If an instalment plan quotes a monthly interest rate of 4%, the effective annual/yearly interest rate would be _____________. 4% Between 4% and 48% 48% More than 48%

Complete Question

For each of the following scenarios, describe the force providing the centripetal force for the motion:

a. a car making a turn

b. a child swinging around a pole

c. a person sitting on a bench facing the center of a carousel

d. a rock swinging on a string

e. the Earth orbiting the Sun.

Answer:

Considering a

    The force providing the centripetal force is the frictional force on the tires \

          i.e  [tex]\mu mg = \frac{mv^2}{r}[/tex]

    where [tex]\mu[/tex] is the coefficient of static friction

Considering b

   The force providing the centripetal force is the force experienced by the boys  hand on the pole

Considering c

     The force providing the centripetal force is the normal from the bench due to the boys weight

Considering d

     The force providing the centripetal force is the tension on the string

Considering e

      The force providing the centripetal force is the force of gravity between the earth and the sun

Explanation:

Answer:

St Lawrence Sea way

Explanation:

The great lake connects the middle of North America which is at the Canada-United states border connecting to the Atlantic Ocean through the St Lawrence River.

Answer:

Explanation:

Using the law of conservation of momentum which states that 'the sum of momentum of bodies before collision is equal to their sum after collision. The bodies will move together with a common velocity after collision.

Momentum = Mass * Velocity

Before collision;

Momentum of receiver m1u1= 0 kgm/s (since the receiver is standing still)

Momentum of the tackler

m2u2 = 2.60*122 = 317.2 kgm/s

where m2 and u2 are the mass and velocity of the tacker respectively.

Sum of momentum before collision = 0+317.2 = 317.2 kgm/s

After collision

Momentum of the bodies = (m1+m2)v

v = their common velocity

m1 = mass of the receiver

Momentum of the bodies = (122+m1)(1.30)

Momentum of the bodies = 158.6+1.30m1

According to the law above;

317.2 = 158.6+1.30m1

317.2-158.6 = 1.30m1

158.6 = 1.30m1

m1 = 158.6/1.30

m1 = 122kg

The mas of the receiver is 122kg

Answer:

3×10^7 m/s or 0.10c (e)

Explanation: If the actual value of the speed of light were to be put into consideration.

Given that the speed of light is c = 3.0×10^8m/s

The alien spaceship is approaching at the rate of 10% of the speed of light.

10% of 3.0×10^8m/s

10/100 × 3.0×10^8m/s

0.1 ×3.0×10^8m/s

3×10^7 m/s. Which is the same thing as 0.1 of c = 0.1×c

Answer: 1.00c

Explanation: I got it correct on the homework

Answer:

r = 20 m

Explanation:

The formula for the angular momentum of a rotating body is given as:

L = mvr

where,

L = Angular Momentum = 10000 kgm²/s

m = mass

v = speed = 2 m/s

r = radius of merry-go-round

Therefore,

10000 kg.m²/s = mr(2 m/s)

m r = (10000 kg.m²/s)/(2 m/s)

m r = 5000 kg.m   ------------- equation 1

Now, the moment of inertia of a solid uniform disc about its axis through its center is given as:

I = (1/2) m r²

where,

I = moment of inertia = 50000 kg.m²

Therefore,

50000 kg.m² = (1/2)(m r)(r)

using equation 1, we get:

50000 kg.m² = (1/2)(5000 kg.m)(r)

(50000 kg.m²)/(2500 kg.m) = r

r = 20 m

Answer:

The speed of the freight car decreases.

Explanation:

According to the law of conservation of momentum indicates that for colliding in an isolated system, the total momentum pre and post collision is same for the two objects this is done because the momentum that one item has lost is same for the momentum that the other received

In the given situation, the freight car travels at constant speed along a frictionless railroad line. The top floor freight car is open. Then a huge load of coal is dumped inside the car.

