The diagram shows the top view of a 65-kg student at point A on an amusement park ride. The ride spins the student in a horizontal circle with a radius of 2.5 m, at a constant speed of 8.6 m/s. The floor is lowered and the student remains against the wall without falling to the floor.
The magnitude of the centripetal force acting on the student at point A is approximately
Question 1 options:

A.) 1923 N
B.) 47 N
C.) 1398 N
D.) 223 N

The net force on the cart is 100 N in the direction of its motion, so by Newton's second law we can find the acceleration a applied to it:

100 N = (50 kg) a

a = (100 N) / (50 kg)

a = 2 m/s²

The cart starts at rest and travels a distance of 10 m, so that its final velocity v satisfies

v ² - 0² = 2 (2 m/s²) (10 m)

v ²= 40 m²/s²

and so the cart ends up with kinetic energy

KE = 1/2 m v ² = 1/2 (50 kg) (40 m²/s²) = 1000 J

Answer:

The velocity of the blood in the thinner arteries is 0.1 times that of the thicker artery.

Explanation:

To find the velocity of the blood we need to use the continuity equation:

[tex] n_{1}A_{1}v_{1} = n_{2}A_{2}v_{2} [/tex]   (1)  

Where:

n: is the number of branches

A: is the cross-sectional area

v: is the velocity

For artery 1, we have:

n₁ = 1, A₁ = 1 cm², v₁ = v

For the 20 arteries (2), we have:

n₂ = 20, A₂ = 0.5 cm², v₂ =?

By using equation (1):

[tex] n_{1}A_{1}v_{1} = n_{2}A_{2}v_{2} [/tex]

[tex] 1 cm^{2}*v = 20*0.5 cm^{2}*v_{2} [/tex]

[tex] v_{2} = \frac{1 cm^{2}*v}{20*0.5 cm^{2}} = \frac{v}{10} = 0.1v [/tex]

Therefore, the velocity of the blood in the thinner arteries is 0.1 times that of the thicker artery.

I hope it helps you!      

Answer:

The answer is "[tex]5.06 \times 10^{-6} \ kg \ m^2[/tex]"

Explanation:

[tex]\to E_1=0..............(i)\\\\\to E_2= \frac{mV^2}{2} +\frac{Iw^2}{2} - mgh.............(ii)\\\\ \Delta E=0\\\\\to mgh= \frac{mV^2}{2} +\frac{Iw^2}{2} \\\\ \to 2 \ mgh= mV^2 +Iw^2\\\\ \to 2 \ mgh- mV^2 =Iw^2\\\\ \to m(2gh- V^2) =Iw^2\\\\ \to I= \frac{m(2gh- V^2)}{w^2}[/tex]

       [tex]= 5.06 \times 10^{-6} \ kg \ m^2[/tex]

Answer:

Im going to take an educated guess and say its B.

Explanation:

A Full moon is located farthest from the Sun. Thus, the correct option for this question is A.

A full moon may be defined as one of the types of lunar phases when the Moon significantly appears fully illuminated from Earth's perspective. This event typically occurred when Earth is located between the Sun and the Moon.

When the earth is located between the sum and the moon, it is the time when the moon is located farthest from the sun. And it directs to an event of a full moon. The average time duration of the full moon and the repetition of the same phase normally requires 29.53 days.

Therefore, the Full moon is located farthest from the Sun. Thus, the correct option for this question is A.

To learn more about the Full moon, refer to the link:

https://brainly.com/question/15032553

#SPJ2

Answer:

a)    n = 9.9       b)      E₁₀ = 19.25 eV

Explanation:

Solving the Scrodinger equation for the electronegative box we get

         Eₙ = (h² / 8m L²2) n²

where l is the distance L = 1.40 nm = 1.40 10⁻⁹ m and n the quantum number

 In this case En = 19 eV let us reduce to the SI system

          En = 19 eV (1.6 10⁻¹⁹ J / 1 eV) = 30.4 10⁻¹⁹ J

          n = √ (In 8 m L² / h²)

let's calculate

          n = √ (8  9.1 10⁻³¹ (1.4 10⁻⁹)²  30.4 10⁻¹⁹ / (6.63 10⁻³⁴)²

          n = √ (98)            n = 9.9

since n must be an integer, we approximate them to 10

b) We substitute for the calculation of energy

        In = (h² / 8mL2² n²

        In = (6.63 10⁻³⁴) 2 / (8 9.1 10⁻³¹  (1.4 10⁻⁹)² 10²

        E₁₀ = 3.08 10⁻¹⁸ J

we reduce eV

      E₁₀ = 3.08 10⁻¹⁸ j (1ev / 1.6 10⁻¹⁹J)

