Ask question

Ask question

All categories

2 years ago

8

The waves in the pool where you are floating have a crest height of about 1 foot. Bobby does a cannonball dive off the side of t

he pool and sends waves 4
feet high toward your raft. The waves that Bobby produced have a greater
O amplitude.
compression.
frequency
wavelength

1 answer:

sukhopar [10]2 years ago

6 0

Answer:

I TNINK ITS C

Explanation:

You might be interested in

Describe an object's velocity when an acceleration-time graph is zero?

svp [43]

Anything times zero is zero

7 0

3 years ago

Read 2 more answers

Bobby places a 4.75 cm tall light bulb a distance of 33.2 cm from a concave mirror. If the mirror has a focal length of 28.2, th

Fittoniya [83]

The image distance can be determined using the mirror equation: 1/f = 1/d_o + 1/d_i, where, f is the focal length, d_o is the object distance, and d_i is the image distance. Given that f = 28.2 and d_o = 33.2 cm, the value of d_i is calculated to be 187.248 cm. On the other hand, the image height is obtained using the magnification equation wherein, h_i/h_o = -d_i/d_o, where h_i is the image height and h_o is the object height. Using the given values, h_i is equal to -26.79 cm. Note that the negative sign indicates that the image is inverted.

3 0

3 years ago

Read 2 more answers

A nonuniform, but spherically symmetric, distribution of charge has a charge density ρ(r) given as follows: ρ(r)=ρ0(1−r/r) for r

Nadusha1986 [10]

A)<span>
dQ = ρ(r) * A * dr = ρ0(1 - r/R) (4πr²)dr = 4π * ρ0(r² - r³/R) dr
which when integrated from 0 to r is
total charge = 4π * ρ0 (r³/3 + r^4/(4R))
and when r = R our total charge is
total charge = 4π*ρ0(R³/3 + R³/4) = 4π*ρ0*R³/12 = π*ρ0*R³ / 3
and after substituting ρ0 = 3Q / πR³ we have
total charge = Q ◄

B) E = kQ/d²
since the distribution is symmetric spherically

C) dE = k*dq/r² = k*4π*ρ0(r² - r³/R)dr / r² = k*4π*ρ0(1 - r/R)dr
so
E(r) = k*4π*ρ0*(r - r²/(2R)) from zero to r is
and after substituting for ρ0 is
E(r) = k*4π*3Q(r - r²/(2R)) / πR³ = 12kQ(r/R³ - r²/(2R^4))
which could be expressed other ways.

D) dE/dr = 0 = 12kQ(1/R³ - r/R^4) means that
r = R for a min/max (and we know it's a max since r = 0 is a min).

<span>E) E = 12kQ(R/R³ - R²/(2R^4)) = 12kQ / 2R² = 6kQ / R² </span></span>

4 0

3 years ago

The law of conservation of momentum states that the total momentum of interacting objects does not change . This means the total

pickupchik [31]

Answer:

The momentum of an object is equal to the product of its mass and its velocity.

Explanation:

Consider an object of mass $m$ travelling at a velocity $\vec{v}$ . The momentum $\vec{p}$ of this object would be:

$\vec{p} = m \cdot \vec{v}$ .

For the law of conservation of momentum, consider two objects: object $\rm a$ and object $\rm b$ . Assume that these two objects collided with each other.

Let $m_{\rm a}$ and $m_{\rm b}$ denote the mass of the two objects.
Let $\vec{v}_{\rm a}(\text{initial})$ and $\vec{v}_{\rm b}(\text{initial})$ denote the velocity of the two object right before the interaction.
Let $\vec{v}_{\rm a}(\text{final})$ and $\vec{v}_{\rm b}(\text{final})$ denote the velocity of the two objects right after the interaction.

The momentum of the two objects right before the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ , respectively.
The momentum of the two objects right after the collision would be $m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final})$ and $m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ , respectively.

The sum of the momentum of the two objects would be:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial})$ right before the collision, and
$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ right after the collision.

Assume that the system of these two objects is isolated. By the law of conservation of momentum, the sum of the momentum of these two objects should be the same before and after the collision. That is:

$m_{\rm a}\cdot \vec{v}_{\rm a}(\text{initial}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{initial}) = m_{\rm a}\cdot \vec{v}_{\rm a}(\text{final}) + m_{\rm b}\cdot \vec{v}_{\rm b}(\text{final})$ .

4 0

3 years ago

If you were to apply a force of 15N to a medicine ball with a mass of 10kg, what would its acceleration be

tresset_1 [31]

One of the equations for force is: Force = mass × acceleration.

The force and mass are given and so:

F=15N

m=10kg

By plugging in these values we obtain:

$F=ma\\\\15=10(a)\\$

$\frac{15}{10}=\frac{10(a)}{10}\\\\a=\frac{15}{10}=1.5m/s^2$

So the acceleration is a=1.5m/s^2

6 0

3 years ago