Which graph would be created by a pendulum with the greatest amplitude?

"down/up the plane" suggests an inclined plane, but no angle is given so I'll call it θ for the time being.

The free body diagram for the crate in either scenario is the same, except for the direction in which static friction is exerted on the crate. With the P = 100 N force holding up the crate, static friction points up the incline and keeps the crate from sliding downward. When P = 350 N, the crate is pushed upward, so static friction points down. (see attached FBDs)

Using Newton's second law, we set up the following equations.

• p = 100 N

∑ F (parallel) = f + p cos(θ) - mg sin(θ) = 0

∑ F (perpendicular) = n - p sin(θ) - mg cos(θ) = 0

• P = 350 N

∑ F (parallel) = P cos(θ) - F - mg sin(θ) = 0

∑ F (perpendicular) = N - P sin(θ) - mg cos(θ) = 0

(where n and N are the magnitudes of the normal force in the respective scenarios; ditto for f and F which denote static friction, so that f = µn and F = µN, with µ = coefficient of static friction)

Solve for n and N :

n = p sin(θ) + mg cos(θ)

N = P sin(θ) - mg cos(θ)

Substitute these into the corresponding equations containing µ, and solve for µ :

µ = (mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ))

µ = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))

Next, you would set these equal and solve for m :

(mg sin(θ) - p cos(θ)) / (mg cos(θ) + p sin(θ)) = (P cos(θ) - mg sin(θ)) / (P sin(θ) + mg cos(θ))

...

Once you find m, you back-substitute and solve for µ, but as you might expect the result will be pretty complicated. If you take a simple angle like θ = 30°, you would end up with

m ≈ 36.5 kg

µ ≈ 0.256

The coefficient of static friction between the plane and the crate is μ = 0.256 and the mass of the crate is m=36.4 kg.

From the given,

The force that opposes the crate by sliding is P = 100N

In X-axis, the sum of forces is zero.

ΣF = 0

Pcosθ - mgsinθ-Ff = 0

Ff = Pcosθ - mgsinθ

In Y-axis

Psinθ - mgcosθ - N = 0

N = Psinθ-mgcosθ

Frictional force, Ff = μN, μ is the coefficient of friction

Ff = μN

Pcos30- mgsin30 + μ( Psin30+mgcos30) = 0

μ = mgsin30-Pcos30/Psin30+mgcos30 ------1

The block is sliding with the horizontal force, F = 350N

X-axis

P₂cosθ - mgsinθ-Ff = 0

Y-axis

P₂sinθ - mgcosθ - N = 0

N = P₂sinθ-mgcosθ

μ = P₂cos30-mgsin30/P₂sin30-mgcos30   -----2

Equate equations 1 and 2

mgsin30-Pcos30/Psin30+mgcos30 =P₂cos30-mgsin30/P₂sin30-mgcos30

4.905m-86.6/50+8.49 = 303.1-4.905m/175+8.49

41.7m² + 123m - 1.516×10⁴ = 0

-41.7m² +2330m -1.516×10⁴(4.905-86.6)(175+8.49) =(303.1-4.905)(50+8.49)

83.4m² - 2207m -3.03×10⁴ = 0

m= 36.4 kg

Hence, the mass of the crate is 36.4 Kg.

Substitute the value of m in equation 1,

μ = 4.905(36.4) - 86.6 / 50 + 8.49

μ  = 0.256

Thus, the coefficient of static friction is 0.256.

To learn more about friction and its types:

https://brainly.com/question/30886698

#SPJ1

Answer "a" would be correct.

