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Anton [14]

Pause?Freeze?
Stop?
Halt?

5 0

3 years ago

A rocket is launched straight up from the earth's surface at a speed of 1.60×104 m/s . what is its speed when it is very far awa

AlladinOne [14]

16,000 m/s
Since it’s speed, and the distance is unknown. Gravity isn’t applying a noticeable force too on the rocket, as if it were, then the rocket would be accelerating negatively.

7 0

3 years ago

sla’s change in velocity is 30 m/s, and Hazel has the same change in velocity. Which best explains why they would have different

sweet [91]

Because acceleration depends not only on the change in velocity.
It also depends on the time during which the change occurs.
The formula is

Acceleration = (change in velocity) divided by (time for the change) .

Maybe Sla changed his velocity in 3 seconds, but Hazel
took all morning to change hers. In that case, even though
the amounts of change were equal, the times were different,
so the quotients of (change/time) were different.

8 0

3 years ago

Blocks A (mass 3.50 kg) and B (mass 6.50 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block

Vedmedyk [2.9K]

Answer:

(a) V (A) = 0.7 m/s,

(b) V (A) = 0.7 m/s,

(c) V (B) = 0.7 m/s

(d) u= - 0.60 m/s

(e) v = 0.75 m/s

Explanation:

Given:

M(A) =3.50 Kg, M(B)=6.50 Kg, V(A) = 2.00 m/s, V(B) = 0 m/s

Sol:

a) law of conservation of momentum

M(a) x V(A) + M(B) x V(B) = ( M(a) + M(B) ) V (let V is Common Velocity of Both block)

so 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = (3.50 Kg + 6.50 Kg ) V

after solving V = 0.7 m/s

After the collision the velocities of the both block will be as the the spring is compressed maximum.

V (A) = 0.7 m/s

b) V(A) = 0.7 m/s ( Part (a) and Part (a) are repeated )

c) as stated above the in the Part (a)

V(B) = 0.7 m/s

d) When the both blocks moved apart after the collision:

Let u=velocity of block A after the collision.

and v = velocity of block B after the collision.

then conservation of momentum

M(a) x V(A) + M(B) x V(B) = M(a) x v + M(B) x u

⇒ 3.50 Kg x 2.00 m/s + 6.50 Kg x 0 m/s = 3.50 Kg x u + 6.50 Kg x v

⇒ 2.00 m/s = u + 1.86 v -----eqn (1) ( dividing both side by 3.50 Kg)

For elastic collision

the velocity relative approach = velocity relative separation

so 2.00 m/s = v-u ----- eqn (2)

⇒v = u + 2.00 m/s

putting this value in eqn (1) we get

2.00 m/s = u + 1.86 (v + 2.00 m/s)

u= - 0.60 m/s

e) putting v= 2.00 m/s in eqn (1)

2.00 m/s = - 2.32 m/s + 1.86 v

v = 0.75 m/s

5 0

3 years ago

Magnus has reached the finals of a strength competition. In the first round, he has to pull a city bus as far as he can. One end

ozzi

Answer:

The total work that the rope does to Mangnus is - 5780 Jules.

Explanation:

By definition, the work is defined as:

$W=F.d$

Where F and d are the force and the total displacement. Note that in the definition the product is a scalar product since F and d are both vectors.

Take into account that according to third Newton's law the force that the rope does to Magnus is opposite to the force that Magnus does to the rope, therefore the scalar product will be negative due the rope's force goes against to Magnus displacement.

For calculating the work, we take 2500 N as the value for the force and 2.312 meters as the value for the displacement:

$W=-2500 N * 2.312 m$

$W=-5780 Nm = -5780 J$

7 0

3 years ago