hii guys!! ~ i've comeback again!! could you please help me with an assignment?? i have a couple of i know questions!! thank you for helping me, hehe.

tenise can run the 100 meter dash in 9.77 seconds. what is her speed??

Answer:

C. [tex]W = 115.12\,Btu[/tex]

Explanation:

Thermodynamically speaking, a Carnot engine represents an entirely reversible thermal process and its energy efficiency represents the maximum theoretical efficiency that thermal machines can reach. The efficiency of the ideal thermal process ([tex]\eta[/tex]), no unit, is:

[tex]\eta = \left(1-\frac{T_{L}}{T_{H}} \right)[/tex] (1)

Where:

[tex]T_{L}[/tex] - Temperature of the cold reservoir, measured in Rankine.

[tex]T_{H}[/tex] - Temperature of the hot reservoir, measured in Rankine.

If we know that [tex]T_{H} = 1809.67\,R[/tex] and [tex]T_{L} = 584.67\,R[/tex], then the energy efficiency of the ideal thermal process is:

[tex]\eta = 0.678[/tex]

By First Law of Thermodynamics, we calculate the work output:

[tex]W = Q_{H}-Q_{L}[/tex]

[tex]W = \frac{W}{\eta} -Q_{L}[/tex] (By definition of efficiency)

[tex]Q_{L} = \frac{W}{\eta}-W[/tex]

[tex]Q_{L} = \left(\frac{1}{\eta}-1 \right)\cdot W[/tex](2)

Where:

[tex]Q_{H}[/tex] - Heat received by the engine, measured in Btu.

[tex]Q_{L}[/tex] - Heat rejected by the engine, measured in Btu.

[tex]W[/tex] - Work output, measured in Btu.

If we know that [tex]\eta = 0.678[/tex] and [tex]Q_{L} = 55\,Btu[/tex], then the work output of the Carnot engine is:

[tex]W = \frac{Q_{L}}{\frac{1}{\eta}-1 }[/tex]

[tex]W = 115.807\,Btu[/tex]

The work output of the Carnot engine is 115.807 Btu. (Answer: C)

Answer:a) - 0.4905 m/s²   b) distance = 35.48 m

Explanation:

Given that  

The initial velocity of the skater = 5.90 m/s

 kinetic friction coefficient = 0.0500

final velocity = 0 m/s(since it  comes to rest)

deceleration cause by the kinetic friction = ?

we know that  

F = μN

and N= mg

Therefore;

F = μ m g....................(1)

also  that

F = m a........................(2)

with our common Force, F, equating  (1) and (2), we have that

m a = - μ m g

a = - μ g

a = - 0.05 × 9.81

a = - 0.4905 m/s²

The deceleration cause by the kinetic friction is a = - 0.4905 m/s²

b)

The distance the skater  travels before stopping

is given as

    Vf² = v₀² - 2 a x

  final velocity = 0 m/s(since it  comes to rest)  

Therefore We have that

 0 = v₀² - 2 a x

 x = - v₀² / 2 a

 x = 5.90² / (2 x 0.4905  )

34.81/0.981  

x = 35.48 m

Or

using

v²-u² = 2aS final velocity = 0 m/s(since it  comes to rest)  

0²-5.90² = -2×0.4905×S

34.81=0.981S

S= 34.81/0.981

S=35.48m

Answer:

a) Find the work done by the resistive force during the roundtrip.

W=-30kJ

b) Is the resistive force a conservative force? explain.

The resistive force is not a conservative force since the work done during the round trip is not zero

Explanation:

The worf done on object y a constant force F is given by:

W= (F  cos ∅)S

Where S is the displacement and ∅ is the angle between the force and the displacement.

The displacement of the bicycle during each part of the trip is s=5000m and teh magnitude of teh resistance force is F=3.0N

∅1=180° he angle between the displacement and the force

W1=W2

W1 = (3.0 cos180) 5000m

W1=-15.O kJ

W=W1+W2

W=-30kJ

The resistive force is not a conservative force since the work done during the round trip is not zero

(a) The work done by the resistive force is 15,000 J

(b) The work done the resistive force is non-conservative since the resultant resistive force in not zero.

Work is said to be when an applied force displaces an object from its initial position.

