula

M = 100 × 0.1 = 10 kg

Weight = mg

Where g = 9.81 m/s

Substitute the M and g into the formula

Weight = 10 × 9.81 = 98.1 N

Therefore, the weight of the brick is 98.1 N

4 0
2 years ago
A beam of electrons is sent horizontally down the axis of a tube to strike a fluorescent screen at the end of the tube. On the w
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Answer:

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So as a result of the magnetic field directed vertically downwards which is at a right angle with the electron beams, the electrons will move to the right and the spot will be deflected to the right of the screen when looking from the electron source.

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A 4.0-mF capacitor initially charged to 50 V and a 6.0-mF capacitor charged to 30 V are connected to each other with the positiv
Juli2301 [7.4K]

Answer:

<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>

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