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otez555 [7]
3 years ago
5

The radius of a small ball is around 3.79747 cm. The radius of a basketball is about 3.16 times larger. What is the ratio of the

surface areas of the small ball and a basketball? 2. What is the ratio of their volumes?
Physics
1 answer:
Svetradugi [14.3K]3 years ago
3 0

Explanation:

The ratio of the areas is the square of the ratio of the radii.

A/A = 3.16² = 9.99

The ratio of the volumes is the cube of the ratio of the radii.

V/V = 3.16³ = 31.6

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Which statement correctly compares the valence electrons for atoms of francium (Fr) and barium (Ba)?
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Answer:

Francium has fewer valence electrons, but they are in a higher energy level

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A clay ball with a mass of 0.48kg has an initial speed of 4.08m/s it strikes a 3.04kg clay ball at rest and the two balls stick
Rom4ik [11]
Final velocity = 0, thus final kinetic energy is 0
Initial kinetic energy:
0.5mv²
= 0.5 x 0.48 x 4.08²
= 4.0 J
Decrease in kinetic energy = 4 - 0 = 4 Joules
5 0
3 years ago
The specific heat of substance A is greater than that of substance B. Both A and B are at the same initial temperature when equa
Sonja [21]

Answer:

m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

Explanation:

For this case we know the following info:

c_{pA} > c_{pB}

Where c means specific heat for the substance A and B.

We also know that the initial temperatures for both sustances are equal:

T_{iA}= T_{iB}

We assume that we don't have melting or vaporization in the 2 substances. So we just have presence of sensible heat given by this formula:

Q = m c_p \Delta T

And for this case we know that Both A and B are at the same initial temperature when equal amounts of energy are added to them, so then we have this:

Q_A = Q_B

And if we replace the formula for sensible heat we got:

m_A c_{pA} \Delta T_A = m_B c_{pB} \Delta T_B

And if we replace for the change of the temperature we got:

m_A c_{pA} (T_{fA} -T_{iA}) = m_B c_{pB} (T_{fB}- T_{iB})

And since T_{iA}= T_{iB}= T we have this:

m_A c_{pA} (T_{fA} -T) = m_B c_{pB} (T_{fB}- T)

For this case, if we try to find the final temperature of A and B, we see that we will obtain an expression in terms of specific heats and masses, from the information given we know the relationship between specific heats, but we don't know the relationship that exists among the masses, then the best option for this case is:

d) More information is needed

(The relation between the masses is not given)

4 0
3 years ago
A Chinook (King) salmon (Genus Oncorynchus) can jump out of water with a speed of 6.75 m / s . If the salmon is in a stream with
Pepsi [2]

Answer:

The maximum height that the fish can jump is 2.19 m.

Explanation:

Hi there!

Please, see the attached figure for a better understanding of the problem.

The motion of the salmon is a parabolic one because when it jumps, it already has a horizontal velocity (see figure).

The position and velocity vectors of the salmon at a time t, can be calculated as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

v = (v0x, v0y + g · t)

Where:

r = position of the salmon at time t.

x0 = initial horizotal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity.

g = acceleration of gravity.

Looking at the figure, notice that at the maximum height, the vertical velocity is zero (because the velocity vector is horizontal). Using the equation of the vertical component of the velocity, we can obtain the time at which the salmon is at its maximum height:

vy = v0y + g · t

To find the initial vertical velocity, v0y, let´s look at the figure. Notice that the initial velocity is the hypotenuse of the triangle formed with the horizontal velocity and the vertical velocity. Then:

v0² = v0x² + v0y²

Solving for v0y:

v0y = √(v0² - v0x²)

v0y = √((6.75 m/s)² - (1.65 m/s)²)

v0y = 6.55 m/s

Now, using the equation of the vertical component of the velocity at the maximum height (vy = 0):

vy = v0y + g · t

0 = 6.55 m/s + (-9.8 m/s²) · t

-6.55 m/s / -9.8 m/s² = t

t = 0.67 s

Now, using the equation of the vertical position at t = 0.67 s, we can find the maximum height:

y = y0 + v0y · t + 1/2 · g · t²

y = 0 m + 6.55 m/s · 0.67 s + 1/2 · (-9.8 m/s²) · (0.67 s)²

y = 2.19 m

The maximum height that the fish can jump is 2.19 m.

4 0
3 years ago
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