      E₁₀ = 1.925 101 eV

      E₁₀ = 19.25 eV

the result with significant figures is

        E₁₀ = 19.25 eV

Answer:

matter

Explanation:

Hope I helped :)

Answer:

48m

Explanation:

Given the following data;

Initial velocity = 0m/s

Final velocity = 6m/s

Time, t = 6 secs

Time, T2 = 5 secs

Mathematically, acceleration is given by the equation;

[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]

Substituting into the equation;

[tex]a = \frac{6 - 0}{6}[/tex]

[tex]a = \frac{6}{6}[/tex]

Acceleration, a = 1m/s²

To find the distance covered in the first phase;

Solving for distance, we would use the second equation of motion;

[tex] S = ut + \frac {1}{2}at^{2}[/tex]

Substituting the values into the equation;

[tex] S = 0(6) + \frac {1}{2}*1*(6)^{2}[/tex]

[tex] S = 0 + \frac {1}{2}*1*36[/tex]

[tex] S = 0.5 *36[/tex]

Distance, S1 = 18m

For the second phase, time T2 = 5 secs;

Mathematically, speed is given by the equation;

[tex]Speed = \frac{distance}{time}[/tex]

Making distance the subject of formula, we have;

[tex]Distance, S = speed * time[/tex]

Substituting into the above equation;

[tex]Distance, S = 6 * 5[/tex]

Distance, S2 = 30m

Total distance = S1 + S2 = 18m + 30m = 48m

Total distance = 48m

Therefore, the total distance traveled by the biker is 48m.

Answer:

The magnitude of the impulse delivered to the baseball is 7.0 Ns

Explanation:

Given;

mass of the foul ball, m = 0.14 kg

initial velocity, u = 40 m/s

final velocity, v = 30 m/s in perpendicular direction

Impulse is given as change in momentum;

initial momentum in horizontal direction, Pi = mu

Pi = 0.14 x 40 = 5.6 Ns

final momentum in perpendicular direction, Pf = mv

Pf = 0.14 x 30

Pf = 4.2 Ns

The resultant impulse is given by;

J² = 5.6² + 4.2²

J² = 49

J = √49

J = 7.0 Ns

Therefore, the magnitude of the impulse delivered to the baseball is 7.0 Ns

Answer:

[tex] R = 2.1R_{\bigodot} [/tex]

Explanation:                          

The radius (R) is related to temperature (T) and the luminosity (L) as follows:    

[tex] L = T^{4} \times R^{2} [/tex]      

By solving the above equation for R we have:

[tex] R = \frac{\sqrt{L}}{T^{2}} [/tex]

With [tex]L = 70L_{\bigodot}[/tex] and [tex]T = 2T_{\bigodot}[/tex] we have:

[tex] R = \frac{\sqrt{70L_{\bigodot}}}{(2T_{\bigodot})^{2}} [/tex]

[tex] R = \frac{\sqrt{70}}{(2)^{2}} [/tex]

[tex] R = 2.1R_{\bigodot} [/tex]

Therefore, the radius is 2.1 times that of the Sun.

I hope it helps you!

Answer:

Pulleys, levers, inclined planes, wedges, screws, and wheel and axle.

Explanation:

Answer:

Explanation:

according to Newtons second law;

F = ma

F = m(v-u)/t

m is the mass = 60+5 = 65kg

v is the final velocity = 0m/s

initial velocity = 8m/s

time t = 4.0s

F = 65(8)/4

F = 65*2

F = 130N

Hence he applied a force of 130N

Answer:

2.) Use a wrench with a longer handle.

Explanation:

It is suggested to use a wrench with longer handle if you cannot exert enough force to loosen a bolt with a wrench.

In doing this work, the torque which is the force that can turn the screw is not enough;

      Torque  = force x distance

If we increase the distance or the length of the handle, we can generate more torque to overcome the force needed.

Answer:

B. soft tissue injuries

Explanation:

i took the test on edge and its really the only one that would make sense if you payed attention to the lesson  :)

Answer:

F' = 100 F/r²

Explanation:

The gravitational force of attraction between two objects is given by the Newton's Gravitational Formula. The Newton's Gravitational Formula is as follows:

F = Gm₁m₂/r²

where,

F = Force between objects

G = Universal Gravitational Constant

m₁ = mass of first object

m₂ = mass of second object

r = distance between objects = 10 m

Therefore,

F = Gm₁m₂/10²

Gm₁m₂ = 100F   --------------------- equation (1)

Now, we consider these objects at any distance r apart. So, the force becomes:

F' = Gm₁m₂/r²

using equation (1), we get:

F' = 100 F/r²

So, if the force (F) between objects 10 m apart is known, we can find it at any distance from the above formula.

