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enot [183]
3 years ago
7

A bus leaves at 9 am with a group of tourists. They travel 350 km before they stop for lunch. Then they travel an additional 250

km until the end of their trip at 3 pm. What was the average speed of the bus?
Physics
1 answer:
Anettt [7]3 years ago
8 0

Average speed = Distance traveled / time taken

In this case Time taken = Difference in hours between 3 PM and 9 AM

                                        = 6 hours

Total distance traveled = 350 km + 250 km

                                       = 600 kilometers

So average speed = 600/6 = 100 km/hr

Average speed of bus = 27.78 m/s

So the bus's average speed = 27.78 m/s or 100 km/hr.

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What is 16.558 m/s rounded to three significant figures?
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Answer:

Option C. 16.6 m/s

Explanation:

To round this 16.558 m/s to 3sf, we need to count the number beginning from 1. When we get to the 3rd number( ie 5), we'll examine the fourth number(i.e 5)to see if it less than five or greater. If it less than five, then we'll discard it. But if it five or greater, we'll approximate it and add it to the 3rd number.

So.

16.558 m/s = 16.6m/s to 3sf

3 0
3 years ago
In which state of matter are water molecules measured as having the lowest temperature?
natali 33 [55]
Molecules in the solid phase have the least amount of energy, while gas particles have the greatest amount of energy. The temperature of a substance is a measure of the average kinetic energy of the particles.
4 0
3 years ago
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Explanation:

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g calculate the effectiveness radiation dosage in sieverts for a 79 kg person who is exposed 6.8x10^9
Elza [17]

Answer:

The answer is "\bold{dosage = 0.031 rem}"

Explanation:

please find the complete question in the attached file.

Given value:

m = 79\  kg  \\\\n = 3.4 \times  10^9 \\\\E = 5.5  \times  10^{-13} \\\\ RBE = 15

\to E = n E\\

        = 3.4  \times  10^9  \times  5.5  \times  10^{-13} \\\\      = 1.87 \times  10^{-3}

\to E(absorbed) = 1.87  \times 10^{-3}  \times  0.87 = 1.63  \times  10^{-3}

calculating the radiation absorbed per kg:

= \frac{1.63  \times  10^{-3}}{79}  \\\\ = 2.06  \times  10^{-5} \\\\ = 0.00206 \  rad

\to Dosage = 0.00206  \times  15 \\

                 = 0.031 \ \ rem

4 0
3 years ago
A particle with a charge of 5.13 μC has a velocity of 8.64 x106 m/s in a direction perpendicular to a magnetic fieldof 1.99 x 10
Svetach [21]

Answer:

F= 0.009 N

Explanation:

Given that

Charge ,q= 5.13 μC

Velocity ,V= 8.64 x 10⁶ m/s

Magnetic field , B = 1.99 x 10⁻⁴ T

The force on a charge q moving with velocity v is given as follows

F= q V B

Now by putting the values in the above equation we get

[tex]F= 5.13\times 10^{-6}\times 8.64\times 10^{6}\times 1.99\times 10^{-4}\ N [\tex]

F=0.00882 N

F= 0.009 N

Therefore the force on the particle will be 0.009 N.

5 0
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