Answer:

d

Explanation:

There's an acceleration from gravity, thus the velocity is becoming faster and faster as it reaches the ground. Thus its D

Brainliest please~

Answer:

W = Q * V     work done on charge Q

A. W = .5 C * 1.5 V = .75 Joules

B. P = W / t = .75 J / 1 sec = .75 Watts

Answer:

Centripetal Acceleration 18.75 m/s^2, Rotational Kinetic Energy 843.75 J

Explanation:

a Linear acceleration (we cant find tangential acceleration with the givens so we will find centripetal)

a= ω^2*r

ω= 300rev/min

convert into rev/s

300/60= 5rev/s

a= 18.75m/s^2

b) use Krot= 1/2 Iω^2

plug in gives

1/2(30*2.25)(25)= 843.75 J

Answer:

0.857 cm

Explanation:

We are given that:

The focal length for a convex lens to be (f) = 1.5cm

The object distance (u) = - 2.0 cm

We are to determine the image distance (v) = ??? cm

By applying the lens formula:

[tex]\dfrac{1}{f} = \dfrac{1}{u}+\dfrac{1}{v}[/tex]

By rearrangement and making (v) the subject of the above formula:

[tex]v = \dfrac{uf}{u-f}[/tex]

replacing the given values:

[tex]v = \dfrac{(-2.0)(1.5)}{(-2.0 -1.5)}[/tex]

[tex]v = \dfrac{-3.0}{(-3.5)}[/tex]

v = 0.857 cm

Answer:

Option A. –5 m/s²

Explanation:

From the question given above, the following data were obtained:

Initial velocity (v₁) = 9 m/s

Initial time (t₁) = 0 s

Final velocity (v₂) = –6 m/s

Final time (t₂) = 3 s

Acceleration (a) =?

Next, we shall determine the change in the velocity and time. This can be obtained as follow:

For velocity:

Initial velocity (v₁) = 9 m/s

Final velocity (v₂) = –6 m/s

Change in velocity (Δv) =?

ΔV = v₂ – v₁

ΔV = –6 – 9

ΔV = –15 m/s

For time:

Initial time (t₁) = 0 s

Final time (t₂) = 3 s

Change in time (Δt) =?

Δt = t₂ – t₁

Δt = 3 – 0

Δt = 3 s

Finally, we shall determine the acceleration of the object. This can be obtained as follow:

Change in velocity (Δv) = –15 m/s

Change in time (Δt) = 3 s

Acceleration (a) =?

a = Δv / Δt

a = –15 / 3

a = –5 m/s²

Thus, the acceleration of the object is

–5 m/s².

Answer:

5.25 m/s to the east

Explanation:

Applying,

MV = mv.............. Equation 1

Where M = mass of the launcher, V = recoil velocity of the launcher, m = mass of the balloon, v = velocity of the balloon

make V the subject of the equation

V = mv/M............ Equation 2

From the question,

M = 2 kg, m = 0.75 kg, v = 14 m/s

Substitute these values into equation 2

V = (0.75×14)/2

V = 5.25 m/s to the east

The complete question is as follows: At 700 K, [tex]CCl_{4}[/tex] decomposes to carbon and chlorine. The Kp for the decomposition is 0.76.

Find the starting pressure of [tex]CCl_{4}[/tex] at this temperature that will produce a total pressure of 1.1 atm at equilibrium.

Answer: The starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.

Explanation:

The equation for decomposition of [tex]CCl_{4}[/tex] is as follows.

[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]

Let us assume that initial concentration of [tex]CCl_{4}[/tex] is 'a'. Hence, the initial and equilibrium concentrations will be as follows.

                   [tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]

Initial:            a                0          0

Equilibrium:  (a - x)          0          2x

Total pressure = (a - x) + 2x = a + x

As it is given that the total pressure is 1.1 atm.

So, a + x = 1.1

a = 1.1 - x

Now, expression for equilibrium constant for this equation is as follows.

[tex]K_{p} = \frac{P^{2}_{Cl_{2}}}{P_{CCl_{4}}}\\0.76 = \frac{(2x)^{2}}{(a - x)}\\0.76 = \frac{4x^{2}}{1.1 - x - x}\\0.76 = \frac{4x^{2}}{1.1 - 2x}\\x = 0.31 atm[/tex]

Hence, the value of 'a' is calculated as follows.

a + x = 1.1 atm

a = 1.1 atm - x

  = 1.1 atm - 0.31 atm

  = 0.79 atm

Thus, we can conclude that starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.

