Why can you see your image in a pond and why does the image faint when the water is not still? anyone?

On a frictionless track,cart 1 is moving with a constant,rightward(+) velocity of 1.0m/s.Cart 2 is also moving rightward with co

mamaluj [8]

Answer:

The final velocity of cart 1 is 3m/s

Explanation:

From principle of conservation of linear momentum, which states that sum of the momentum before collision is equal to the sum of the momentum after collision.

Momentum, P is given as mass x velocity.

ΔP = Δmv = m₁u₁ +m₂u₂ = m₁v₁ + m₂v₂

Assumptions:

If the two carts are moving on frictionless track, then limiting frictional forces due to their weights are negligible.

After the elastic collision, the two carts will move separately with different velocity

u₁ + u₂ = v₁ + v₂;

where;

u₁ and u₂ are the initial velocity for cart 1 and cart 2 respectively

v₁ and v₂ are the final velocity for cart 1 and cart 2 respectively

1 m/s + 5 m/s = v₁ + 3m/s

6 m/s = v₁ + 3m/s

v₁ = 6 m/s - 3m/s = 3m/s

Therefore, the final velocity of cart 1 is 3m/s

4 0

3 years ago

Pages 64 to 65 of your reading material trace an extended problem that involes lifting a bag of sugar up to a shelf at first it

nadezda [96]

Answer:

d) the amount of work is the same whether the bag is moved all at once or in two stages, provided the total height lifted is the same in either case.

Explanation:

While moving the bag to the shelf in one shot we can say that the total work done is given as

$W = mg(2H)$

here we know that

2H = total height raised by the bag

now when we raise the bag to first shelf and then move it to next shelf

then we will have

$W = W_1 + W_2[tex][tex]W = mgH + mgH$

$W = 2mgH$

so the correct answer will be

d) the amount of work is the same whether the bag is moved all at once or in two stages, provided the total height lifted is the same in either case.

7 0

3 years ago

A rocket initially at rest accelerates at a rate of 99.0 meters/second2. Calculate the distance covered by the rocket if it atta

Jobisdone [24]

Answer: The correct answer is option b.

Explanation: We are given that the rocket is at rest initially final velocity is 445m/s.

The acceleration of the rocket is $99.0m/s^2$

To calculate the distance of rocket, we use third equation of motion, which is:

$v^2-u^2=2as$

where, v = final velocity = 445m/s

u = initial velocity = 0m/s

a = acceleration = $99m/s^2$

s = distance = ? m

Putting values in above equation, we get:

$(445)^2-(0)^2=2(99)s\\\\s=1\times 10^3meters$

3 0

3 years ago

A cheetah and a gazelle are grazing in the savannah. The gazelle is 275 meters away from the gazelle safe zone and the cheetah i

KATRIN_1 [288]

Answer:

1) Yes, the gazelle gets to safety

2) The speed with which the gazelle needs to run, to beat the cheater by 2 seconds is approximately 29.79 m/s

Explanation:

1) The distance of the gazelle, from the gazelle safe zone = 275 m

The distance of the cheetah from the gazelle = 455 m

The speed of the gazelle = 25 m/s

The speed of the cheetah = 65 m/s

Therefore, we have;

Let the time the gazelle reaches the safe zone = t, which gives;

t = 275 m/(25 m/s) = 11

t = 11 seconds

Let the time the cheetah reaches the gazelle = t₁, we have;

455 + 25 × t₁ = 65 × t₁

t₁ = 455/40 = 11.375

t₁ = 11.375 seconds

The gazelle reaches the gazelle safe zone before the cheetah reaches the gazelle

Therefore, the gazelle gets to safety

2) In order for the gazelle to beat the gazelle by 2 seconds, we have;

The time for the cheetah to reach the safe zone = (275 + 455)/65 = 11.23 seconds

Therefore, we have;

In order for the gazelle to beat the gazelle by 2 seconds the time the cheetah reaches the safe zone = 11.23 - 2 = 9.23 s

The speed of the gazelle is then 275/9.23 ≈ 29.79 m/s

6 0

3 years ago

Two people are standing on rollerskates. One is more massive than the other. They push against each other and move away. How do

Gala2k [10]

If they push away, C. The more massive person moves slower.

6 0

3 years ago

Why can you see your image in a pond and why does the image faint when the water is not still? anyone?​

Why can you see your image in a pond and why does the image faint when the water is not still? anyone?