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2 years ago

Why is evaporation of water greater in summer?

2 answers:

7 0

Evaporation of water is greater during the summer, because of the sever change of temperature (Depending where you live) and the increase of light rays on the earths surface, and bodies of water.

prohojiy [21]2 years ago

3 0

More sun greater temperatures <span />

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The mass of a planet is 3.7 x 1024 kg. If the planet has a radius of 9.2 x 106 m what is the acceleration of gravity for a perso

nalin [4]

Explanation:

It s given that,

Mass of a planet, $M=3.7\times 10^{24}\ kg$

Radius of a planet, $R=9.2\times 10^{6}\ m$

(1) We need to find the acceleration due to gravity for a person on the surface of the planet. Its formula is given by :

$g=\dfrac{GM}{R^2}$

$g=\dfrac{6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{(9.2\times 10^{6}\ m)^2}$

$g=2.91\ m/s^2$

(2) The escape velocity is given by :

$v=\sqrt{\dfrac{2GM}{R}}$

$v=\sqrt{{\dfrac{2\times 6.67\times 10^{-11}\ Nm^2/kg^2\times 3.7\times 10^{24}\ kg}{9.2\times 10^{6}\ m}}$

v = 7324.61 m/s

Hence, this is the required solution.

3 0

3 years ago

Psychological profiling is also known as criminal profiling. True or False?

Nikolay [14]

So my opinion that is false bc. the psychological profiling not is necessary being as criminal profiling ,the criminal mean a greater,intensive word in the way of ,,killer" ,when the ,,psychological" profile is total different way

hope helped

8 0

2 years ago

Read 2 more answers

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical ener

defon

Complete question:

A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find

(a) the force constant of the spring and (b) the amplitude of the motion.

Answer:

(a) the force constant of the spring = 47 N/m

(b) the amplitude of the motion = 0.292 m

Explanation:

Given;

mass of the spring, m = 200g = 0.2 kg

period of oscillation, T = 0.410 s

total mechanical energy of the spring, E = 2 J

The angular speed is calculated as follows;

$\omega = \frac{2\pi}{T} \\\\\omega = \frac{2\pi}{0.41} \\\\\omega = 15.33 \ rad/s$

(a) the force constant of the spring

$\omega = \sqrt{\frac{k}{m} } \\\\\omega^2 = \frac{k}{m} \\\\k = m \omega^2\\\\k = (0.2)(15.33)^2\\\\k = 47 \ N/m$

(b) the amplitude of the motion

E = ¹/₂kA²

2E = kA²

A² = 2E/k

$A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2\times 2}{47} }\\\\A = 0.292 \ m$

7 0

3 years ago

A train has a length of 81.1 m and starts from rest with a constant acceleration at time t = 0 s. At this instant, a car just re

arlik [135]

Answer: a) vcar= 7 m/s ; b) a train= 0.65 m/s^2

Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.

We have for the car

distance = v car* t, considering the length of train (81.1 m) travel by the car during the first 11.6 s

the v car = distance/time= 81.1 m/11.6s= 7 m/s

In order to calculate the acceleration we have to use the kinematic equation for the train from the rest

distance train = (a* t^2)/2

distance train : distance travel by the car at constant speed

so distance train= (vcar*36.35)m=421 m

the a traiin= (2* 421 m)/(36s)^2=0.65 m/s^2

4 0

2 years ago

Consider a projectile of mass 20 kg launched with a speed 9 m/s at an elevation angle of 45 degrees. Taking the launch point as

viva [34]

Answer:

a) L=0. b) L = 262 k ^ Kg m²/s and c) L = 1020.7 k^ kg m²/s

Explanation:

It is angular momentum given by

L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

One of the easiest ways to make this vector product is with the use of determinants

${array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]$

Let's apply this relationship to our case

Let's start by breaking down the speed

v₀ₓ = v₀ cosn 45

voy =v₀ sin 45

v₀ₓ = 9 cos 45

voy = 9 without 45

v₀ₓ = 6.36 m / s

voy = 6.36 m / s

a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

vfy² = voy²- 2 g y

y = voy² / 2g

y = (6.36)²/2 9.8

y = 2.06 m

Let's calculate the angular momentum

L= $\left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]$

L = -px y k ^

L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^ Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

R = vo² sin 2θ / g

R = 9² sin (2 45) /9.8

R = 8.26 m

L = $\left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]$

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^ kg m² /s

5 0

3 years ago