A force =(16.8N/s)t is applied to an object of m=45.0kg. Ignoring friction, how far does the object travel from rest in 5.00s?

Answer:

The ratio of gravitational force to electrical force is 3.19 x 10^-36

Explanation:

mass of an alpha particle = 6.64 x [tex]10^{-27}[/tex] kg

charge on an alpha particle = +2e = +2(1.6 x [tex]10^{-19}[/tex] C) = 3.2 x [tex]10^{-19}[/tex] C

distance between particles = d

For gravitational attraction:

The force of gravitational attraction F = [tex]\frac{Gm^{2} }{r^{2} }[/tex]

where G = gravitational constant = 6.67 x [tex]10^{-11}[/tex] m^3 kg^-1 s^-2

r = the distance between the particles = d

m = the mass of each particle

therefore, gravitational force = [tex]\frac{6.67*10^{-11}*(6.64*10^{-27} )^{2} }{d^{2} }[/tex] = [tex]\frac{2.94*10^{-63} }{d^{2} }[/tex]  Newton

For electrical repulsion:

Electrical force between the particles = [tex]\frac{-kQ^{2} }{r^{2} }[/tex]

where k is the Coulomb's constant = 9.0 x [tex]10^{9}[/tex] N•m^2/C^2

r = distance between the particles = d

Q = charge on each particle

therefore, electrical force = [tex]\frac{-9*10^{9}*(3.2*10^{-19} )^{2} }{d^{2} }[/tex] = [tex]\frac{-9.216*10^{-28} }{d^{2} }[/tex] Newton

the negative sign implies that there is a repulsion on the particles due to their like charges.

Ratio of the magnitude of gravitation to electrical force = [tex]\frac{2.94*10^{-63} }{9.216*10^{-28} }[/tex]

==> 3.19 x 10^-36

Answer:

13,750 N

Yes

Explanation:

Given:

v₀ = 90 km/h = 25 m/s

v = 0 m/s

t = 4 s

Find: a and Δx

a = Δv / Δt

a = (0 m/s − 25 m/s) / (4 s)

a = -6.25 m/s²

F = ma

F = (2200 kg) (-6.25 m/s²)

F = -13,750 N

Δx = ½ (v + v₀) t

Δx = ½ (0 m/s + 25 m/s) (4 s)

Δx = 50 m

Answer:

Do u have a picture of the graph?

Explanation:

I can solve it with refraction

Answer:

a=4m/s²

Explanation:

F=ma

0.4=0.1a

Answer:

a=4m/s

Explanation:

F=ma

0.4=0.1a

[tex] \frac{0.4}{0.1} = \frac{0.1}{0.1} [/tex]

a =4m/ s

Answer:

t = 0.31s

Explanation:

In order to calculate the time that the object takes to travel from the point with its maximum speed to the point with the maximum acceleration, you first use the following formulas, for the maximum speed and the maximum acceleration:

[tex]v_{max}=\omega A\\\\a_{max}=\omega^2A[/tex]

A: amplitude

v_max = 1.38m/s

a_max = 6.83m/s^2

w: angular frequency

From the previous equations you can obtain the angular frequency w.

You divide vmax and amax, and solve for w:

[tex]\frac{v_{max}}{a_{max}}=\frac{\omega A}{\omega^2 A}=\frac{1}{\omega}\\\\\omega=\frac{a_{max}}{v_{max}}=\frac{6.83m/s^2}{1.38m/s^2}=4.94\frac{rad}{s}[/tex]

Next, you take into account that the maximum speed is obtained when the object passes trough the equilibrium point, and the maximum acceleration for the maximum elongation, that is, the amplitude. In such a trajectory the time is T/4 being T the period.

You calculate the period  by using the information about the angular frequency:

[tex]T=\frac{2\pi}{\omega}=\frac{2\pi}{4.94rad/s}=1.26s[/tex]

Then the required time is:

[tex]t=\frac{T}{4}=\frac{1.26s}{4}=0.31s[/tex]

Answer:

please check attachment and follow the steps.

Answer:

v = 6t² − 2t^³/₂

s = 2t³ − ⅘t^⁵/₂ + 15

Explanation:

a = 12t − 3t^½

Integrate to find velocity.

v = ∫ a dt

v = ∫ (12t − 3t^½) dt

v = 6t² − 2t^³/₂ + C

Use initial condition to find C.

0 = 6(0)² − 2(0)^³/₂ + C

C = 0

v = 6t² − 2t^³/₂

Integrate to find position.

s = ∫ v dt

s = ∫ (6t² − 2t^³/₂) dt

s = 2t³ − ⅘t^⁵/₂ + C

Use initial condition to find C.

