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Rina8888 [55]
3 years ago
7

What is number 30,31 in this picture?

Physics
1 answer:
Maslowich3 years ago
6 0
On question 30, that is a displacement- time graph (DT). On this type of graph the gradient is equal to the velocity. B has the steepest gradient, then A and finally C

Now velocity is a vector quantity so it has a direction and speed ( speed doesn't have a fixed direction.)

on the DT graph im going to assume that movement B is a positive velocity with A and C being negative. 
So by ranking these: A is the most negative, C is the least negative and B has to be the greatest as it is the only positive velocity. 

Q31, The same type of graph is present, by looking at the gradients we can rank the largest and smallest velocities- speeds in the case of the question. 
i'll skip my working out as its the same as before:

C, B, A and then D

the same idea as on Q30 applies to Q31 part b, 

D,C,B then A 
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slega [8]

Answer:

The answer of this question is =1.258*10-4

4 0
3 years ago
PLZ HELP
lianna [129]
The air movements toward the equator are called trade winds, which are warm, steady breezes that blowalmost continuously. The Coriolis Effect makes the trade winds appear to be curving to the west, whether they are traveling to the equator from the south or north. Answer trade wind
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3 years ago
In a 400-m relay race the anchorman (the person who runs the last 100 m) for team A can run 100 m in 9.8 s. His rival, the ancho
melisa1 [442]

Answer:

largest lead = 3 m

Explanation:

Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.

So, first we need to calculate the velocities of both the anchorman  

given data:

Distance = d = 100 m

Time arrival for A = 9.8 s

Time arrival for B = 10.1 s

Velocity of anchorman A = D / Time arrival for A

=100/ 9.8 = 10.2 m/s

Velocity of anchorman B = D / Time arrival for B

=100/10.1 = 9.9 m/s

As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use

d = vt

= 9.9 x 9.8 = 97 m  

So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.

largest lead = 100 - 97 = 3 m

So if his lead no more than 3 m anchorman A win the race.

5 0
3 years ago
What is the speed of a wave that has a frequency of 6 Hz and a wavelength of 4 m?
notsponge [240]

Answer:

a ) 24 m/s

Explanation:

Given,

Frequency ( f ) = 6 Hz

Wavelength ( λ ) = 4 m

To find : Speed ( v ) = ?

Formula : -

v = f x λ

v

= 4 x 6

= 24 m/s

Therefore, the speed of a wave that has a frequency of 6 Hz and a wavelength of 4 m

is 24 m/s.

4 0
3 years ago
Find the ratio of effusion rates of hydrogen gas and krypton gas.
algol13

Answer:

RE of Hydrogen = 6.47 x RE of Krypton

Explanation:

Actually the correct formula for comparing rate of effusion (RE) of two gases is:

RE of Gas A

------------------- = √ ( Molar mass of B / Molar mass of A)

RE of Gas B

You can designate which of the two gases you have (hydrogen and krypton) will be your gas A and gas B. So for this particular problem, let us make hydrogen as gas A and Krypton as gas B. So the equation becomes:

RE of Hydrogen

------------------------- = √ (Molar mass of Krypton / Molar mass of Hydrogen)

RE of Krypton

Get the molar masses of Hydrogen and Krypton in the periodi table:

RE of Hydrogen

------------------------- = √ (83.798 g/mol / 2 g/mol)

RE of Krypton

RE of Hydrogen

------------------------- = 6.47  ====> this can also be written as:

RE of Krypton

RE of Hydrogen = 6.47 x RE of Krypton

It means that the rate of effusion of Hydrogen gas will be 6.47 faster than the rate of effusion of Krypton gas. With the type of question you have, it doesn't matter which gases goes on your numerator and denominator. What's important is that you show the rate of effusion of a gas with respect to the other. But if that's concerns you the most, then take the gas which was stated first as your gas A and the latter as your gas B unless the problem tells you which one will be on top and which is in the bottom.

3 0
3 years ago
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