All categories

3 years ago

A plane flies for 4.2 hours at the velocity of 1250km/hr. What distance will the plane travel?

Physics

1 answer:

Mariulka [41]3 years ago

3 0

If it goes 1250km every hour, times 4.2 hours
1250*4.2
5250 km

You might be interested in

The transfer of energy carried in rays is called convection true or false

Oxana [17]

False that is radiation

4 0

3 years ago

The human body is an organism made up of cells, tissues, organs, and organ system. Here are examples of three types of muscle ti

REY [17]

I would think that would be C.

8 0

4 years ago

A body is traveling at 5.0 m/s along the positive direction of an x axis; no net force acts on the body. An internal explosion s

loris [4]

To solve this problem it is necessary to apply the concepts related to the conservation of momentum and conservation of kinetic energy.

By definition kinetic energy is defined as

$KE = \frac{1}{2} mv^2$

Where,

m = Mass

v = Velocity

On the other hand we have the conservation of the moment, which for this case would be defined as

$m*V_i = m_1V_1+m_2V_2$

Here,

m = Total mass (8Kg at this case)

$m_1=m_2 =$ Mass each part

$V_i =$ Initial velocity

$V_2 =$ Final velocity particle 2

$V_1 =$ Final velocity particle 1

The initial kinetic energy would be given by,

$KE_i=\frac{1}{2}mv^2$

$KE_i = \frac{1}{2}8*5^2$

$KE_i = 100J$

In the end the energy increased 100J, that is,

$KE_f = KE_i KE_{increased}$

$KE_f = 100+100 = 200J$

By conservation of the moment then,

$m*V_i = m_1V_1+m_2V_2$

Replacing we have,

$(8)*5 = 4*V_1+4*V_2$

$40 = 4(V_1+V_2)$

$V_1+V_2 = 10$

$V_2 = 10-V_1$ (1)

In the final point the cinematic energy of EACH particle would be given by

$KE_f = \frac{1}{2}mv^2$

$KE_f = \frac{1}{2}4*(V_1^2+V_2^2)$

$200J=\frac{1}{2}4*(V_1^2+V_2^2)$ (2)

So we have a system of 2x2 equations

$V_2 = 10-V_1$

$200J=\frac{1}{2}4*(V_1^2+V_2^2)$

Replacing (1) in (2) and solving we have to,

200J=\frac{1}{2}4*(V_1^2+(10-V_1)^2)

PART A: $V_1 = 10m/s$

Then replacing in (1) we have that

PART B: $V_2 = 0m/s$

8 0

3 years ago

A change in position of an object is its:

finlep [7]

Answer:

it's B I think I did this the other day tbh sorry if it's wrong btw hope this helps

8 0

3 years ago

Suppose you walk 21.5 m straight east and then 12.5 m straight north. How far are you from your starting point (in m)

stira [4]

Answer:

24.9 m

Explanation:

A triangle is a polygon with three sides and three angles. There are different types of triangles such as the equilateral triangle, isosceles triangle, scalene triangle, right angled triangle.

You walk 21.5 m straight east and then 12.5 m straight north. This forms a right angled triangle with a horizontal distance of 21.5 m, a vertical distance of 12.5 m and the hypotenuse is the distance between the ending and starting point. Let x represent the distance between the ending and starting point. Therefore using Pythagoras theorem:

x² = 21.5² + 12.5²

x² = 462.25 + 156.25

x² = 618.5

Taking square root of both sides:

√x² = √618.5

x = √618.5

x = 24.9 m = distance between the ending and starting point.

4 0

3 years ago