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2 years ago

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Frank is leaving Gainesville at 7:30 am and needs to be in Tampa by noon. He is cycling at an average speed of 18.3 mph. At what

time should he arrive in Tampa?

1 answer:

garri49 [273]2 years ago

4 0

If Frank leaves Gainesville at 7:30 am, the time he should arrive in Tampa is 12.00 pm after spending 4 hours 30 minutes on the road.

<h3>What is average speed? </h3>

The average speed of an object is the ratio of total distance to total time of motion.

V = total distance/total time

For Frank to be in Tampa by noon, he must spend atleast 4 hours 30 minutes on the road.

18.3 mph = d/4.5 h

d = 82.35 miles

The distance between Gainesville and Tampa is 82.35 miles.

Thus, we can conclude that, if Frank leaves Gainesville at 7:30 am, the time he should arrive in Tampa is 12.00 pm after spending 4 hours 30 minutes on the road.

Learn more about average speed here: brainly.com/question/6504879

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A racecar accelerates from rest at 6.5 m/s2 for 4.1 s. How fast will it be going at the end of that time?

Pie

Answer:

The final velocity of the car is 26.65 m/s.

Explanation:

Given;

acceleration of the racecar, a = 6.5 m/s²

initial velocity of the car, u = 0

time of motion, t = 4.1 s

The final velocity of the car is given by;

v = u + at

where;

v is the final velocity of the car

suvstitute the givens

v = 0 + (6.5)(4.1)

v = 26.65 m/s.

Therefore, the final velocity of the car is 26.65 m/s.

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3 years ago

Please tell me if this is Newton's first law, second, or third law of motion. There can be more than one of the same answer very

AfilCa [17]

Answer: Hope This Helps!

Explanation:

1: Newton’s first law of motion can explain how a magician pulls a tablecloth from underneath the dishes. A negligible horizontal force is applied during the process. As per Newton’s first law of motion, the dishes and glasses remain in their state of motion (rest); as a result, they remain undisturbed.

2: Newton's First Law of Motion is defined as "An object at rest and an object in motion will stay in motion with the same speed and in the same direction unless acted upon by an unbalanced force."In soccer, when the soccer ball is in the soccer field and it is not moving, that means that it is at rest and there is no force acting upon it. When there is a person that is ready to play soccer and wants to kick the ball and play, then the unbalanced force would be the power from the person's foot.

3: Newtons third law can explain, as the cannonball is pushed forwards by the expanding high-pressure gases created by the exploding gunpowder, it pushes back on these gases. The gases push back on the cannon itself, causing it to roll backwards. Alternative answer: the cannon pushes forward on the cannonball. the reaction force is the cannonball pushing backwards on the cannon.

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3 years ago

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

Suppose you observe two stars and you know they have the same luminosity. If one star is twice as far away as the other, the mor

rosijanka [135]

Answer:

The farther star will appear 4 times fainter than the star that is near to the observer.

Explanation:

Since it is given that the luminosity of the 2 stars is same thus they radiate the same energy per unit time

Consider a spherical wave front of energy 'E' that leaves both the stars (Both radiate 'E' as they have same luminosity)

This Energy is spread over the whole surface area of sphere Thus when the wave front is at a distance 'r' the energy per unit surface area is given by

$e_{1}=\frac{E}{4\pi r^{2}}$

For the star that is twice away from the earth the distance is '2r' thus we will receive an energy given by

$e_{2}=\frac{E}{4\pi (2r)^{2}}=\frac{E}{8\pi r^{2}}=\frac{e_{1}}{4}$

Hence we sense it as 4 times fainter than the nearer star.

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3 years ago

You have a set of calipers that can measure thicknesses of a few inches with an uncertainty of ± 0.005 inches. You measure the t

Bond [772]

Answer:

a) x = (0.0114 ± 0.0001) in , b) the number of decks is 5

Explanation:

a) The thickness of the deck of cards (d) is measured and the thickness of a card (x) is calculated

x = d / 52

x = 0.590 / 52

x = 0.011346 in

Let's look for uncertainty

Δx = dx /dd Δd

Δx = 1/52 Δd

Δx = 1/52 0.005

Δx = 0.0001 in

The result of the calculation is

x = (0.0114 ± 0.0001) in

b) You want to reduce the error to Δx = 0.00002, the number of cards to be measured is

#_cards = n 52

The formula for thickness is

x = d / n 52

Uncertainty

Δx = 1 / n 52 Δd

n = 1/52 Δd / Δx

n = 1/52 0.005 / 0.00002

n = 4.8

Since the number of decks must be an integer the number of decks is 5

3 0

3 years ago