Answer:

The number of revolutions is 44.6.

Explanation:

We can find the revolutions of the wheel with the following equation:

[tex]\theta = \omega_{0}t + \frac{1}{2}\alpha t^{2}[/tex]

Where:

[tex]\omega_{0}[/tex]: is the initial angular velocity = 13 rad/s              

t: is the time = 8 s

α: is the angular acceleration

We can find the angular acceleration with the initial and final angular velocities:

[tex] \omega_{f} = \omega_{0} + \alpha t [/tex]

Where:

[tex] \omega_{f} [/tex]: is the final angular velocity = 57 rad/s

[tex] \alpha = \frac{\omega_{f} - \omega_{0}}{t} = \frac{57 rad/s - 13 rad/s}{8 s} = 5.5 rad/s^{2} [/tex]

Hence, the number of revolutions is:

[tex] \theta = \omega_{0}t + \frac{1}{2}\alpha t^{2} = 13 rad/s*8 s + \frac{1}{2}*5.5 rad/s^{2}*(8 s)^{2} = 280 rad*\frac{1 rev}{2\pi rad} = 44.6 rev [/tex]

Therefore, the number of revolutions is 44.6.

I hope it helps you!

Answer:

 E = 2,964 10⁻¹⁹ J

Explanation:

The energy of the photons is given by the Planck relation

          E = h f

the speed of light is related to wavelength and frequency

          c = λ f

we substitute

          E = h c /λ

let's reduce the magnitude to the SI system

          λ = 671 nm = 671 10⁻⁹ m

let's calculate

          E = 6.63 10⁻³⁴ 3 10⁸ /671 10⁻⁹

          E = 2,964 10⁻¹⁹ J

Answer:

1.66 N

Explanation:

The force of gravity of the moon on the body is given by

F = GMm/R² where G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg², M = mass of moon = 7.2 × 10²² kg, m = mass of body = 1 kg and R = radius of moon = 1.7 × 10⁶ m

Substituting the values of the variables into the equation, we have

F = GMm/R²

F = 6.67 × 10⁻¹¹ Nm²/kg² × 7.2 × 10²² kg × 1 kg/(1.7 × 10⁶ m)²

F = 48.024 × 10¹¹ Nm²/2.89 × 10¹² m²

F = 16.62 × 10⁻¹ N

F = 1.662 N

F ≅ 1.66 N

So, the gravity on the moon is 1.66 N

Answer:

hope it helps

much as you can

Answer:

The correct option is (E).

Explanation:

Given that,

Mass of object 1, m₁ = 1 kg

Mass of object 2, m₂ = 2 kg

They collides after the collision. We need to find the speed of the two boxes after the collision.

The initial speeds of both boxes is not given. So, we can't put the values of their speeds in the momentum conservation equation.

So, the information is not enough.

Answer:

a)   a = 91.4 m / s²,  b)    t = 0.175 s, c)  

Explanation:

a) This is a kinematics exercise

           v² = vox ² + 2a (x-xo)

           a = v² - 0/2 (x-0)

let's calculate

          a = 16² / 2 1.4

          a = 91.4 m / s²

b) the shooting time

          v = vox + a t

          t = v-vox / a

          t = 16 / 91.4

          t = 0.175 s

c) let's use Newton's second law

          F = ma

          F = 7.9 91.4

          F = 733 N

Answer:

F/2

Explanation:

In the first case, the two charges are Q1 and Q2 and the distance between them is r. K is the Coulomb's constant

Hence;

F= KQ1Q2/r^2 ------(1)

Where the charge on Q1 is doubled and the distance separating the charges is also doubled;

F= K2Q1 Q2/(2r)^2

F2= 2KQ1Q2/4r^2 ----(2)

F2= F/2

Comparing (1) and (2)

The magnitude of force acting on each of the two particles is;

F= F/2

Explanation:

Conservation of energy, principle of physics according to which the energy of interacting bodies or particles in a closed system remains constant

hope it is helpful to you

Answer:

The change in entropy is 1.6 W/K.