Therefore the speed of the freight car decreased by applying the law of conservation of momentum i

Answer:

[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]

Explanation:

You have three forces F1, F2 an F3 that produce the following  acceleration:

a = −(8.00m/s^2)ˆi + (6.00m/s^2)ˆj

you know that force F1 and F2 are:

F1 = (30.0N)ˆi + (16.0N)ˆj

F2 = −(12.0N)ˆi + (8.00N)ˆj

and the force F3 is unknown:

F3 = F3x ˆi + F3y ˆj

The second Newton law is given by the following equation:

[tex]\vec{F}=m\vec{a}[/tex]

F: the total force = F1 +F2 + F3

m: mass of the object = 2 kg

By the properties of vectors you have:

[tex]\vec{F_1}+\vec{F_2}+\vec{F_3}=m\vec{a}\\\\(30.0-12.0+F_{3x})N\hat{i}+(16.0+8.00+F_{3y})N\hat{j}=(2.0kg)[(-8.00m/s^2)\hat{i}+(6.00m/s^2)\hat{j}]\\\\(18.0+F_{3x})N\hat{i}+(24.0+F_{3y})\hat{j}=(-16.00N)\hat{i}+(12.0N)\hat{j}[/tex]

Both x and y component must be equal in the previous equality, then you have:

[tex]18.0N+F_{3x}=-16.00N\\\\F_{3x}=-34.00N\\\\24.0N+F_{3y}=12.0N\\\\F_{3y}=-12.00N[/tex]

Hence, the vector F3 is:

[tex]\vec{F_3}=(-34.0N)\hat{i}+(-12.0N)\hat{j}[/tex]

Answer:

It will lose electrical potential energy.

Explanation:

A photon held at rest in a uniform electrical field will lose electrical potential energy when it is released this is because the electrical potential energy is the energy posses by the photon at rest or by virtue of the position is converted to kinetic energy which is energy posses by a body in motion.

Since the photon is released and set in motion , it now has kinetic energy and has lost the potential energy because it is set in motion.

Answer:

In beta-minus decay, a neutron breaks down to a proton and an electron, and the electron is emitted from the nucleus. In beta-plus decay, a proton breaks down to a neutron and a positron, and the positron is emitted from the nucleus.

Explanation:

Hope this helps!

Answer:

Exothermic

Explanation:

Depending on the unit you are in, the answer may vary.

This is an exothermic reaction because it produces heat (the beaker gets warm).

Answer:

The entropy change of the Universe that occurs is 19.346 J/K

Explanation:

Given;

temperature of the sun, [tex]T_s[/tex] = 5,300 K

temperature of the Earth, [tex]T_E[/tex] = 293 K

radiation energy transferred by the sun to the earth, E = 6000 J

The sun loses Q of heat and therefore decreases its entropy by the amount

[tex]\delta S_{sun} = \frac{-Q}{T_s}[/tex]

The earth gains Q of heat and therefore increases its entropy by the amount

[tex]\delta S_{Earth} = \frac{-Q}{T_E}[/tex]

The total entropy change is:

[tex]\delta S_{Earth} + \delta S_{sun} = \frac{Q}{T_E} -\frac{Q}{T_S} \\\\ = Q(\frac{1}{T_E} -\frac{1}{T_S} )\\\\= 6000(\frac{1}{293} -\frac{1}{5300} )\\\\=6000(0.0032243)\\\\= 19.346 \ J/K[/tex]

Therefore, the entropy change of the Universe that occurs is 19.346 J/K

Answer:

D

Explanation:

Now the net force is the applied force minus the frictional force; this is expressed mathematically as:

Fnet= Fappplied - Ffrictional

Now the frictional force is given as ;

Coefficient of friction × normal reaction

Normal reaction is the weight of the human acting in opposite direction.

Normal reaction of the human is ;

35 × 9.8 = 343N { note that weight = m× g and g= 9.8m/S2, a known standard }

Hence the Frictional force =343×0.4 =137.20N

Hence Fnet = 200-137.20 = 62.8N

Fnet = 63N to the nearest whole

The net force on the couch as it slides is  63N.

What is frictional force?

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.

The friction force is given by

f = coefficient of friction x Normal force

Given, the coefficient of kinetic friction between a couch and the floor is 0.4. If the couch has a mass of 35 kg and you push it with a force of 200 N.

Normal reaction is the weight of the human acting in opposite direction.

Normal reaction N =35 × 9.8 = 343N

Frictional force f =0.4 x 343

                          f =137.20N

The net force will be

Fnet= Fappplied - Ffrictional

Fnet = 200-137.20 = 62.8N

Fnet = 63N

Thus,  the net force on the couch as it slides is  63N.

Learn more about friction force.

https://brainly.com/question/1714663

#SPJ2

Answer:

  c.  in the absence of an unbalanced force, an object at rest will stay at rest and an object in motion will stay in motion.

Explanation:

First law: things keep doing what they are doing, unless force is applied.

