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2 years ago

What type of weathering is RUSTING?

Physics

2 answers:

Zolol [24]2 years ago

6 0

When a metal rusts, it’s called oxidation.

Troyanec [42]2 years ago

4 0

Oxidation

Oxidation is another kind of chemical weathering that occurs when oxygen combines with another substance and creates compounds called oxides. Rust, for example, is iron oxide.

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A 20.5 kg ball moving at 38.5 m/s on a horizontal, frictionless surface runs into a light spring of force

tekilochka [14]

Answer:

7.68 m

Explanation:

Kinetic energy in ball = elastic energy in spring

KE = EE

½ mv² = ½ kx²

mv² = kx²

x = v √(m / k)

x = (38.5 m/s) √(20.5 kg / 515 N/m)

x = 7.68 m

5 0

2 years ago

A ball of radius r rolls on the inside of a track of radius R. If the ball starts from rest at the vertical edge of the track, f

meriva

Answer:

$v=\sqrt{\dfrac{10g(R-r)}{7}}$

Explanation:

Given that

Radius of track = R

Radius of ball = r

The ball can be treated as solid sphere, so

The moment of inertia of ball

$I=\dfrac{2}{5}mr^2$

When the ball reach at the lowest position then it will have both angular and linear speed.

Condition for rolling without slipping v= ωr

Form energy conservation

$mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

v= ωr

$I=\dfrac{2}{5}mr^2$

$mgR=mgr+\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mr^2\omega^2$

$mg(R-r)=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times \dfrac{2}{5}mv^2$

$2mg(R-r)=mv^2+\dfrac{2}{5}mv^2$

$2g(R-r)=\dfrac{7}{5}v^2$

$v=\sqrt{\dfrac{10g(R-r)}{7}}$

3 0

3 years ago

a cyclist must travel 800 km. how many days will the trip take if the cyclist travels 8/h day at an average speed of 16km/h

sergey [27]

It would b 6 days and 6 hours
You would times 16 by 8 which gives 128km a day. To find how many days you would divide 800 by how many kilometres they cover a day which is 128
So 800/128 =6.25 which converts to 6 days and 6 hours.

7 0

3 years ago

Read 2 more answers

How many miles must you travel to make the Earth fall the same amount?

Ede4ka [16]

Let's take the analogy of the baseball pitcher a step farther. When a baseball is thrown in a straight line, we already said that the ball would fall to Earth because of gravity and atmospheric drag. Let's pretend again that there is no atmosphere, so there is no drag to slow the baseball down. Now, let's assume that the person throwing the ball throws it so fast that as the ball falls towards the Earth, it also travels so far, before falling even a little, that the Earth's surface curves away from the ball's path.

In other words, the baseball falls as it did before, but the ball is moving so fast that the curvature of the Earth becomes a factor and the Earth "falls away" from the ball. So, theoretically, if a pitcher on a 100 foot (30.48 m) high hill threw a ball straight and fast enough,the ball would circle the Earth at exactly 100 feet and hit the pitcher in the back of the head once it circled the globe! The bad news for the person throwing the ball is that the ball will be traveling at the same speed as when they threw it, which is about 8 km/s or several times faster than a rifle bullet. This would be very bad news if it came back and hit the pitcher, but we'll get to that in a minute.

4 0

3 years ago

Una pelota de béisbol sale despedida de un bate con una velocidad horizontal de 20 m/s. En un tiempo de 0.25 s, ¿a qué distancia

Tpy6a [65]

Answer:

Translation:

A baseball is thrown from a bat with a horizontal speed of 20 m / s. In a time of 0.25 s, how far will you have traveled horizontally and how much will you have fallen vertically?

Explanation:

This question deals directly with projectile motions and we can use modifications of equation of motion since it's a horizontal motion.

(a). How far will it have travelled;

X = Vo * t

X = 20 * 2.5 = 50m

(b). How much will it have fallen vertically?

y = Voy + ½gt²

But Voy = 0

y = ½ * 9.8 * 2.5²

y = 30.625m

4 0

3 years ago