15 = 2(0)³ − ⅘(0)^⁵/₂ + C

15 = C

s = 2t³ − ⅘t^⁵/₂ + 15

Answer:

Explanation:

Kinetic energy at the height = 1/2 m v²

= 1/2 x 750 x 20²

= 150000 J

Its potential energy = mgh

= 750 x 9.8 x 5

=36750 J

Total energy = 186750 J

Its total kinetic energy will be equal to 186750 J , according to conservation of mechanical energy

If v be its velocity at the bottom

1/2 m v² = 186750

v = √498

= 22.31 m /s

Answer: Net electrostatic force on C is 24.2×[tex]10^{-2}[/tex] Newtons.

Explanation: Coulomb's Law is used to determine Electrostatic Force. Its formula is:

F = k.[tex]\frac{q_{0}.q_{1}}{r^{2}}[/tex]

where:

k is electrostatic constant (k = 8.987×[tex]10^{9}[/tex] Nm²/C²);

q is the charge of the object in Coulumb;

r is the distance between charges;

The net force is the sum of all the forces acting on C, so:

Force B on C:

They are both positive, so there is a relpusive force acting between them on the y-axis.

[tex]F_{BC} = 8,987.10^{9}.\frac{4.35.10^{-3}.9.67.10^{-4}}{(6.14.10^{2})^{2}}[/tex]

[tex]F_{BC} = 10.03.10^{-2}[/tex] N

Force D on C:

There is an atractive force between them on the x-axis.

[tex]F_{CD} = 8.987.10^{9}.\frac{9.67.10^{-4}.1.92.10^{-3}}{(1.42.10^{3})^{2}}[/tex]

[tex]F_{CD} = 13.64.10^{-4}[/tex] N

Force A on C:

First, find the distance between objects:

The distance is a diagonal line that divides the rectangle into a right triangle. Distance is square of the hypotenuse .

[tex]r^{2} = (6.14.10^2)^{2} + (1.42.10^{3})^{2}[/tex]

[tex]r^{2} = 37.72.10^{4}[/tex]

and hypotenuse: r = [tex]6.14.10^2[/tex]m

There is an atractive force between charges, but there are components of the force in x- and y-axis. So, because of that, force will be:

[tex]F_{CA} = F_{CA}[/tex].sinα + [tex]F_{CA}.[/tex]cosα

[tex]F_{CA} = 8.987.10^{9}.\frac{3.12.10^{-3}.9.67.10^{-4}}{37.72.10^{4}}[/tex]

[tex]F_{CA} = 7.2.10^{-2}[/tex]

The trigonometric relations is taken from the rectangle:

sinα = [tex]\frac{6.14.10^{2}}{6.14.10^{2}}[/tex]

cosα = [tex]\frac{1.42.10^{3}}{6.14.10^{2}}[/tex]

[tex]F_{CA}.[/tex]cosα = [tex]7.2.10^{-2}(\frac{1.42.10^{3}}{6.14.10^{2}})[/tex] = 0.17

[tex]F_{CA}.[/tex]sinα = [tex]7.2.10^{-2}.(\frac{6.14.10^{2}}{6.14.10^{2}} )[/tex] = 0.072

[tex]F_{CA} =[/tex] 0.17î + 0.072^j

Now, sum up all the terms in its respective axis:

X: [tex]13.64.10^{-4} + 0.17 =[/tex] 0.1714

Y: [tex]10.03.10^{-2} + 7.2.10^{-2}[/tex] = 0.1723

These forms another right triangle, whose hypotenuse is the net electrostatic force:

[tex]F_{net} = \sqrt{(0.1714)^{2} + (0.1723)^2}[/tex]

[tex]F_{net} = 24.3.10^{-2}[/tex] N

The net electrostatic force acting on C has magnitude [tex]F_{net} = 24.3.10^{-2}[/tex] N.

Answer:

16200 km/s

270 km/min

4.5 km/h

Explanation:

Acceleration Formula: Average Acceleration = Δv/Δt (change in velocity over change in time)

Simply plug in our known variables and solve:

a = (45.0 - 0)/10

a = 45.0/10

a = 4.5 km/h

Answer:

[tex]\boxed{\mathrm{4.5 \: kmph/s \: or \: 1.25 \: m/s^2 }}[/tex]

Explanation:

[tex]\displaystyle \mathrm{acceleration = \frac{change \: in \: velocity}{time \: taken}}[/tex]

[tex]\displaystyle \mathrm{a = \frac{v - u}{t}}[/tex]

[tex]\displaystyle \mathrm{v=final \: velocity}\\\displaystyle \mathrm{u=initial \: velocity}[/tex]

[tex]\displaystyle \mathrm{a = \frac{45- 0}{10}}[/tex]

[tex]\displaystyle \mathrm{a = \frac{45}{10}}[/tex]

[tex]\displaystyle \mathrm{a = 4.5}[/tex]