The work done by the resistive force is calculated as follows;

W = FΔr

W = 3 x (5,000 - 0)

W = 15,000 J

Thus, the work done the resistive force is non-conservative since the resultant resistive force in not zero.

Learn more about conservative force here: https://brainly.com/question/15357875

Answer:

okay I'm a bit confused but I like the little emoji dudw

Answer:

?

Explanation:

.

Answer:

11.98 m

Explanation:

Given that:

mass of the child [tex]m_c[/tex] = 53 kg

mass of the boat [tex]m_b[/tex] = 107 kg

[tex]\text{length of the boat L = 7 m}[/tex]

the distance of the boat from pies l = 7.3 m

initial momentum [tex]P_i = 0[/tex]

Final momentum [tex]P_f = mc \dfrac{L}{f}- (m_c +m_b) \dfrac{x}{l}[/tex]

where;

x = distance moved by boat towards left

t = time taken for the child to travel to the far end of the boat

[tex]P_i =P_f[/tex]

∴

[tex]m_c \dfrac{L}{t}=(m_c +m_b) \dfrac{x}{t}[/tex]

Then;

[tex]x = \dfrac{m_cL}{m_c+m_b}[/tex]

[tex]x = \dfrac{53 \times 7}{53+107}[/tex]

x = 2.32 m

The distance of the child from the pier is:

d = L +(l - x)

d = 7 m + ( 7.3 m - 2.32 m)

d = 7 m + 4.98 m

d = 11.98 m

Answer:

290

Explanation:

I tried not so certain abt the answer

Answer:

Explanation:

Givens

Mass: 1450 kg

vi = 0

a = 5 m/s^2

vf=?

Formula

KE = 1/2 m v^2

d = vi * t + 1/2 a t^2

a = (vf - vi)/t

Solution

The final KE = 0 because the final velocity = 0 (The concrete wall saw to that).

Second formula

d = vi*t  + 1/2 a t^2

20  = 0 + 1/2 * 5 * t^2

40 = 5*t^2                         Divide by 5

40/5 = t^2

8 = t^2

2√2 = t

Third formula

5 = (vf - 0)/2√2                Multiply by 2√2

5* 2√2 = vf

10√2= vf

First formula

KE = 1/2 m * (10√2)^2

KE = 1/2 1450 * 100*2

KE = 1450 * 100              2 and 1/2 cancel out.

KE = 145000

This means that the change in KE is 145000 Joules or 145 kJ

Answer:

The tension in the rope if it is parallel to the slope is 24.31 N.

Explanation:

Given;

mass of the sled, m = 9 kg

angle of inclination of the slope, θ = 16⁰

The tension in the rope if it is parallel to the slope is calculated from the parallel component of the tension;

[tex]T_|_| = mgSin \theta\\\\T_|_| = 9 \times 9.8 \times sin(16^0)\\\\T_|_| = 24.31 \ N[/tex]

Therefore, the tension in the rope if it is parallel to the slope is 24.31 N.

Answer:

the force acting on the car is 3600 N

Explanation:

The computation of the force acting on the car is shown below:

As we know that

Force = mass × acceleration

= 1200 kg × 3.0 ms/^2

= 3600 N

hence, the force acting on the car is 3600 N

Answer:

21560 J

Explanation:

Work = mg*h = 1100*9.8*2 = 21560 J

21560 J work must be done to raise an 1100 kg car 2m above the ground.

Work = mass * gravity * height

= 1100 * 9.8 * 2

= 21560 J

In precis, work is carried out whilst pressure acts upon an item to purpose a displacement. 3 portions must be regarded in the way to calculate the quantity of work. The ones 3 portions are force, displacement, and the perspective between the pressure and the displacement.

Paints carried out are elaborated in this type of manner that includes both forces exerted on the body and the whole displacement of the body. This block is preceded by a steady force F. The purpose of this pressure is to move the body a certain distance d in an immediate route in the route of the pressure.

