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3 years ago

Which mountains are YOUNGEST?

Physics

2 answers:

Lisa [10]3 years ago

4 0

The Answer is definitely B

Hoochie [10]3 years ago

3 0

The answer is B,
because it is more round and lower than the other mountains.
The steeper and more pointy and higher a mountain is, the older it is.

Hope this helps!
-Jana

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Lions can run at speeds up to approximately 80.0 km / h. A hungry 109 kg lion running northward at top speed attacks and holds o

sukhopar [10]

Answer:

17.34 m/s

Explanation:

Given:

Mass of lion (m₁) = 109 kg

Initial speed of lion (v₁) = 80.0 km/h (Northward direction)

Mass of gazelle (m₂) = 39.0 kg

Initial speed of gazelle (v₂) = 78.5 km/h (Eastward direction)

Final velocity of both lion and gazelle is, $v_f=?$

First, let us convert the speeds from km/h to m/s using the conversion factor.

We know that, 1 km/h = 5/18 m/s

Therefore,

$v_1= 80.0\ km/h=80\times \frac{5}{18}=22.22\ m/s\\\\v_2=78.5\ km/h=78.5\times \frac{5}{18}=21.81\ m/s$

Now, the concept of conservation of total momentum is used here as this is a case of perfectly inelastic collision. In inelastic collision, the masses move together with same velocity after collision.

Here, as the lion and gazelle are moving in directions at right angles to each other, the vector sum of their momentums will give the net initial momentum of the system.

So, initial momentum is given as:

$P_i=\sqrt{P_1^2+P_2^2}\\\\Where,\\\\P_1\to initial\ momentum\ of\ lion\\P_2\to initial\ momentum\ of\ gazelle$

Now, we calculate P₁ and P₂.

$P_1=m_1v_1=(109\ kg)(22.22\ m/s) = 2421.98\ Ns\\\\P_2=m_2v_2=(39\ kg)(21.81\ m/s) = 850.59\ Ns$

Therefore, the net initial momentum of the system is given as:

$P_i=\sqrt{(2421.98)^2+(850.59)^2}=2567\ Ns$

The final momentum of the system is given as:

$P_f=(m_1+m_2)(v_f)\\\\P_f=(109+39)v_f\\\\P_f=148v_f$

From the law of conservation of momentum, the final momentum is equal to the initial momentum. So,

$P_f=P_i\\\\148v_f=2567\\\\v_f=\frac{2567}{148}=17.34\ m/s$

Therefore, the final speed of the lion-gazelle system is 17.34 m/s

3 0

3 years ago

What is the relationship between karst development and topography?

mylen [45]

The karst topography is typically defined as a geographic location characterized by a rugged terrain containing landscapes like underground rivers, fissures, and cracks. It is mainly due to the dissolution of the bedrock due to a much heavier precipitation taking place in the geographic location.

5 0

3 years ago

Four fixed point charges are at the corners of a square with sides of length L. Q1 is positive and at (OL) Q2 is positive and at

Ne4ueva [31]

Answer:

A) See Annex

B) Fq₁₂ = K * Q₁*Q₂ /16 [N] (repulsion force)

C) Fq₃₂ = K * Q₃*Q₂ /16 [N] (repulsion force)

D) Fq₄₂ = K * Q₄*Q₂ /32 [N] (attraction force)

E) Net force (its components)

Fnx = (2,59/64 )* K*Q² [N] in direction of original Fq₃₂

Fny =(2,59/64 )* K*Q² [N] in direction of original Fq₁₂

Explanation:

For calculation of d (diagonal of the square, we apply Pythagoras Theorem)

d² = L² + L² ⇒ d² = 2*L² ⇒ d = √2*L² ⇒ d= (√2 )*L

d = 4√2 units of length (we will assume meters, to work with MKS system of units)

B) Force of Q₁ exerts on charge Q₂

Fq₁₂ = K * Q₁*Q₂ /(L)² Fq₁₂ = K * Q₁*Q₂ /16 (repulsion force in the direction indicated in annex)

C) Force of Q₃ exerts on charge Q₂

Fq₃₂ = K * Q₃*Q₂ /(L)² Fq₃₂ = K * Q₃*Q₂ /16 (repulsion force in the direction indicated in annex)

D) Force of -Q₄ exerts on charge Q₂

Fq₄₂ = K * Q₄*Q₂ / (d)² Fq₄₂ = K * Q₄*Q₂ /32 (Attraction force in the direction indicated in annex)

E) Net force in the case all charges have the same magnitude Q (keeping the negative sign in Q₄)

Let´s take the force that Q₄ exerts on Q₂ and Q₂ = Q ( magnitude) and

Q₄ = -Q

Then the force is:

F₄₂ = K * Q*Q / 32 F₄₂ = K* Q²/32 [N]

We should get its components

F₄₂(x) = [K*Q²/32 ]* √2/2 and so is F₄₂(y) = [K*Q²/32 ]* √2/2

Note that this components have opposite direction than forces Fq₁₂ and

Fq₃₂ respectively, and that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively

In new conditions

Fq₁₂ = K * Q₁*Q₂ /16 becomes Fq₁₂ = K * Q²/ 16 [N] and

Fq₃₂ = K* Q₃*Q₂ /16 becomes Fq₃₂ = K* Q² /16 [N]

Note that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively

Then over x-axis we subtract Fq₃₂ - F₄₂(x) = Fnx

and over y-axis, we subtract Fq₁₂ - F₄₂(y) = Fny

And we get:

Fnx = K* Q² /16 - [K*Q²/32 ]* √2/2 ⇒ Fnx = K*Q² [1/16 - √2/64]

Fnx = (2,59/64 )* K*Q²

Fny has the same magnitude then

Fny =(2,59/64 )* K*Q²

The fact that Fq₁₂ and Fq₃₂ are bigger than F₄₂(x) and F₄₂(y) respectively, means that Fnx and Fny remains as repulsion forces

5 0

3 years ago

The parents of a young child carefully placed plastic plug protectors in all the electrical outlets in their house. Which statem

DanielleElmas [232]

I think the answer is c or b but i p ick c

4 0

3 years ago

Read 2 more answers

How do u find net force?

Damm [24]

1Draw a quick sketch of the object.
2Draw an arrow showing every force acting on the object.
3<span>To calculate the net force, add any vectors acting on the same axis (x and y), making sure to pay attention to the directions.
Hope this helps :)
</span>

7 0

3 years ago