The force between objects that are any distance apart is expressed as [tex]P'=\frac{100P}{r^2}[/tex]

According to the gravitational law, the force acting on an object is directly proportional to the product of their masses and inversely proportional to the square of their distance apart. Mathematically,

[tex]P=\frac{GMm}{r^2}[/tex]

M and m are the masses

r is the distance between the masses

If the force between objects that are 10 meters apart, hence;

[tex]P=\frac{GMm}{10^2}\\P=\frac{GMm}{100}\\GMm = 100P[/tex]

To find the force between objects that are any distance apart, we will use the same formula above to have;

[tex]P'=\frac{GMm}{r^2}\\[/tex]

Substitute the result above into the expression to have:

[tex]P'=\frac{100P}{r^2}[/tex]

Hence the force between objects that are any distance apart is expressed as [tex]P'=\frac{100P}{r^2}[/tex]

Learn more on gravitational law here: https://brainly.com/question/11760568

The correct answer is A continúe moving with constant velocity

Answer:

v₃ = 1.625 [m/s]

Explanation:

To solve this problem we must use the definition of linear momentum conservation, which tells us that momentum is conserved before and after a collision.

Since the collision is inelastic, the two bodies are joined after the collision.

P = m*v [kg*m/s]

m = mass [kg]

v = velocity [m/s]

where:

P = lineal momentum [kg*m/s]

Now, it is important to clarify that in the following equation we will take the left side of the equation as the momentum before the collision and the right side of the equal sign as the momentum after the collision.

Pbefore = Pafter

(m₁*v₁) + (m₂*v₂) = (m₁+m₂)*v₃

where:

m₁ = mass one = 5 [kg]

v₁ = velocity of the mass one = 2 [m/s]

m₂ = mass two = 3 [kg]

v₂ = velocity of the mass two = 1 [m/s]

v₃ = velocity of the combined masses after the collision [m/s]

Now replacing we have:

(5*2) + (3*1) = (5 + 3)*v₃

10 + 3 = 8*v₃

v₃ = 13/8

v₃ = 1.625 [m/s]

Answer:

If the car travels 150 kilometers to the Southeast, and then 50 kilometers to the North, it is understood that the car does not travel the same way out as it does back. Therefore, the car's displacement will be the sum of both distances traveled, that is, 150 + 50, which is equal to 200. Therefore, the car traveled about 200 kilometers from its starting point.

Answer:

z_c = ⅜R

Explanation:

If we assume that the hemisphere has uniform density, we can express the centre of mass as;

z_c = (ρ/M)∫∫∫ z•dV

We know that density(ρ) = mass(M)/volume(V)

Thus, Vρ = M

And volume of hemisphere = 2πr³/3

Thus;

2Vρπr³/3 = M

So;

z_c = (ρ/(2Vρπr³/3))∫∫∫ z•dV

Where r = R in this case.

ρ will cancel out to give;

z_c = (3/(2πr³))∫∫∫_V (z•dV)

In spherical coordinates,

r is radius

Φ = angle between the point and the z − axis

θ = azimuthal angle

Therefore, the integral becomes what it is in the attached image.

I've completed the explanation as well in the attachment.

Answer:

An antimony can form 3 bonds

Let me know if this helps

Thanks!

:)

Explanation:

Answer:

i think the answer is solar,

Explanation:

because it is a natural resource im sorry if im wrong

Answer:

32.67km

Explanation:

~= round off

11.2km/h~3.11m/s

175minutes=10500s

10500*3.11~32, 666.67m

=32.66667km

~ 32.67km

Answer:

Solution A has a pH of 6 and solution B has a pH of 8. Which of the following is true regarding the concentration of hydrogen ions in each solution? A) A has 100 times greater H+ concentration than B. B) B has 100 times greater H+ concentration than A. C) A has 7/9 of the H+ concentration of B. D) A has 9/7 of the H+ concentration of B. E) none of these

Explanation:

Hey im super sorry if i get this wrong :)

Answer:

[tex]I=182.97\ kg-m^2[/tex]

Explanation:

Given that,

Radius of a spherical shell, r = 0.7 m

Torque acting on the shell, [tex]\tau=860\ N[/tex]

Angular acceleration of the shell, [tex]\alpha =4.7\ m/s^2[/tex]

We need to find the rotational inertia of the shell about the axis of rotation. The relation between the torque and the angular acceleration is given by :

[tex]\tau=I\alpha[/tex]

I is the rotational inertia of the shell

[tex]I=\dfrac{\tau}{\alpha }\\\\I=\dfrac{860}{4.7}\\\\I=182.97\ kg-m^2[/tex]

So, the rotational inertia of the shell is [tex]182.97\ kg-m^2[/tex].