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1 year ago

11

A migrating bird can travel at a speed of 30 m/s. How far will it travel in 25 minutes at this speed? Give your answer in metres

(m), and in kilometres (km).

1 answer:

Nady [450]1 year ago

7 0

Answer:

It will travel 45 km (45000 m)

Explanation:

25 minutes is equal to 1500 seconds (25*60=1500)

distance=speed*time

distance=30m/s * 1500s = 45000 m = 45 km

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What is the structure located inside the nucleus of a cell that contains an organism’s genetic code?

schepotkina [342]

This belongs in the bio section but the structure are the chromosomes that contain the DNA strands themselves containing the genetic information of any living things

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3 years ago

Raul wants to increase his upper body and cardiovascular strength. He plans to go to the gym and use his backyard swimming pool

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Answer: An exercise, activity, or workout that targets a specific training need Progression - Gradually increasing your skill level and ability over time Overload -Working your body harder than usual to improve how it functions Reversibility - Losing what you have gained by taking time off Tedium - getting bored by doing the same thing

Explanation:

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3 years ago

Boyle's law says that the volume of a gas varies inversely with the pressure. When the volume of a certain gas is 4l , the press

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Boyle's law says that the volume of a gas varies inversely with the pressure. When the volume of a certain gas is 4l , the pressure is 720 kpa (kilopascals). What is the pressure when the volume is 10l ?

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2 years ago

In 2 1/2 hours an airplane travels 1150 km against the wind. It takes 50 min to travel 450 km with the wind. Find the speed of t

Tamiku [17]

Answer:

Velocity of airplane is 500 km/h

Velocity of wind is 40 km/h

Explanation:

$V_a$ = Velocity of airplane in still air

$V_w$ = Velocity of wind

Time taken by plane to travel 1150 km against the wind is 2.5 hours

$V_a-V_w=\frac {1150}{2.5}\\\Rightarrow V_a-V_w=460\quad (1)$

Time taken by plane to travel 450 km against the wind is 50 minutes = 50/60 hours

$V_a+V_w=\frac {450}{50}\times 60\\\Rightarrow V_a-V_w=540\quad (2)$

Subtracting the two equations we get

$V_a-V_w-V_a-V_w=460-540\\\Rightarrow -2V_w=-80\\\Rightarrow V_w=40\ km/h$

Applying the value of velocity of wind to the first equation

$V_a-40=460\\\Rightarrow V_a =500\ km/h$

∴ Velocity of airplane in still air is 500 km/h and Velocity of wind is 40 km/h

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3 years ago

The smallest unit of charge is − 1.6 × 10 − 19 C, which is the charge in coulombs of a single electron. Robert Millikan was able

vovangra [49]

Answer:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

Explanation:

<u>Charge of an Electron</u>

Since Robert Millikan determined the charge of a single electron is

$q_e=-1.6\cdot 10^{-19}\ C$

Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is $5\times -1.6\cdot 10^{-19}\ C=-8 \cdot 10^{-19}\ C$

Let's test the possible charges listed in the question:

$-8.0 \times 10 ^{-19 }$ . We have just found it's a possible charge of a particle

$-3.2 \times 10 ^{-19 }$ . Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets

$-1.2 \times 10 ^{-19 }$ this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge

$-5.6 \times 10 ^{-19 },\ -9.4 \times 10 ^{-19 }$ cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6

Finally, the charge $-4.8 \times 10 ^{-19 }\ C$ is four times the charge of the electron, so it is a possible value for the charge of an oil droplet

Summarizing, the following are the possible values for the charge of an oil droplet:

$-8.0 \times 10 ^{-19 }\ C,\ -3.2 \times 10 ^{-19 }\ C, -4.8 \times 10 ^{-19 }\ C$

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2 years ago