Explanation:

Thickness, d = 0.5 cm

Area, A = 1 m^2

T = 25°C

T' = - 10°C

Coefficient of thermal conductivity of glass, K = 0.8 W/mK

The change in entropy is given by

S = Q/T

Here,

[tex]S =\frac{Q}{T}\\\\S = \frac{K A (T - T')}{d(T - T')}\\\\S = \frac{0.8\times 1}{0.5} = 1.6 W/K[/tex]

Answer:

c

Explanation:

darker colors absorb app light

Answer:

C. Is absorbing all wavelengths of light.

Explanation:

Black isn't a color, but rather the absence of color. We see a T-shirt as black because it isn't reflecting any light toward our eyes. A black T-shirt absorbs all of the wavelengths of light, causing it to absorb more energy and become warmer than white, which reflects light.

Answer: See attached pic. Hope this helps.

Explanation:

Answer:

The the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Explanation:

We are given that

Angular acceleration, [tex]\alpha=3.3 rad/s^2[/tex]

Diameter of the wheel, d=21 cm

Radius of wheel, [tex]r=\frac{d}{2}=\frac{21}{2}[/tex] cm

Radius of wheel, [tex]r=\frac{21\times 10^{-2}}{2} m[/tex]

1m=100 cm

Magnitude of total linear acceleration, a=[tex]1.7 m/s^2[/tex]

We have to find the linear speed  of a  at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

Tangential acceleration,[tex]a_t=\alpha r[/tex]

[tex]a_t=3.3\times \frac{21\times 10^{-2}}{2}[/tex]

[tex]a_t=34.65\times 10^{-2}m/s^2[/tex]

Radial acceleration,[tex]a_r=\frac{v^2}{r}[/tex]

We know that

[tex]a=\sqrt{a^2_t+a^2_r}[/tex]

Using the formula

[tex]1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}[/tex]

Squaring on both sides

we get

[tex]2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}[/tex]

[tex]\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}[/tex]

[tex]v^4=r^2\times 2.7699[/tex]

[tex]v^4=(10.5\times 10^{-2})^2\times 2.7699[/tex]

[tex]v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}[/tex]

[tex]v=0.418 m/s[/tex]

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

Answer:

30J

Explanation:

Given data

The total quantity of heat recieved= 50J

Quantity of heat used to do work= 20J

Hence the net change is

ΔU= Total Heat - Net work

ΔU= 50-20

ΔU= 30J

Hence the change in the internal energy is 30J

Explanation:

No matter where you are in the universe, your mass is always the same: mass is a measure of the amount of matter which makes up an object. Weight, however, changes because it is a measure of the force between an object and body on which an object resides (whether that body is the Earth, the Moon, Mars, et cetera).

Explanation:

Hence, weight of a body will change from one place to another place because the value of g is different in different places. For example, the value of g on moon is 1/6 times of the value of g on earth. As mass is independent of g , so it will not change from place to place.

The final speed of the ball is 91.72 m/s.

Given the following data:

To find the final speed of the ball, we would use the following formula:

[tex]F = \frac{M(V - U)}{t}[/tex]

Where:

Substituting the parameters into the formula, we have;

[tex]15000 = \frac{0.145(V \;- \;40)}{0.0005}\\\\15000(0.0005) = 0.145(V \;- \;40)\\\\7.5 = 0.145V - 5.8\\\\0.145V = 7.5 + 5.8\\\\0.145V = 13.3\\\\V = \frac{13.3}{0.145}[/tex]

Final speed, V = 91.72 m/s

Therefore, the final speed of the ball is 91.72 m/s.

Read more here: https://brainly.com/question/24029674

Answer:

Following are the solution to the given question:

Explanation:

Please find the complete question in the attached file.

