All categories

3 years ago

A plane starting from rest accelerates to 40 m/s in 10 s. How far did the plane travel during this time?

Physics

1 answer:

andrew-mc [135]3 years ago

5 0

Answer:

400 meters

Explanation:

using the equation:

speed = distance/time
it would look like this when plugged in: 40=d/10
d (distance) is the unknown variable, so you have to solve for that
multiply both sides by 10 since you want to separate d to solve its number.
10d/10=40 × 10
the two tens on the d part of the equation would cancel out, leaving only d.
Therefore, d = 400

Hopefully this helps.

You might be interested in

When x rays are passed through successive aluminium sheets, their hardness ___.

Sidana [21]

Answer:

decreases

Explanation:

on each collision, the electrons that make up x-rays get absorbed.

4 0

3 years ago

When a small object is launched from the surface of a fictitious planet with a speed of 52.9 m/s, its final speed when it is ver

NISA [10]

Answer:

The escape speed of the planet is 41.29 m/s.

Explanation:

Given that,

Speed = 52.9 m/s

Final speed = 32.3 m/s

We need to calculate the launched with excess kinetic energy

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

$K.E=\dfrac{1}{2}\times m\times(32.3)^2$

We need to calculate the escape speed of the planet

Using formula of kinetic energy

$\text{escape kinetic energy}=\text{launch kinetic energy}-\text{excess kinetic energy}$

$\dfrac{1}{2}mv^2=\dfrac{1}{2}mv^2-\dfrac{1}{2}mv^2$

$\dfrac{1}{2}\times v^2=\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2$

$v=\sqrt{2\times(\dfrac{1}{2}\times(52.9)^2-\dfrac{1}{2}\times(32.3)^2)}$

$v=41.29\ m/s$

Hence, The escape speed of the planet is 41.29 m/s.

4 0

3 years ago

What is the distance covered when the speed is 960km/hr and time is 1 hour and 50 minutes?

OLEGan [10]

The distance is 1760 km

The speed is 960 km hour I.e 960 divide by 60 = 16 km/minute
hence in time 1 hour 50 minutes I.e 110 minutes
distance travelled is 110 - 16 = 1760 km

3 0

3 years ago

Read 2 more answers

An astronaut landed on a far away planet that has a sea of water. To determine the gravitational acceleration on the planet's su

sergiy2304 [10]

To solve this problem it is necessary to apply the concepts related to hydrostatic pressure or pressure due to a fluid.

Mathematically this pressure is given under the formula

$P_h = \rho g h$

Where,

$\rho$ = Density

h = Height

g = Gravitational acceleration

Rearranging in terms of g

$g = \frac{P_h}{\rho h}$

our values are given as

$P_h = 1.1 atm (\frac{101325Pa}{1atm}) = 111457.5Pa$

$\rho = 1000kg/m^3$

$h = 12.3m$

Replacing we have

$g = \frac{111457.5}{(1000)(12.3)}$

$g = 9.061m/s^2$

Therefore the gravitational acceleration on the planet's surface is $9.061m/s^2$ (Almost the gravity of the Earth)

3 0

3 years ago

An object becomes positively charged when it

goldfiish [28.3K]

Answer:

A. loses an electron, it becomes positively charged.

5 0

3 years ago

Read 2 more answers