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2 years ago

If two balls have the same volume,

1 answer:

8 0

Here, we are required to find the relationship between balls of different mass(a measure of weight) and different volumes.

1. Ball A will have the greater density
2. Ball C and Ball D have the same density.
3. Ball Q will have the greater density.
4. Ball X and Y will have the same density

The density of an object is given as its mass per unit volume of the object.

Mathematically;.

Density = Mass/Volume.

For Case 1:

Va = Vb and Ma = 2Mb
D(b) = (Mb)/(Vb) and D(a) = 2(Mb)/Vb
Therefore, the density of ball A,
D(a) = 2D(b).
Therefore, ball A has the greater density.

For Case 2:

Vc = 3Vd,

Vd = (1/3)Vc

Md = (1/3)Mc

D(c) = (Mc)/(Vc) and D(d) = (1/3)Md/(1/3)Vd

D(c) = D(d).

Therefore, ball C and D have the same density

For Case 3:

Vp = 2Vq and Mp = Mq
D(p) = (Mq)/2(Vq) and D(q) = (Mq)/Vq
Therefore, the density of ball P is half the density of ball Q
Therefore, ball Q has the greater density.

For case 4:

Mx = (1/2)My
Vx = Vy

Therefore, Ball X and Ball Y have the same density.

Read more:

brainly.com/question/18110802

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The potential energy of a 1.7 x 10-3 kg particle is described by U (x )space equals space minus (17 space J )space cos open squa

hichkok12 [17]

Answer:

Explanation:

Given a particle of mass

M = 1.7 × 10^-3 kg

Given a potential as a function of x

U(x) = -17 J Cos[x/0.35 m]

U(x) = -17 Cos(x/0.35)

Angular frequency at x = 0

Let find the force at x = 0

F = dU/dx

F = -17 × -Sin(x/0.35) / 0.35

F = 48.57 Sin(x/0.35)

At x = 0

Sin(0) =0

Then,

F = 0 N

So, from hooke's law

F = -kx

Then,

0 = -kx

This shows that k = 0

Then, angular frequency can be calculated using

ω = √(k/m)

So, since k = 0 at x = 0

Then,

ω = √0/m

ω = √0

ω = 0 rad/s

So, the angular frequency is 0 rad/s

4 0

3 years ago

The half-life of the radioactive element beryllium-13 is 5 × 10-10 seconds, and half-life of the radioactive element beryllium-1

telo118 [61]

<h2>Answer: The half-life of beryllium-15 is 400 times greater than the half-life of beryllium-13.</h2>

Explanation:

The half-life $h$ of a radioactive isotope refers to its decay period, which is the average lifetime of an atom before it disintegrates.

In this case, we are given the half life of two elements:

beryllium-13: $h_{B-13}=5(10)^{-10}s=0.0000000005s$

beryllium-15: $h_{B-15}=2(10)^{-7}s=0.0000002s$

As we can see, the half-life of beryllium-15 is greater than the half-life of beryllium-13, but how great?

We can find it out by the following expression:

$h_{B-15}=X.h_{B-13}$

Where $X$ is the amount we want to find:

$X=\frac{h_{B-15}}{h_{B-13}}$

$X=\frac{2(10)^{-7}s}{5(10)^{-10}s}$

Finally:

$X=400$

Therefore:

The half-life of beryllium-15 is <u>400 times greater than</u> the half-life of beryllium-13.

8 0

3 years ago

Later, while taking your groceries back to the car, you accidentally let go of your cart. It rolls straight down a grassy slope

VladimirAG [237]

Car is moving on the glassy slope with constant speed

Now we know that

$a = \frac{dv}{dt}$

so acceleration is rate of change in velocity

as we know that velocity is constant here so acceleration is zero

so here

$a = 0$

now as we know by Newton's II law

$F = ma$

since a = 0

$F = 0$

so net force will be ZERO on it during this motion

6 0

3 years ago

We know that every object exerts an attraction on every other object and the heavier the object ___ the attraction.

hichkok12 [17]

I don't know what the exact word is, but I do know that the bigger an objects mass is the more it will attract other objects toward it, mainly smaller objects with less mass. it might be gravity or something around those lines....is it a multiple choice question?

6 0

3 years ago

A 40-kg uniform semicircular sign 1.6 m in diameter is supported by two wires as shown. What

garri49 [273]

Answer:

T1 = 131.4 [N]

T2 = 261 [N]

Explanation:

To solve this problem we must make a sketch of how will be the semicircle, for this reason we conducted an internet search, to find the scheme of the problem. This scheme is attached in the first image.

Then we make a free body diagram, with this free body diagram, we raise the forces that act on the body. Since it is a problem involving static equilibrium, the sum of forces in any direction and moments must be equal to zero.

By performing a sum of forces on the Y axis equal to zero we can find an equation that relates the forces of tension T1 & T2.

The second equation can be determined by summing moments equal to zero, around the point of application of the T1 force. In this way we find the T2 force.

The value of T2, is replaced in the first equation and we can find the value for T1.

Therefore

T1 = 131.4 [N]

T2 = 261 [N]

The free body diagram and the developed equations can be seen in the second attached image.

4 0

3 years ago