Answer:

Stretch in the spring = 0.1643 (Approx)

Explanation:

Given:

Mass of the sled (m) = 9 kg

Acceleration of the sled (a) = 2.10 m/s ²

Spring constant (k) = 115 N/m

Computation:

Tension force in the spring  (T) = ma

Tension force in the spring  (T) = 9 × 2.10

Tension force in the spring  (T) = 18.9 N

Tension force in the spring = Spring constant (k) × Stretch in the spring

18.9 N = 115 N  × Stretch in the spring

Stretch in the spring = 18.9 / 115

Stretch in the spring = 0.1643 (Approx)

-- F = m a ... ==>  a = F/m

-- The tension in the rope is 362 N.  That same force acts on the asteroid and on the tug, pulling them together.

-- The asteroid's acceleration is 362N / 6240 kg = 0.058 m/s², headed for a point on the rope somewhere between the asteroid and the tug.

-- The tug's acceleration is 362 N / 3220 kg = 0.112 m/s², also headed for a point on the rope somewhere between the tug and the asteroid.

-- So now we have a gap between them, initially 311 m long, closing with a speed that starts at zero and accelerates at  0.170 m/s² .

-- D = (1/2) a T²

311 m = (1/2) (0.170 m/s²) (T²)

T²  =  311 m / 0.085 m/s²

T = √(311/0.085)  seconds

T = 60.41 seconds

The answer I get is so durn near 60 seconds (1 minute) that it suggests two things to me:  ==> That's where the weird numbers of 362N and 311m came from, and ==> there's a good chance that my answer is correct.

Note:  It's important to me that you know that 5 points for this one is really cheap and chintzy, and the reason I decided to try it was only to see whether I could.

Answer:

The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].

Explanation:

As CD is acceleration uniformly, the following equation of motion can be used to determine the angular acceleration:

[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot \Delta n[/tex]

Where:

[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.

[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.

[tex]\ddot n[/tex] - Angular acceleration, measured in revolution per square minute.

[tex]\Delta n[/tex] - Change in angular position, measured in revolutions.

The angular acceleration is cleared and calculated:

[tex]\ddot n = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \Delta n}[/tex]

Given that [tex]\dot n_{o} = 0\,\frac{rev}{min}[/tex], [tex]\dot n = 430\,\frac{rev}{min}[/tex] and [tex]\Delta n = 4\, rev[/tex], the angular acceleration is:

[tex]\ddot n = \frac{\left(430\,\frac{rev}{min} \right)^{2}-\left(0\,\frac{rev}{min} \right)^{2}}{2\cdot (4\,rev)}[/tex]

[tex]\ddot n = 23112.5\,\frac{rev}{min^{2}}[/tex]

The angular accelaration measured in radians per square second is:

[tex]\alpha = \left(23112.5\,\frac{rev}{min^{2}} \right)\cdot \left(2\pi\,\frac{rad}{rev}\right)\cdot \left(\frac{1}{3600}\,\frac{min^{2}}{s^{2}} \right)[/tex]

[tex]\alpha \approx 40.339\,\frac{rad}{s^{2}}[/tex]

Net torque experimented by the CD during its accleration is equal to the product of its moment of inertia with respect to its axis of rotation and angular acceleration:

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

Where:

[tex]I[/tex] - Moment of inertia, measured in [tex]kg \cdot m^{2}[/tex].

[tex]\alpha[/tex] - Angular acceleration, measured in radians per square second.

In addition, a CD has a form of a uniform disk, whose moment of inertia is:

[tex]I = \frac{1}{2}\cdot m \cdot r^{2}[/tex]

Where:

[tex]m[/tex] - Mass of the CD, measured in kilograms.

[tex]r[/tex] - Radius of the CD, measured in meters.

If [tex]m = 0.017\,kg[/tex] and [tex]r = 0.07\,m[/tex], then:

[tex]I = \frac{1}{2}\cdot (0.017\,kg)\cdot (0.07\,m)^{2}[/tex]

[tex]I = 4.165\times 10^{-5}\,kg\cdot m^{2}[/tex]

Now, the net torque exerted on CD is:

[tex]\tau = (4.165\times 10^{-5}\,kg\cdot m^{2})\cdot \left(40.339\,\frac{rad}{s^{2}} \right)[/tex]

[tex]\tau = 1.680\times 10^{-3}\,N\cdot m[/tex]

The net torque exerted on CD is [tex]1.680 \times 10^{-3}\,N\cdot m[/tex].