[tex]\mathrm{4.5 \: kmph/s = 1.25 \: m/s^2 }[/tex]

Answer:

The new voltage between the plates of the capacitor is 18 V

Explanation:

The charge on parallel plate capacitor is calculated as;

q = CV

Where;

V is the battery voltage

C is the capacitance of the capacitor, calculated as;

[tex]C = \frac{\epsilon _0A}{d} \\\\q =CV = (\frac{\epsilon _0A}{d})V = \frac{\epsilon _0A V}{d}[/tex]

[tex]q = \frac{\epsilon _0A V}{d}[/tex]

where;

ε₀ is permittivity of free space

A is the area of the capacitor

d is the space between the parallel plate capacitors

If only the space between the capacitors is doubled and every other parameter is kept constant, the new voltage will be calculated as;

[tex]q = \frac{\epsilon _0A V}{d} \\\\\frac{\epsilon _0A V}{d} = \frac{\epsilon _0A V}{d} \\\\\frac{V_1}{d_1} = \frac{V_2}{d_2} \\\\V_2 = \frac{V_1d_2}{d_1} \\\\(d_2 = 2d_1)\\\\V_2 = \frac{V_1*2d_1}{d_1} \\\\(V_1 = 9V)\\\\V_2 = \frac{9*2d_1}{d_1} \\\\V_2 = 9*2\\\\V_2 = 18 \ V[/tex]

Therefore, the new voltage between the plates of the capacitor is 18 V

Answer:

The speed is  [tex]v =10.27 *10^{7} \ m/s[/tex]

Explanation:

From the question we are told that

      The  voltage  is  [tex]V = 30 kV = 30*10^{3} V[/tex]

      The  initial velocity of the electron is  [tex]u = 0 \ m/s[/tex]

Generally according to the law of energy conservation

    Electric potential Energy  =  Kinetic energy of the electron

So  

      [tex]PE = KE[/tex]

Where  

      [tex]KE = \frac{1}{2} * m* v^2[/tex]

Here  m is the mass of the electron with a value of  [tex]m = 9.11 *10^{-31} \ kg[/tex]

     and  

         [tex]PE = e * V[/tex]

      Here  e is the charge on the electron with a value  [tex]e = 1.60 *10^{-19} \ C[/tex]

=>    [tex]e * V = \frac{1}{2} * m * v^2[/tex]

=>      [tex]v = \sqrt{ \frac{2 * e * V}{m} }[/tex]

substituting values  

           [tex]v = \sqrt{ \frac{2 * (1.60*10^{-19}) * 30*10^{3}}{9.11 *10^{-31}} }[/tex]

          [tex]v =10.27 *10^{7} \ m/s[/tex]

Answer:

The answer is "[tex]\bold{7.30 \times 10^{-6}}[/tex]"

Explanation:

length of the copper wire:

L= 62.9 cm

r is the radius of the loop then:

[tex]r=\frac{L}{2 \pi}\\[/tex]

  [tex]=\frac{62.9}{2\times 3.14}\\\\=\frac{62.9}{6.28}\\\\=10.01\\[/tex]

area of the loop Is:

[tex]A_L= \pi r^2[/tex]

     [tex]=100.2001\times 3.14\\\\=314.628[/tex]

change in magnetic field is:

[tex]=\frac{dB}{dt} \\\\ = 0.01\ \frac{T}{s}[/tex]

then the induced emf is:  [tex]e = A_L \times \frac{dB}{dt}[/tex]

                                              [tex]=314.628 \times 0.01\\\\=3.14\times 10^{-5}V[/tex]

resistivity of the copper wire is: [tex]\rho =[/tex]  1.69 × 10-8Ω·m

diameter d = 1.15mm

radius (r) = 0.5mm

               [tex]= 0.5 \times 10^{-3} \ m[/tex]

hence the resistance of the wire is:

[tex]R=\frac{\rho L}{\pi r^2}\\[/tex]

   [tex]=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times (0.5 \times 10^{-3})^2}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.5 \times 0.5 \times 10^{-6}}\\\\=\frac{1.69 \times 10^{-8}(62.9)}{3.14 \times 0.25 \times 10^{-6}}\\\\=135.41 \times 10^{-2}\\=1.35\times 10^{-4}\\[/tex]

Power:

[tex]P=\frac{e^2}{R}[/tex]

[tex]=\frac{3.14\times 10^{-5}\times 3.14\times 10^{-5}}{1.35 \times 10^{-4}}\\\\=7.30 \times 10^{-6}[/tex]

The final answer is: [tex]\boxed{7.30 \times 10^{-6} \ W}[/tex]

Answer:

Explanation:

Given;

initial angular displacement, θ₁ = 0 radians

final angular displacement, θ₂ = 4.5 radians

Angular velocity is calculated as;

