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Svetllana [295]

3 years ago

12

How can a cyclist minimize friction as he or she rides? A. by increasing the weight of the bike B. by increasing the tread on hi

s or her tires C. by decreasing the air resistance D. by decreasing aerodynamic design

2 answers:

erastova [34]3 years ago

7 0

The answer is B becausew the others would add friction

jek_recluse [69]3 years ago

6 0

B I believe is the answer!

Hope this helps and have a great day!!!

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A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find th

Igoryamba

A particle moves along a straight line with equation of motion s = f(t), where s is measured in meters and t in seconds. Find the velocity and the speed when t = 4. f(t) = 12t² + 35 t + 1

Answer:

Velocity = 131 m/s

Speed = 131 m/s

Explanation:

Equation of motion, s = f(t) = 12t² + 35 t + 1

To get velocity of the particle, let us find the first derivative of s

v (t) = ds/dt = 24t + 35

At t = 4

v(4) = 24(4) + 35

v(4) = 131 m/s

Speed is the magnitude of velocity. Since the velocity is already positive, speed is also 131 m/s

5 0

3 years ago

at location a, what are the directions of the electric fields contributed by the electron. calculate the magnitudes of the elect

Lisa [10]

We can use the equation E = k | Q | r 2 E = k | Q | r2 to find the magnitude of the electric field. The direction of the electric field is determined by the sign of the charge,

<h3>What is electric and magnetic field ?</h3>

With the use of electricity and other types of artificial and natural illumination, invisible energy fields known as electric and magnetic fields (EMFs) and radiation are created.

While the magnetic field is discernible by the force it exerts on other magnetic particles and moving electric charges, the electric field is actually the force per unit charge experienced by a non-moving point charge at any given location inside the field.

Learn more about Electromagnetic field here:

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4 0

1 year ago

Is this right? Please help me

morpeh [17]

Answer:

yes

Explanation:

4 0

3 years ago

Read 2 more answers

Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

$V = \frac{4}{3} \pi R^3$

Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0

3 years ago

Inserting the formulas you found for Xman(t) and Xbus(t) into the conditionXman(tcatch)=Xbus(tcatch) , you obtain the following-

lara [203]

Answer:

c > √(2ab)

Explanation:

In this exercise we are asked to find the condition for c in such a way that the results have been real

The given equation is

½ a t² - c t + b = 0

we can see that this is a quadratic equation whose solution is

t = [c ±√(c² - 4 (½ a) b)] / 2

for the results to be real, the square root must be real, so the radicand must be greater than zero

c² -2a b > 0

c > √(2ab)

3 0

3 years ago