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Deffense [45]
3 years ago
12

If someone told you that they traveled 50 km east, are they describing the distance they traveled or their displacement? How do

you know? *
Physics
1 answer:
tamaranim1 [39]3 years ago
4 0

Answer:

they are describing their displacement since displacement is nothing but distance along with the direction of motion

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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
Sunny_sXe [5.5K]

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

7 0
4 years ago
The dosage of technetium-99m for a lung scan is 20. μCi /kg of body mass. How many millicuries of technetium-99m should be given
expeople1 [14]

Explanation:

Below is an attachment containing the solution.

7 0
3 years ago
Robby skateboards 0.50 blocks to his friend's house in 1.2 minutes. What is his speed?
lianna [129]
The answer is 0.42 blocks per minute, Good Luck!
4 0
3 years ago
What is the number of electrons to come be held in the first orbit closest to the nucleus
Mariulka [41]
In the first shell there can be a maximum of 2 electrons, then 8o in the second one and 8 in the third one.
6 0
3 years ago
If the length of a building is 2 1 2 times the width and each dimension is increased by 7 ft, then the perimeter is 266 ft. Find
N76 [4]

Answer:

The original dimensions of the building is 95 ft × 38 ft.

Explanation:

Let the original length be 'l' and original width be 'w'.

Given:

Original length (l) = 2\frac{1}{2}\times original\ width

Original width = 'w'.

So, l=2\frac{1}{2}w=\frac{5}{2}w

Now, as per question:

Length and width is increased by 7 ft.

So, new length (l') = l+7=\frac{5w}{2}+7

New width (w') = w+7

New perimeter (P) = 266 ft

Perimeter is given as:

P=2(l' +w')\\\\266=2(\frac{5w}{2}+w)\\\\\frac{266}{2}=\frac{5w+2w}{2}\\\\266=7w\\\\w=\frac{266}{7}=38\ ft

Therefore, original width = 38 ft.

Original length is, l=\frac{5\times 38}{2}=\frac{190}{2}=95\ ft

Hence, the original dimensions of the building is 95 ft × 38 ft.

7 0
4 years ago
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