Answer:

a population increase

Explanation:

During the 20th century, the world population increased from 1.65 billion to 6 billion. In 1970, the world's population was half that of today. In less than 15 years, 47% of the population will live in areas already under heavy water stress. In Africa, between 75 and 250 million people will face growing shortages in 2020 due to climate change. The scarcity of some arid and semi-arid regions will have a decisive impact on migration.

Answer:

1.48 g

Explanation:

A = A₀ (½)^(t / T)

where A is the final amount,

A₀ is the initial amount,

t is time,

and T is the half life.

A = (5.00 g) (½)^(5.00 a / 2.85 a)

A = 1.48 g

Answer:

B. F/2

Explanation:

The radiation force per unit area (radiation pressure Prad) exerted by an electromagnetic wave on a perfectly absorbing body has been found by experiment to be equal to the energy density of the wave

i.e Prad = u

For a reflecting body, this force exerted per unit area has been found to be twice the energy density of the wave.

i.e Prad = 2u.

Therefore, if the force exerted on a perfectly reflective body is F, then the force exerted on a perfectly absorbing body will be F/2

Answer:

a = 1.72 m/s²

Explanation:

The given kinematic equation is the 2nd equation of motion. The equation is as follows:

xf = xi + (Vi)(t) + (1/2)(a)t²

where,

xf = the final position =  5000 m

xi = the initial position = 1000 m

Vi = the initial velocity = 15 m/s

t = the time taken = 60 s

a = acceleration = ?

Therefore,

5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²

5000 m = 1000 m + 900 m + a(1800 s²)

5000 m = 1900 m + a(1800 s²)

5000 m - 1900 m = a(1800 s²)

a(1800 s²) = 3100 m

a = 3100 m/1800 s²

a = 1.72 m/s²

Answer:

The final speed of the ball is 9 m/s.

Explanation:

We have,

A steel ball is dropped from a diving platform. It is required to find the velocity of the ball 0.9 seconds after its released. It will move under the action of gravity. Using equation of motion to find it as :

[tex]v=u+at[/tex]

u = 0 (at rest), a = g

[tex]v=gt\\\\v=10\times 0.9\\\\v=9\ m/s[/tex]

So, the final speed of the ball is 9 m/s.

the answer is 1.5 hope this helps

Answer:

1.5

Explanation:

0.5=σ/(15/0.5)

σ=3/2 or 1.5

Answer:

(a) f = 185 Hz

(b) v = 266.4 m/s

Explanation:

(a) The lowest frequency can be calculated by using the following formula for the calculation of the modes (resonant frequencies) in a string:

[tex]f_n=\frac{nv}{2L}[/tex]

[tex]f_n=nf[/tex]

n: order of the mode

v: velocity of the waves in the string

L: length of the string = 72.0cm = 0.72m

fn: frequency of the n-th mode

With the information about two consecutive modes you can find the lowest resonant frequency. First you find the resonant mode n:

[tex]f_n=nf\\\\f_{n-1}=(n-1)f\\\\\frac{f_n}{f_{n-1}}=\frac{n}{n-1}[/tex]

you solve the previous equation for n:

[tex](n-1)f_n=nf_{n-1}\\\\555n-555=370n\\\\n=3[/tex]

With this information you can calculate the lowest resonant frequency:

[tex]f_n=nf\\\\f=\frac{f_n}{n}=\frac{555}{3}=185Hz[/tex]

b) You have information about two consecutive modes fn, fn-1. Then, you can calculate the velocity of the waves:

[tex]f_{n}-f_{n-1}=n\frac{v}{2L}-(n-1)\frac{v}{2L}\\\\f_n-f_{n-1}=\frac{v}{2L}\\\\v=2L(f_n-f_{n-1})[/tex]

fn = 555 Hz

fn-1: 370 Hz

[tex]v=2(0.72m)(555-370)Hz=266.4\frac{m}{s}[/tex]´

hence, the velocityof the waves in the string is 266.4 m/s

Answer:

i) angle of incidence;i = 29.43°

ii) δm = 38.92°

Explanation:

Prism is equilateral so angle of prism (A) = 60°

Refractive index of glass; n_glass = 1.52

A) Let's assume the incident angle = i and Critical angle = θc

We know that, sin θc = 1/n

Thus;

sin θc = 1/n_glass

θc = sin^(-1) (1/n_glass)

θc = sin^(-1) (1/1.52)

θc = 41.14°

Now, the angle of prism will be the sum of external angle that is critical angle and reflected angle.

