All categories

3 years ago

How long does it take an airplane to fly 5000 m if it maintains a speed of 240 meters per second?

Physics

1 answer:

kolbaska11 [484]3 years ago

8 0

Answer:

20 5/6 sec

Explanation:

To find the solution we divide 5000 by 240

However, when you see a problem, always try to simplify

5000/240=500/24=250/12=125/6

Now the division is much easier

20 5/6 sec

You might be interested in

An iceberg at its melting point

iren2701 [21]

The mass of the iceberg is 71.9 kg

Explanation:

The amount of thermal energy needed to completely melt a substance at its melting point is given by

$Q=\lambda m$

where

$\lambda$ is the latent heat of fusion

m is the mass of the substance

In this problem, we have a block of ice at its melting point (zero degrees). The amount of heat given to the block is

$Q=2.40\cdot 10^7 J$

And the latent heat of fusion of ice is

$\lambda = 334 J/g$

So, we can re-arrange the equation to find m, the amount of ice that will melt:

$m=\frac{Q}{\lambda}=\frac{2.40\cdot 10^7}{334}=71856 g = 71.9 kg$

Learn more about specific heat:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

5 0

3 years ago

A paint can is sitting on a ladder 20 m high. It has a mass of 7 kg. What is the

loris [4]

Answer:

<h3>The answer is 1400 J</h3>

Explanation:

The potential energy of a body can be found by using the formula

PE = mgh

where

m is the mass

h is the height

g is the acceleration due to gravity which is 10 m/s²

From the question we have

PE = 20 × 10 × 7

We have the final answer as

Hope this helps you

3 0

3 years ago

In addition to their remarkable top speeds of almost 60 mph, cheetahs have impressive cornering abilities. In one study, the max

Sav [38]

Answer:

the minimum value of the coefficient of static friction between the ground and the cheetah's feet is 1.94

Explanation:

Given that ;

the top speed of Cheetahs is almost 60 mph

In cornering abilities ; the maximum centripetal acceleration of a cheetah was measured to be = 19 m/s^2

The objective of this question is to determine the what minimum value of the coefficient of static friction between the ground and the cheetah's feet is necessary to provide this acceleration?

From the knowledge of Newton's Law;

we knew that ;

Force F = mass m × acceleration a

Also;

The net force $F_{net}$ = frictional force $\mu_k mg$

so we can say that;

m×a = $\mu_k mg$

where;

the coefficient of static friction $\mu_k$ is:

$\mu_k = \dfrac{m*a}{m*g}$

$\mu_k = \dfrac{a}{g}$

$\mu_k = \dfrac{19 \ m/s^2}{9.81 \ m/s^2}$

$\mathbf{\mu_k}$ = 1.94

Hence; the minimum value of the coefficient of static friction between the ground and the cheetah's feet is 1.94

5 0

3 years ago

Now switch to “Position vs. Time #2” on the Choose a graph menu. Once again, experiment until you are able to reproduce the grap

seraphim [82]

Answer:

The caterpillar had to move across the horizontal segments. I hope this helps :)

Explanation:

6 0

3 years ago

Read 2 more answers

In a thunderstorm, charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considered to be d

amm1812

Answer:

1) q = 83.4 C

2)52.13 × 10^(19) electrons

Explanation:

1) To calculate the total charge q on the cloud when the breakdown of the surrounding air begins, we will use the formula;

E = kq/r²

Making q the subject, we have;

q = Er²/k

Where k is a constant = 1/(4πε_o)

We are given;

ε_o = 8.85 × 10^(-12)) C²/N.m²

Thus;

k = 1/(4 × π × 8.85 × 10^(-12)) N.m²/C²

k = 8.99 × 10^(9)

Also,we are given E = 3 × 10^(6) N/C

Diameter = 1km = 1000 m

Radius(r) = diameter/2 = 1000/2 = 500 m

Thus;

q = (3 × 10^(6) × 500²)/(8.99 × 10^(9))

q = 83.4 C

2) To get the number of Excess electrons, we will divide the charge gotten in Part 1 above by the charge of a single electron.

Now, charge of a single electron = 1.6 × 10^(-19) C

Thus, number of Excess elecrons = 83.4/(1.6 × 10^(-19)) = 52.13 × 10^(19) electrons

5 0

3 years ago