The cost after 30 days is 60 dollars. As energy remains constant, the cost per hour over 30 days will be decreased.

[tex]\to \frac{\$60}{\frac{30 \ days}{24\ hours}} = \$0.08 / kwh.[/tex]

Thus, [tex]\frac{\$0.08}{\$0.12} = 0.694 \ kW \times 0.694 \ kW \times 1000 = 694 \ W.[/tex]

The electricity used is continuously 694W over 30 days.

If just resistor loads (no reagents) were assumed,

[tex]\to I = \frac{P}{V}= \frac{694\ W}{120\ V} = 5.78\ A[/tex]

Energy usage reduction percentage = [tex](\frac{60\ W}{694\ W} \times 100\%)[/tex]

This bulb accounts for [tex]8.64\%[/tex] of the energy used, hence it saves when you switch it off.

Answer:

[tex]d_2=1.5*10^-3m[/tex]

Explanation:

From the question we are told that:

Initial Wavelength [tex]\lambda_1=550nm=550*10^{-9}[/tex]

Space 1 [tex]d_1=2.3*10^{-3}[/tex]

Final wavelength [tex]\lambda_2=360*10^{-9}[/tex]

Generally the equation for Fringe space at [tex]\lambda _2[/tex] is mathematically given by

 [tex]d_2=\frac{d_1}{\lambdaI_1}*\lambda_2[/tex]

 [tex]d_2=\frac{2.3*10^{-3}}{550*10^{-9}}*360*10^{-9}[/tex]

 [tex]d_2=1.5*10^-3m[/tex]

Answer:

 [tex]k_{eq} = \frac{k}{N}[/tex]

Explanation:

For this exercise let's use hooke's law

         F = - k x

where x is the displacement from the equilibrium position.

        x = [tex]- \frac{F}{k}[/tex]

if we have several springs in series, the total displacement is the sum of the displacement for each spring, F the external force applied to the springs

       x_ {total} = ∑ x_i

we substitute

       x_ {total} =  ∑ -F / ki

       F / k_ {eq} =  -F  [tex]\sum \frac{1}{k_i}[/tex]

      [tex]\frac{1}{k_{eq}} = \frac{1}{k_i}[/tex] 1 / k_ {eq} =  ∑ 1 / k_i

if all the springs are the same

     k_i = k

     [tex]\frac{1}{k_{eq}} = \frac{1}{k} \sum 1 \\[/tex]

     [tex]\frac{1}{k_{eq} } = \frac{N}{k}[/tex]

     [tex]k_{eq} = \frac{k}{N}[/tex]

Answer:

(7.115 ; 13.405)

Explanation:

Given :

Sample size, n = 22

Mean bill, μ = 10.26

Standard deviation, s = 5.21

To obtain the 99% confidence interval for the mean bill of all orders ;

Mean ± margin of error

Margin of Error = Tcritical * s/√n

Tcritical at 99%, df = n-1, 22 - 1 = 21

Tcritical = 2.831

Margin of Error = 2.831 * (5.21/√22) = 3.145

Confidence interval = 10.26 ± 3.145

Lower boundary = 10.26 - 3.145 = 7.115

Upper boundary = 10.26 + 3.145 = 13.405

Confidence interval :

(7.115 ; 13.405)

Answer:

Force is zero.

Explanation:

According to the Newton's second law, when an object is moving with an acceleration the force acting on the object is directly proportional to the rate of change of momentum of the object.

F = m a

if the object is moving with uniform velocity, the acceleration is zero, and thus, the force is also zero.  

Answer: Near the moon’s surface, a thrust over 11,250 N but under 13,500 N would make it travel at a constant vertical velocity.

Explanation: .Newton’s first law of motion states that an object in motion continues to move in a straight line at a constant velocity unless acted upon by an unbalanced force. In accordance with this law, the lunar lander moves in a downward direction toward the surface of the moon under the influence of force due to gravity. A thrust somewhere between 11,250 and 13,500 balances this gravitational force out.