Learn more about work here: https://brainly.com/question/1382377

#SPJ2

Answer:

3525.19 kg

Explanation:

The computation of the mass of the car is shown below:

As we know that

Fc = m × V^2 ÷ R

m = Fc × R ÷ V^2

Provided that:

Fc = 34.652 kN = 34652 N

R = Radius = 24.98 m

V = speed = 15.67 m/s

So,

m = 34652 × 24.98 ÷ 15.67^2

 = 3525.19 kg

Answer:

Explanation:

Given the following data;

Velocity = 51 m/s

Mass = 7,692 kg

To find the momentum;

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

[tex] Momentum = mass * velocity [/tex]

Substituting into the equation, we have;

Momentum = 7692 × 51

Momentum = 392292 Kgm/s

Answer & Explanation:

The energy of a wave depends upon its amplitude, frequency and wave speed.

Assuming that both these waves are light waves, they have the same speed and amplitude can be assumed to be constant.

Wave 2 has a smaller wavelength and hence a higher frequency as c = λf, where

c is the speed of light,

λ is the wavelength and

f is the frequency of the wave.

Since wave 2 has a higher frequency, it has more energy.

Answer:

M1.5Iab-Ib

Explanation:

Answer:

a

Explanation:

I think its a its speed increases.....

Answer:

2.5 cm

Explanation:

Using the relation :

Refractive index = Real Depth / Apparent depth

Refractive index = 1.6

Real depth = 4cm

Virtual depth = apparent depth = x

1.6 = 4cm / x

1.6x = 4

x = 4 / 1.6

x = 2.5

Hence, virtual depth = 2.5cm

Answer:

The speed of this particle is constantly [tex]c[/tex].

Explanation:

Position vector of this particle at time [tex]t[/tex]:

[tex]\displaystyle \mathbf{r}(t) = b\, \cos(Q)\, \mathbf{i} + b\, \sin(Q) \, \mathbf{j} + c\, t\, \mathbf{k}[/tex].

Write [tex]\mathbf{r}(t)[/tex] as a column vector to distinguish between the components:

[tex]\mathbf{r}(t) = \begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}[/tex].

Both [tex]b[/tex] and [tex]Q[/tex] are constants. Therefore, [tex]b\, \cos(Q)[/tex] and [tex]b \sin (Q)[/tex] would also be constants with respect to [tex]t[/tex]. Hence, [tex]\displaystyle \frac{d}{dt}[b\, \cos(Q)] = 0[/tex] and [tex]\displaystyle \frac{d}{dt}[b\, \sin(Q)] = 0[/tex].

Differentiate [tex]\mathbf{r}(t)[/tex] (component-wise) with respect to time [tex]t[/tex] to find the velocity vector of this particle at time [tex]t\![/tex]:

[tex]\begin{aligned}\mathbf{v}(t) &= \frac{\rm d}{{\rm d} t} [\mathbf{r}(t)] \\ &=\frac{\rm d}{{\rm d} t} \left(\begin{bmatrix}b\, \cos(Q) \\ b\, \sin(Q) \\ c\, t\end{bmatrix}\right) \\ &= \begin{bmatrix}\displaystyle \frac{d}{dt}[b\, \cos(Q)] \\[0.5em] \displaystyle \frac{d}{dt}[b\, \sin(Q)]\\[0.5em]\displaystyle \frac{d}{dt}[c \cdot t]\end{bmatrix} = \begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\end{aligned}[/tex].

The speed [tex]v[/tex] (a scalar) of a particle is the magnitude of its velocity :

[tex]\begin{aligned}v(t) &= \| \mathbf{v}(t) \| \\ &= \left\|\begin{bmatrix}0 \\ 0 \\ c\end{bmatrix}\right\| \\ &= \sqrt{0^2 + 0^2 + c^2} = c\end{aligned}[/tex].

Therefore, the speed of this particle is constantly [tex]c[/tex] (a constant.)

Answer:

R = 1.2295 10⁵  m

Explanation:

After reading your problem they give us the diameter of the lens d = 4.50 cm = 0.0450 m, therefore if we use the Rayleigh criterion for the resolution in the diffraction phenomenon, we have that the minimum separation occurs in the first minimum of diffraction of one of the bodies m = 1 coincides with the central maximum of the other body

            θ = 1.22 λ / D

where the constant 1.22 leaves the resolution in polar coordinates and D is the lens aperture

how angles are measured in radians

          θ = y / R

where y is the separation of the two bodies (bulbs) y = 2 m and R the distance from the bulbs to the lens

            [tex]\frac{y}{R} = 1.22 \frac{ \lambda}{D}[/tex]

            R = [tex]\frac{ y \ D}{1.22 \lambda}[/tex]

let's calculate

            R = [tex]\frac{ 2 \ 0.045}{ 1.22 \ 600 \ 10^{-9}}[/tex]

            R = 1.2295 10⁵  m

Answer:

a control group

Explanation:

To have a succsessful experiment you need something to compare your variable to.

