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amm1812
3 years ago
12

Which part of a stream's sediment load moves the slowest?

Physics
1 answer:
Rudiy273 years ago
4 0
The bed load moves the slowest from all the parts of the stream's sediment. It consists of particles suspended that are suspended and float around the bed. This part is the slowest in motion, as it rolls, and moves with the flow. The particles near the bed are not dissolved so they settle at the bottom and move with the stream.
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How does the vertical component of a projectile’s motion compare with the motion of vertical free fall when air resistance is ne
likoan [24]

Answer:

Explanation:

The vertical component of velocity remains same as the free fall. The vertical motion of the projectile is same as the free fall motion.

5 0
3 years ago
Juan rides his horse with a constant<br> speed of 18 km/h. How far can he travel<br> in 1/2 hour?
Brut [27]

Answer:

He traveled 9km

Explanation:

To do this problem you need to use the equation which is Speed= distance/time and this problem gives you the speed which is 18 km/h and it gives you the time 1/2 hour so you write the equation 18= d/ 1/2 which his distance is 9km

3 0
3 years ago
State the formula for calculating power
Alex
Where power<span> P is in watts, voltage V is in volts and current I is in amperes (DC).</span>Power Formula<span> 2 – Mechanical </span>power equation<span>: </span>Power<span> P = E ⁄ t where </span>power<span> P is in watts, </span>Power<span> P = work / time (W ⁄ t). Energy E is in joules, and time t is in seconds.</span>
7 0
2 years ago
1. A kangaroo hops 84 m to the east in 7 seconds.
DENIUS [597]

Explanation:

Given parameters:

Distance hopped  = 84m

Displacement  = 84m due east

Time  = 7s

Unknown:

Speed of kangaroo  = ?

Velocity of kangaroo  = ?

Solution:

To solve this problem,

    Speed  = \frac{distance}{time }   = \frac{84}{7}  = 12m/s

  Velocity  = \frac{displacement}{time}   = \frac{84}{7}   = 12m/s due east

3 0
3 years ago
You are holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. what
pychu [463]

By holding a positive charge and there are positive charges of equal magnitude 1 m to your north and 1 m to your east. Therefore, the direction of the force on the charge you are holding will be to the southwest.

Let I hold the charge , q at the centre of given co-ordinate system and two positive charge of equal magnitude Q are placed 1 m to my North and 1 m to my South .

now, both the charge are same nature e.g., positive . Let my charge is also positive (well, you can assume negative too , I am considering positive because it makes me easy to solve) then, both charge repel to my charge.

charge Q placed on east is repelling my charge q toward west . similarly charge Q placed on North is repelling my charge q toward south.

Now , use vector for solve it.

vector F_{net} = vector Fe + vector Fn,

⇒ |F_{net}| = \sqrt{}  F^{2} _{e } + F^{2}_n

⇒ Fe = Fs = KqQ/(1m)² = KqQ

⇒ F_{net} = √{Fe² + Fs²} = √{(kqQ)²+(KqQ)²}

⇒ F_{net}= √2KqQ

Hence, net force act on q {my charge } is √2KqQ and the direction of force is S - W (southwest )direction.

To learn more about positive charges here

brainly.com/question/2903220

#SPJ4

8 0
2 years ago
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