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IgorLugansk [536]
3 years ago
6

An airplane flies at 40 m/s at an altitude of 500 m. The pilot drops a package that falls to the ground, How long does it take f

or the package to reach the ground?
Physics
1 answer:
Irina-Kira [14]3 years ago
3 0

Answer:

10.1 s

Explanation:

The time of flight of the package is entirely determined by its vertical motion, which is a free fall motion (it is acted upon the force of gravity only). Therefore, it is a uniformly accelerated motion, so we can use the suvat equation:

s=ut+\frac{1}{2}gt^2

where

s is the vertical distance covered

u is the initial vertical velocity

t is the time of flight

g=9.8 m/s^2 is the acceleration due to gravity

For the package in this problem, we have:

s = 500 m is the vertical distance covered

u = 0 is the initial vertical velocity (initially it has only horizontal motion)

So, solving for t, we find the time of flight:

t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(500)}{9.8}}=10.1 s

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An electron moves through a uniform electric field vector E = (2.80î + 5.20ĵ) V/m and a uniform magnetic field vector B = 0.400k
alina1380 [7]

Answer:

1.758820×10^11(-2.5i-0.8j) m/s^2

Explanation:

From the question, the parameters given are; E=(2.80i+ 5.20j) v/m, a uniform magnetic field,B= 0.400K T, acceleration, a= ??? and velocity vector, v= 11.0i metre per seconds (m/s)...

We can solve this problem using the formula below;

Ma= q[E+V × B] ---------------(1).

Note: q is negative, m= mass of electron.

Making acceleration,a the subject of the formula and substituting the parameters into equation (1);

a= -e/m × (2.5i + 5.2j +11.0i × 0.400K)

a= -e/m × (2.5i+5.2j-4.4j)

a= e/m × (-2.5i - 0.8j)

e/m= 1.758820×10^11 c/kg

Therefore, slotting in the value of charge to mass(e/m) ratio;

a= 1.7588×10^11×(-2.5i-0.8j) m/s^2

7 0
3 years ago
The climate of a town on the coast will have less dramatic changes than a town further inland. Is this statement true or false?
zhenek [66]

Answer:

ture

Explanation:

4 0
2 years ago
What is the speed of a wave if the wavelength is 100m and the period is 20s
Sergio039 [100]
I don’t think I’m right but I want to say 500 m/s
4 0
3 years ago
Read 2 more answers
Suppose that the space shuttle Columbia accelerates at 14.0 m/s2 for 8.50 minutes after takeoff.
givi [52]

Answer:

A. speed = 7.14 Km/s

B. distance = 1820.7 Km

Explanation:

Given that: a = 14.0 m/s^{2}, t = 8.50 minutes.

But,

t = 8.50 = 8.50 x 60

  = 510 seconds

A. By applying the first equation of motion, the speed of the shuttle at the end of 8.50 minutes can be determined by;

v = u + at

where: v is the final velocity, u is the initial velocity, a is the acceleration and t is the time.

u = 0

So that,

v = 14 x 510

 = 7140 m/s

The speed of the shuttle at the end of 8.50 minute is 7.14 Km/s.

B. the distance traveled can be determined by applying second equation of motion.

s = ut + \frac{1}{2}at^{2}

where: s is the distance, u is the initial velocity, a is the acceleration and t is the time.

u = 0

s = \frac{1}{2}at^{2}

  = \frac{1}{2} x 14 x (510)^{2}

 = 7 x 260100

 = 1820700 m

The distance that the shuttle has traveled during the given time is  1820.7 Km.

5 0
2 years ago
About 65 million years ago an asteroid struck Earth in the area of the Yucatán Peninsula and wiped out the dinosaurs and many ot
9966 [12]

Answer:

2.44156\times 10^{13}\ m^3

29010.53917 m

Explanation:

\rho = Density of asteroid = 2 g/cm³

V = Volume

d = Diameter = 10 km

r = Radius = \dfrac{d}{2}=\dfrac{10}{2}=5\ km

v = Velocity = 11 km/s

H_v = Heat vaporization of water = 2.26\times 10^6\ J/kg

\Delta T = Change in temperature = 100-20

Mass is given by

m=\rho V\\\Rightarrow m=\rho\dfrac{4}{3}\pi r^3\\\Rightarrow m=2000\dfrac{4}{3}\times \pi\times 5000^3\\\Rightarrow m=1.0472\times 10^{15}\ kg

The kinetic energy is

K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}1.0472\times 10^{15}\times 11000^2\\\Rightarrow K=6.33556\times 10^{22}\ J

Heat is given by

Q=mc\Delta T+mH_v\\\Rightarrow 6.33556\times 10^{22}=m\times (4186\times (100-20)+2.26\times 10^6)\\\Rightarrow m=\dfrac{ 6.33556\times 10^{22}}{4186\times (100-20)+2.26\times 10^6}\\\Rightarrow m=2.44156\times 10^{16}\ kg

Mass of water is 2.44156\times 10^{16}\ kg

Volume is \dfrac{2.44156\times 10^{16}}{10^3}=2.44156\times 10^{13}\ m^3

Amount of water is 2.44156\times 10^{13}\ m^3

If it were a cube

h=V^{\dfrac{1}{3}}\\\Rightarrow h=(2.44156\times 10^{13})^{\dfrac{1}{3}}\\\Rightarrow h=29010.53917\ m

The height of the water would be 29010.53917 m

4 0
3 years ago
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