Answer:

68000

Explanation:

1 MHz = 1000 Hz

just multiply by 1000 :)

Answer:

T = 77.5 days.

Explanation:

       [tex]a = \frac{F}{m} = \frac{56e-3N}{474kg} = 1.18 e-4 N (1)[/tex]

       [tex]v_{f} = a* t (2)[/tex]

       [tex]t = \frac{v_{f}}{a} = \frac{790m/s}{1.18e-4m/s2} = 6694915 s (3)[/tex]

       [tex]t_{days} = \frac{t_{sec} }{1 day} = \frac{6694915s}{86400s} = 77.5 days (4)[/tex]

Answer:

Explanation:

A)

Density= charge/total volume .......eqn(1)

But volume= 4/3πr^3

r= radius= 20 cm= 0.20m

If we substitute into the volume equation, we have

volume= 4/3 * 3.142 *( 0.20)^3

= 0.0335 m^3

The volume= 0.0335 m^3

Charge=71 nC= 71×10^-9

If we substitute into eqn(1) we have

Density= (71 *10^-9C )/0.0335

= 2.11µc/m^3

B) charge enclose= Density × volume

spheres of radii are

5cm

10 cm

20 cm

Volume for 5cm

V= 4/3 * 3.142 *( 0.05)^3 = 0.0005237 m^3

charge enclose=2.11µc/m^3×0.0005237

charge enclose= 2.110 nC

Volume for 10cm

V= 4/3 * 3.142 *( 0.10)^3 = 0.004189 m^3

charge enclose= 2.11µc/m^3 ×0.004189

=8.9 nC

Volume for 20cm

V= 4/3 * 3.142 *( 0.20)^3 = 0.0335 m^3

charge enclose= 71nC

Answer:

3 and 3 and 3

Explanation:

I am sure Hope for brain list

Answer:

The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.

Answer:

A Polarizing sheet transmits only the component of light polarized along a particular direction and absorbs the component perpendicular to that direction.

Consider a light beam in the z direction incident on a Polaroid which has its transmission axis in the y direction. On the average, half of the incident light has its polarization axis in the y direction and half in the x direction. Thus half the intensity is transmitted,and the transmitted light is linearly polarized in the y direction.

Answer:

Explanation:

The magnitude of the force is known as the resultant.

R = √Fx²+Fy²

Fx = 80cos 20 + 40cos70

Fx = 80(0.9397)+40(0.3420)

Fx = 75.176 + 13.68

Fx = 88.856N

Fy = 80sin 20 + 40sin70

Fy = 80(0.3420)+40(0.9397)

Fy = 27.36 + 37.588

Fy = 64.948N

R = √88.586²+64.948²

R = √7,847.48+4,218.24

R = √12,065.72

R = 109.5

R = 110N

Hence the magnitude of the forces is 110N

Answer:

C. V1 < V2

Explanation:

The computation is shown below:

As we know that

Byoyancy force represent the displaced water weight

[tex]B = V_{in},_{w}P_{w}g[/tex]

[tex]V_{in},_{w}[/tex] denotes cork volume that inside the water

[tex]P_{w}[/tex] denotes water density

And, byoyancy force represent the displaced weight of corn syrup

[tex]B = V_{in},_{syr}P_{syr}g[/tex]

[tex]V_{in},_{syr}[/tex] denotes the cork volume that inside the water

[tex]P_{syr}[/tex] denotes syrup density

Now

[tex]P_{syr}>P_{w}\\\\V_{in},_{syr}<V_{in},_{wat}\\\\V_2>V_1 or V_1 <V_2[/tex]

Hence, the option c is correct

Answer:

Thomas Edisons most famous invention was the phonograph

Thomas Edison announces his invention of the phonograph, a way to record and play back sound. Edison stumbled on one of his great inventions—the phonograph—while working on a way to record telephone communication at his laboratory in Menlo Park, New Jersey.

Explanation:

Hope I helped