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s344n2d4d5 [400]
3 years ago
7

To practice Problem-Solving Strategy 8.1 for circular-motion problems. A cyclist competes in a one-lap race around a flat, circu

lar course of radius 140 m . Starting from rest and speeding up at a constant rate throughout the race, the cyclist covers the entire course in 60 s . The mass of the bicycle (including the rider) is 76 kg . What is the magnitude of the net force Fnet acting on the bicycle as it crosses the finish line?
Physics
1 answer:
Anna007 [38]3 years ago
5 0

Answer:

Explanation:

distance travelled

s = 2πR

= 2 X 3.14 X 140

= 880 m

final velocity = v

initial velocity = u

distance travelled = s

time = 60 s

s = ut + 1/2 at²

880 = .5 x a x 60²

a = .244 m/s²

final velocity v = at

= .244 x 60

= 14.66

centripetal acceleration at final moment

v² /R

14.66 X 14.66 / 140

= 1.53 m/s⁻²

1.53 m/s²

this is centripetal acceleration which acts towards the centre.

tangential acceleration calculated a _t = .244

redial acceleration ( centripetal ) = 1.53

Resultant acceleration

R²= 1.53² + .244 ²

R = 1.55 m/s²

total force = 1.55 x 76

= 118 N  

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A block rests on a frictionless horizontal surface and is attached to a spring. When set into simple harmonic motion, the block
MariettaO [177]

Answer:

v₀ = 0.5058 m/s

Explanation:

From the question, for the block to hit the bottle, the elastic potential energy of the spring at the bottle (x = 0.08 m) should be equal to the sum of the elastic potential energy of the spring at x = 0.05 m and the kinetic energy of block at x = 0.05 m

Now, the potential energy of the block at x = 0.08 m is ½kx²

where;

k is the spring constant given by; k = ω²m

ω is the angular velocity of the oscillation

m is the mass of the block.

Thus, potential energy of the spring at the bottle(x = 0.08 m) is;

U = ½ω²m(0.08m)²

Also, potential energy of the spring at the bottle(x = 0.05 m) is;

U = ½ω²m(0.05m)²

and the kinetic energy of the block at x = 0.05 m is;

K = ½mv₀²

Thus;

½ω²m(0.08)² = ½ω²m(0.05)² + ½mv₀²

Inspecting this, ½m will cancel out to give;

ω²(0.08)² = ω²(0.05)² + v₀²

Making v₀ the subject, we have;

v₀ = ω√((0.08)² - (0.05)²)

So,

v₀ = 8.1√((0.08)² - (0.05)²)

v₀ = 0.5058 m/s

7 0
3 years ago
The longer the time between the arrival of the P-wave and S-wave, the _______ is the epicenter.
Brrunno [24]

The longer the time between the arrival of the P-wave and S-wave, the <u>farther away</u> is the epicenter.

<h3>What is epicenter and the relation between P-wave and S-wave?</h3>
  • The point on the earth's surface vertically above the hypocenter (or focus), point in the crust where a seismic rupture begins is said to be epicenter.
  • There are two types of waves during earthquakes, they are:
  1. P - wave
  2. S - wave
  • Each seismograph records the times when the first (P waves) and second (S waves) seismic waves arrive.
  • From the graph, through the information, scientists can determine how fast the waves are traveling.
  • The longer the time between the arrival of the P-wave and S-wave, the farther away is the epicenter.

Hence, Option B is the correct answer.

Learn more about epicenter,

brainly.com/question/28136716

#SPJ1

7 0
2 years ago
A satellite at a particular point along an elliptical orbit has a gravitational potential energy of 4700 MJ with respect to Eart
erastovalidia [21]

during satellite motion we know that total energy is always conserved

so here we will have

KE_i + PE_i = KE_f + PE_f

here we know that

KE_i = 4800 MJ

PE_i = 4700 MJ

now at other position

PE_f = 6000 MJ

now from above equation we have

4800 +4700 = 6000 + KE

now we have

9500 = 6000 + KE

KE = 9500 - 6000 = 3500 MJ

so its kinetic energy will be 3500 MJ

6 0
3 years ago
A. Draw the electric field lines around a negative charge.
Alborosie
<h2>a. Answer:</h2>

We use Electric field lines for visualizing electric  fields, so this helps us to see the problem more real. So an electric field line is an imaginary  line or curve drawn through a region of space such that the tangent at any point comes from the direction of the electric-field vector at that point. The electric field lines around a negative charge is shown in the First figure below.

<h2>b. Answer:</h2>

Electric forces can be found by using the Coulomb Law's that states <em>that The magnitude of the electric force between two point charges is directly proportional  to the product of the charges and inversely proportional to the square  of the distance between them. </em>This can be expressed as follows:

F=k\frac{\left | q_{1}q_{2} \right |}{r^2} \\ \\ Where: \\ \\ k=9\times 10^9Nm^2/c^2 \\ \\ q_{1}=0.00150 C \ and \ q_{2}=0.00240 C \\ \\ r=0.900 m

Then:

F=9\times 10^9\frac{\left | 0.00150 \times 0.00240 \right |}{(0.900)^2} \\ \\ \therefore \boxed{F=40000N}

This force is repulsive because the two charges are positive and recall that two positive charges or two negative charges repel each other while a positive charge  and a negative charge attract each other.

<h2>C. Answer:</h2>

From the statement, we have two charged objects. Let's say that this charges are:

q_{1} \ and \ q_{2}

If the amount of charge on one of the objects is tripled, let's say this is the charge q_{2}, then the new charge is:

q_{N}=3q_{2}

In the formula of Coulomb:

F=k\frac{\left | q_{1}q_{N} \right |}{r^2} \\ \\ \therefore F=k\frac{\left | q_{1}(3q_{2}) \right |}{r^2} \\ \\ \therefore \boxed{F=3k\frac{\left | q_{1}q_{2} \right |}{r^2}}

<em>The conclusion is that if the amount of charge on one of the objects is tripled, the electric force between two charged objects is also tripled</em>

<h2>d. Answer:</h2>

Let's use the Coulomb's Law again to solve this problem. We want to know how the electric force between two charged objects changes if the charges are moved closer together:

F=k\frac{\left | q_{1}q_{N} \right |}{r^2}

<em>By saying that the charges are moved closer together, we want to express that r becomes smaller. Since r is in the denominator, this implies that the electric force between these two charged objects becomes greater.</em>

<h2>e. Answer:</h2>

From the figure, we can see a metal sphere on a stand. There we have both positive and negative charges. We can say that the positive charge of this sphere is +10q and the negative and the negative charge is -10q. Since the electric charge is conserved, then the algebraic sum of all the electric charges in any closed system is constant. In conclusion, <em>the sphere has no net charge.</em>

<h2>f. Answer:</h2>

Here we want to know how the negative charges in the same sphere are redistributed when a positively charged rod is brought near it. Therefore, positive charge on rod  repels positive charges on the sphere, creating  zones of negative and  positive charge as indicated in the second Figure.

7 0
3 years ago
Read 2 more answers
Ice melting is not a chemical change because​
prohojiy [21]

Melting ice is not a chemical reaction, but is a physical change. There is no chemical reaction involved during the melting process, and it is still the same molecular substance.

7 0
3 years ago
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