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ipn [44]
4 years ago
15

Are basins a collection of smaller watersheds

Physics
1 answer:
mihalych1998 [28]4 years ago
7 0

Explanation:

both are areas of land that drain to particular water bodies such as lakes

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A 300g ball and a 100g ball are dropped from a tower. Both balls are the same size and their is no air resistance. Which one hit
kifflom [539]

Answer: The balls would hit the ground at the same time.

Explanation: Since there is no air resistance, we would put Galileo's experiment into motion. Galileo once performed an experiment of dropping two items, with different masses, from the tower of Pisa. Since there was no air resistance, the balls hit the ground at the same time. In this problem, the balls are "free-falling." Freefall is a term used in Physics to describe the motion of a falling object experiencing only the acceleration due to gravity. g=9.8 m/s^2 (acceleration due to gravity.) 9.8 will always be the acceleration in a free fall event like this with no air resistance, since Gravity remains constant.

Newton's second law of motion states that F=ma, (force is equal to mass times acceleration), but since Newton also states that f=mg, we can conclude that ma=mg. Stating that no matter what the mass of the object is, both of the objects will fall at the same time with the same velocity.

3 0
3 years ago
The impact of the type of material of which the slope is made on acceleration
Pachacha [2.7K]

 The impact of the material type with which the slope is made affects the acceleration. Acceleration will be higher and smoother if the material of the slope surface is smoother as opposed to a texture which is not smooth. Smoother surface allows more acceleration because it will have less friction and resistance. Otherwise the friction will slow the object down for example a grassy ground will have more friction than a well maintained marble floor. 

6 0
3 years ago
Taro stated that when someone hits a golf ball with a club, the amount of energy the ball has changes, the amount of energy that
topjm [15]
The total energy of the system remains the same.
6 0
3 years ago
The first excited state of a particular atom in a gas is 5.7 eV above the ground state. A moving electron collides with one of t
Georgia [21]

Answer:

Kinetic energy, E = 9.1 eV

Explanation:

It is given that,

At the first excited state of a particular atom is 5.7 eV above the ground state. The kinetic energy of the atom after the collision is 3.4 eV.

We need to find the kinetic energy of the electron just before the collision. The conservation of momentum will be followed here. The energy gets transferred from moving electrons to these atoms is 5.7.

Let E is the kinetic energy of the electron just before the collision. It is equal to :

E = 5.7 eV + 3.4 eV

E = 9.1 eV

So, the kinetic energy of the electron just before the collision is 9.1 eV. Hence, this is the required solution.

5 0
4 years ago
In the figure, a proton is projected horizontally midway between two parallel plates that are separated by 0.50 cm, and are 5.60
noname [10]

a) Minimum speed of the proton: 6.05\cdot 10^6 m/s

b) Angle of the velocity: \theta=-5.1^{\circ}

Explanation:

a)

The proton experiences a vertical force due to the electric field, given by:

F=qE

where

q=1.6\cdot 10^{-19}C is the proton charge

E=610,000 N/C is the magnitude of the electric field

The vertical acceleration of the proton is therefore

a=\frac{qE}{m}

where

m=1.67\cdot 10^{-27}kg is its mass

Therefore, the vertical position of the proton at time t is

y(t)=\frac{1}{2}at^2=\frac{1}{2}\frac{qE}{m}t^2

where we assumed that the initial vertical velocity is zero (because the proton is fired horizontally) and the initial vertical position, halfway between the two plates, is the origin.

The horizontal motion of the proton instead is uniform, so the horizontal position is given by

x(t)=v_0 t

where v_0 is the initial speed. This equation can be rewritten as

t=\frac{x(t)}{v_0}

And substituting into the eq. for y,

y(t) = \frac{1}{2}\frac{qE}{m} \frac{x^2}{v_0^2}

Solving for the initial speed,

v_0 = \sqrt{\frac{qEx^2}{2my}}

The proton just misses one of the plate when

x = 5.60 cm = 0.056 m (length of the plates)

y = 0.25 cm = 0.0025 m (half the distance between the plates)

Therefore, we find the initial speed:

v=\sqrt{\frac{(1.6\cdot 10^{-19})(610,000)(0.056)^2}{2(1.67\cdot 10^{-27})(0.0025)}}=6.05\cdot 10^6 m/s

b)

In order to find the angle, we just need to analyze the horizontal and vertical component of the final velocity of the proton.

The horizontal velocity is constant so it is:

v_x = v_0 = 6.05\cdot 10^6 m/s

The vertical velocity is given by:

v_y^2 - u_y^2 = 2ay

where:

u_y=0 (initial vertical velocity is zero)

a=\frac{qE}{m} (acceleration)

y = 0.0025 m (vertical displacement)

Solving for v_y,

v_y = \sqrt{2ay}=\sqrt{2\frac{qEy}{m}}=5.4\cdot 10^5 m/s

Therefore, the final angle of the velocity with respect to the horizontal is:

\theta = tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{5.4\cdot 10^5}{6.05\cdot 10^6})=5.1^{\circ}

And since the electric field is downward (the proton just misses the lower field), it means that the angle is below the horizontal:

\theta=-5.1^{\circ}

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

8 0
3 years ago
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