Thus;

A = r + θc

r = A - θc

So;

r = 60° - 41. 14°

r = 18.86°

From, Snell's law. If we apply it to this question, we will have;

(sin i)/(sin r) = n_glass

Where;

i is angle of incidence and r is angle of reflection.

Let's make i the subject;

i = sin^(-1) (n_glass × sin r)

i = sin^(-1) (1.52 × sin 18.86)

i = sin^(-1) 0.4914

i = 29.43°

B) The formula to calculate minimum deviation would be from;

μ = [sin ((A + δm)/2)]/(sin A/2)

Where;

μ is Refractive index

δm is minimum angle of deviation

A is angle of prism

Now Refractive index is given by a formula; μ = (sin i)/(sin r)

So; μ = (sin 29.43)/(sin 18.86)

μ = 1.52

Thus;

1.52 = [sin ((60 + δm)/2)]/(sin 60/2)

1.52 * sin 30 = sin ((60 + δm)/2)

0.76 = sin ((60 + δm)/2)

sin^(-1) 0.76 = ((60 + δm)/2)

49.46 × 2 = (60 + δm)

98.92 - 60 = δm

δm = 38.92°

Answer:

v = 25.45 m/s

Explanation:

In order to calculate the initial speed of the object, you take into account the formula for the maximum height reaches by the object. Such a formula is given by:

[tex]h_{max}=\frac{v_o^2}{g}[/tex]   (1)

vo: initial speed of the object = 18 m/s

g: gravitational acceleration = 9.8 m/s²

Furthermore you use the following formula for the final speed of the object:

[tex]v^2=v_o^2-2gh[/tex]       (2)

h: height

You know that the speed of the object is 18m/s when it reaches one fourth of the maximum height. You use this information, and you replace the equation (1) in to the equation (2), as follow:

[tex]v^2=v_o^2-2g(\frac{h_{max}}{4})=v_o^2-\frac{1}{2}g(\frac{v_o^2}{g})\\\\v^2=v_o^2-\frac{1}{2}v_o^2=\frac{1}{2}v_o^2[/tex]

Then, you solve the previous result for vo:

[tex]v_o=\sqrt{2}v=\sqrt{2}(18m/s)=25.45\frac{m}{s}[/tex]

The initial speed of the object was 25.45 m/s

Answer:  If the woodpecker drums upon a tree 5 times in one second, then the frequency is 5 Hz; each drum must endure for one-fifth a second, so the period is 0.2 s.

Answer:

Explanation:

vertical component of the velocity of arrow

= 26 sin 60 = 22.516 m

height reached by it after 3.99 s

h = ut - 1/2 g t²

= 22.516 x 3.99 - .5 x 9.8 x 3.99²

= 89.83 - 78

11.83 m

Total height of cliff = 1.55 + 11.83

= 13.38 m

c ) maximum height covered s

v² = u² - 2gs

0 = u² - 2gs

s = u² / 2g

= 22.516² / 2 x 9.8

= 25.86

maximum height reached

= 25.86 + 1.55

= 27.41 m

d )

vertical speed after 3.99 s

v = u - gt

= 22.516 - 9.8 x 3.99

= -16.586

Horizontal component will remain unchanged

Horizontal component = 26 cos 60

= 13 m /s

Resultant of two velocities

= √ 13²+ 16.568²

= 21 m /s

Answer:

The distance covered by puck A before collision is  [tex]z = 8.56 \ m[/tex]

Explanation:

From the question we are told that

   The label on the two hockey pucks is  A and  B

    The distance between the  two hockey pucks is D   18.0 m

     The speed of puck A is  [tex]v_A = 3.90 \ m/s[/tex]

        The speed of puck B is  [tex]v_B = 4.30 \ m/s[/tex]

The distance covered by puck A is mathematically represented as

     [tex]z = v_A * t[/tex]

  =>  [tex]t = \frac{z}{v_A}[/tex]

 The distance covered by puck B  is  mathematically represented as

      [tex]18 - z = v_B * t[/tex]

=>   [tex]t = \frac{18 - z}{v_B}[/tex]

Since the time take before collision is the same

        [tex]\frac{18 - z}{V_B} = \frac{z}{v_A}[/tex]

substituting values

          [tex]\frac{18 -z }{4.3} = \frac{z}{3.90}[/tex]

=>      [tex]70.2 - 3.90 z = 4.3 z[/tex]

=>       [tex]z = 8.56 \ m[/tex]