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3 years ago

Kieran is graduating from high school. He wants to be a urologist. What career path would be best for Kieran to follow?

Physics

2 answers:

faust18 [17]3 years ago

8 0

Answer:

four-year degree Right arrow. graduate school Right arrow. license from the NCLEX

valentinak56 [21]3 years ago

7 0

Answer:

four-year degree Right arrow. graduate school Right arrow. residency

Explanation:

edge

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Make a conclusion on the relationship of mass to the rate of the change of motion of the balloon at a constant force.

sasho [114]

At a constant force, the mass of the balloon is inversely proportional to the rate of change motion of the balloon.

The force applied to an object can be determined by applying Newton's second law of motion, the force applied to an object is directly proportional to the product of mass and acceleration of the object.

F = ma

$F = m\frac{\Delta v}{t} \\\\\frac{F}{\frac{\Delta v}{t} } = m$

The mass of an object is inversely proportional to the rate of change motion of the object.

Thus, we can conclude that at constant force, the mass of the balloon is inversely proportional to the rate of change motion of the balloon.

Learn more here:brainly.com/question/19887955

6 0

3 years ago

A car travels due east with a speed of 38.0 km/h. Raindrops are falling at a constant speed vertically with respect to the Earth

DiKsa [7]

Answer: 116.926 km/h

Explanation:

To solve this we need to analise the relation between the car and the Raindrops. The cars moves on the horizontal plane with a constant velocity.

Car's Velocity (Vc) = 38 km/h

The rain is falling perpedincular to the horizontal on the Y-axis. We dont know the velocity.

However, the rain's traces on the side windows makes an angle of 72.0° degrees. ∅ = 72°

There is a relation between this angle and the two velocities. If the car was on rest, we will see that the angle is equal to 90° because the rain is falling perpendicular. In the other end, a static object next to a moving car shows a horizontal trace, so we can use a trigonometric relation on this case.

The following equation can be use to relate the angle and the two vectors.

Tangent (∅) = Opposite (o) / adjacent (a)

Where the Opposite will be the Rain's Vector that define its velocity and the adjacent will be the Car's Velocity Vector.

Tan(72°) = Rain's Velocity / Car's Velocity

We can searching for the Rain's Velocity

Tan(72°) * Vc = Rain's Velocity

Rain's Velocity = 116.926 km/h

3 0

3 years ago

HELP ASAP!!!!!!! PLEASE WILL GIVE BRAINLIEST

Pani-rosa [81]

Answer:

The three major parts of a cell are:

a. Cell membrane

b. Nucleus

c. Cytoplasm

Explanation:

a. Cell membrane:

It provides protection for a cell.

b. Nucleus:

It controls and regulates the activities of cell.

c. Cytoplasm:

It supports and suspends organelles and cellular molecules.

8 0

3 years ago

Read 2 more answers

Two forces,

serg [7]

First compute the resultant force F:

$\mathbf F_1=(5.90\,\mathbf i-5.60\,\mathbf j)\,\mathrm N$

$\mathbf F_2=(4.65\,\mathbf i-5.55\,\mathbf j)\,\mathrm N$

$\implies\mathbf F=\mathbf F_1+\mathbf F_2=(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N$

Then use Newton's second law to determine the acceleration vector $\mathbf a$ for the particle:

$\mathbf F=m\mathbf a$

$(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N=(2.10\,\mathrm{kg})\mathbf a$

$\mathbf a\approx(5.02\,\mathbf i-5.31\,\mathbf j)\dfrac{\rm m}{\mathrm s^2}$

Let $\mathbf x(t)$ and $\mathbf v(t)$ denote the particle's position and velocity vectors, respectively.

(a) Use the fundamental theorem of calculus. The particle starts at rest, so $\mathbf v(0)=0$ . Then the particle's velocity vector at t = 10.4 s is

$\mathbf v(10.4\,\mathrm s)=\mathbf v(0)+\displaystyle\int_0^{10}\mathbf a(u)\,\mathrm du$

$\mathbf v(10.4\,\mathrm s)=\left((5.02\,\mathbf i-5.31\,\mathbf j)u\,\dfrac{\rm m}{\mathrm s^2}\right)\bigg|_{u=0}^{u=10.4}$

$\mathbf v(10.4\,\mathrm s)\approx(52.2\,\mathbf i-55.2\,\mathbf j)\dfrac{\rm m}{\rm s}$

If you don't know calculus, then just use the formula,

$v_f=v_i+at$

So, for instance, the velocity vector at t = 10.4 s has x-component

$v_{f,x}=0+\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)(10.4\,\mathrm s)=52.2\dfrac{\rm m}{\mathrm s^2}$

(b) Compute the angle $\theta$ for $\mathbf v(10.4\,\mathrm s)$ :

$\tan\theta=\dfrac{-55.2}{52.2}\implies\theta\approx-46.6^\circ$

so that the particle is moving at an angle of about 313º counterclockwise from the positive x axis.

(c) We can find the velocity at any time t by generalizing the integral in part (a):

$\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du$

$\implies\mathbf v(t)=\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j$

Then using the fundamental theorem of calculus again, we have

$\mathbf x(10.4\,\mathrm s)=\mathbf x(0)+\displaystyle\int_0^{10.4}\mathbf v(u)\,\mathrm du$

where $\mathbf x(0)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m$ is the particle's initial position. So we get

$\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\displaystyle\int_0^{10.4}\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\right)\,\mathrm du$

$\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\dfrac12\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=10.4}$

$\mathbf x(10.4\,\mathrm s)\approx(542\,\mathbf i-570\,\mathbf j)\,\mathrm m$

So over the first 10.4 s, the particle is displaced by the vector

$\mathbf x(10.4\,\mathrm s)-\mathbf x(0)\approx(270\,\mathbf i-283\,\mathbf j)\,\mathrm m-(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m\approx(272\,\mathbf i-287\,\mathbf j)\,\mathrm m$

or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).

(d) See part (c).

3 0

3 years ago

The wavelength of light mainly affects our perception of

s2008m [1.1K]

Answer:

I would think the answer is color, if the wavelength is within the visible light spectrum. This could be answered in different ways but I'm pretty sure the answer you are looking for is hue/color.

6 0

3 years ago