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tino4ka555 [31]
3 years ago
8

What do you mean by the statement relative density of gold is 19.3

Physics
1 answer:
sergey [27]3 years ago
8 0
The relative density of gold is 19.3 it means the ratio obtained by dividing the density of gold by water at temp of 4 degree celcius is 19.3
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ustapha Jones is speeding on the interstate in his Ferrari at 231km/hr when he passes a police car at rest . If the cop accelera
Lena [83]

Answer:

461 km/h

Explanation:

In order to solve this problem we must first sketch a drawing of what the situation looks like so we can better visualize it. (See attached picture).

We have two situations there, the first one is Mustapha's car that is traveling at a constant speed of 231km/hr.

The second situation is the police that is accelerating from rest until he reaches Mustapha. (We are going to suppose the acceleration is constant and that he will not stop accelerating until he reaches Mustapha). He has an acceleration of 15m/s^{2}.

We want to find what the final velocity of the police is at the time he reaches Mustapha. From this we can imply that the displacement x will be the same for both particles.

So let's model the first situation.

The displacement of Mustapha can be found by using the following equation:

V_{M}=\frac{x}{t}

when solving the equation for the displacement x we get that it will be:

x=V_{M}t

Now let's model the displacement of the cop. Since the cop has a constant acceleration, we can model his displacement with the following formula.

x=V_{0}t+\frac{1}{2}at^{2}

Since the initial velocity of the cop is zero, we can get rid of that part of the equation leaving us with:

x=\frac{1}{2}at^{2}

We can now set both equations equal to each other so we get:

\frac{1}{2}at^{2}=V_{M}t

When solving this for t, we get that:

t=\frac{2V_{M}}{a}  (let's call this equation 1)

Now, we know the cop has constant acceleration, so we can model it with the following formula too:

a=\frac{V_{f}-V_{0}}{t}

since the initial velocity of the cop is zero, we can get rid of that here too, so we get the following formula:

a=\frac{V_{f}}{t}

when solving for the final velocity, we get that:

V_{f}=at  (let's call this equation 2)

when substituting equation 1 into equation 2 we get:

V_{f}=a(\frac{2V_{M}}{a})

we can now cancel a leaving us with:

V_{f}=2V_{M}

This tells us that the final velocity of the cop will not depend on his acceleration. (This is only if the acceleration is constant all the time) So we get that the final velocity of the cop is:

V_{f}=2(231km/hr)

so

V_{f}=461km/hr

which is our answer.

8 0
3 years ago
Andy is walking around the 100 meter track that has the start line on the west side of the field and has
Ad libitum [116K]

Answer:

its distance and displacement

Explanation:

3 0
3 years ago
Two blocks are on a horizontal frictionless surface. Block a is moving with an initial velocity
AfilCa [17]

The final velocity of the block A will be 2.5 m/sec. The principal of the momentum conversation is used in the given problem.

<h3>What is the law of conservation of momentum?</h3>

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

In a given concern, mass m₁ is M, mass m₂ is 3M. Initial speed for the mass m₁ and  m₂ will be u₁=5 and u₂=0 m/s respectively,

According to the law of conservation of momentum

Momentum before collision =Momentum after collision

m₁u₁+m₂u₂=(m₁+m₂)v

M×5+3M×0=[M+3M]v

The final velocity is found as;

V=51.25 m/s

The velocity of block A is found as;

\rm V_f=\frac{(m_1-m_2)u_1}{m_1+m_2} +\frac{2m_2u_2}{m_1+m_2} \\\\  V_f=\frac{(M-3M)5+2\times3M\times0}{m_1+m_2} \\\\ V_f=2.5 m/s

Hence, the final velocity of the block A will be 2.5 m/sec.

To learn more about the law of conservation of momentum, refer;

brainly.com/question/1113396

#SPJ4

5 0
3 years ago
2. What will be the extension of this spring if the load is a) 4N and b) 75 g?
Furkat [3]

Answer:

6

Explanation:

just add

7 0
3 years ago
Help science 70 points
nirvana33 [79]
I didn't want to comment but don't trust those links it's a scam or a virus
3 0
3 years ago
Read 2 more answers
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