[tex]\omega = \frac{\delta \theta}{\delta t}= \frac{4.5 -0}{18} \\\\\omega = 0.25 \ rad/s[/tex]

Angular acceleration is calculated as;

[tex]\alpha = \frac{\delta \omega}{\delta t} = \frac{\omega _f - \omega_i}{t_2 -t_1}[/tex]

where;

[tex]\omega_f[/tex] is the final angular velocity = 0 rad/s

[tex]\omega _i[/tex] is the initial angular velocity = 0.25 rad/s

t₂ is the final time of the motion = 3 seconds

t₁ is the initial time of the motion = 18 seconds

[tex]\alpha =\frac{\omega _f - \omega_i}{t_2 -t_1} \\\\\alpha = \frac{0 - 0.25}{3-18} \\\\\alpha = 0.017 \ rad/s^2[/tex]

Answer:

kinetic energy (K.E) = 5.28 ×10⁻¹⁷            

Explanation:

Given:

Mass of  α particle (m) = 6.50 × 10⁻²⁷ kg

Charge of  α particle (q) = 3.20 × 10⁻¹⁹ C

Potential difference ΔV = 165 V

Find:

kinetic energy (K.E)

Computation:

kinetic energy (K.E) = (ΔV)(q)

kinetic energy (K.E) = (165)(3.20×10⁻¹⁹)

kinetic energy (K.E) = 528 (10⁻¹⁹)

kinetic energy (K.E) = 5.28 ×10⁻¹⁷              

Answer:

a.   q2 = 16.4μC, positive charge

b.   F = 0.900N

c.   downward

Explanation:

a. In order to calculate the charge of the unknown charge you use the following formula, for the electric force between two charges:

[tex]F_e=k\frac{q_1q_2}{r^2}[/tex]            (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C^2

r: distance between the charges = 0.300m

q1: charge 1 = -0.550 μC = 0.550*10^-6C

q2: charge 2 = ?

Fe: electric force = 0.900N

The force exerted in the second charge points upward, then, the sign of the second charge is positive because this charge is getting closer to the first one.

You solve the equation (1) for the second charge ans replace the values of the other parameters:

[tex]q_2=\frac{r^2F_e}{kq_1}=\frac{(0.300m)^2(0.900N)}{(8.98*10^9Nm^2/C^2)(0.550*10^{-6}C)}\\\\q_2=1.64*10^{-5}C\\\\q_2=16.4*10^{-6}C=16.4*10\mu C[/tex]

The values of the second charge is 1.64 μC

b. By the third Newton Law, you have that the force exerted in the second charge is equal to the force exerted by the first charge on the second one.

The force exerted on the first charge is 0.900N

c. The charges are attracting between them, then, the force exerted on the first charge is pointing downward.

Answer:

Fx = 2.5 N

Fy = 5 N

|F| = 5.59 N

Explanation:

Given:-

- The mass of puck, m = 4.0 kg

- The initial velocity of puck, u = 3.00 i m/s

- The final velocity of puck, v = ( 8.00 i + 10.00 j ) m/s

- The time interval for the duration of force, Δt = 8 seconds

Find:-

the components of the force and (b) its magnitude.

Solution:-

- We will set up a coordinate system ( x - y ) plane. With unit vectors i and j along x and y axes respectively.

- To model the situation we will seek help from Newton's second law of motion. Defined by the rate of change of linear momentum of the system.

                                [tex]F_net = \frac{m*( v - u ) }{dt}[/tex]

Where,

                   Fnet: The net force that acts on the puck-rocket system

- Here we will assume that the mass of rocket is negligible compared to the mass of the puck. The only force ( F ) acting on the puck is due to the thrust produced of the rocket. The dry and air frictions are both neglected for the analysis.

- We will apply the newton's second law of motion in component forms. And determine the components of force F, as  ( Fx ) and ( Fy ) as follows:

                         [tex]F_x = \frac{m* ( v_x - u_x)}{dt} \\\\F_x = \frac{4* ( 8 - 3)}{8} \\\\F_x = 2.5 N\\\\F_y = \frac{m* ( v_y - u_y)}{dt} \\\\F_y = \frac{4* ( 10 - 0)}{8} \\\\F_y = 5 N\\\\[/tex]

- We will apply the Pythagorean theorem and determine the magnitude of the thrust force produced by the rocket with which the puck accelerated:

                          [tex]| F | = \sqrt{( F_x)^2 + ( F_y)^2} \\\\| F | = \sqrt{( 2.5)^2 + ( 5)^2} \\\\| F | = \sqrt{31.25} \\\\| F | = 5.590[/tex]            

Answer: the magnitude of the thrust force is F = 5.59 N

Answer:

6 N

Explanation:

From the laws of friction

F = ¶R = 0.3 × 20 = 6 N

The force of friction opposing the block's motion is 6 N.

The given parameters;

The force of friction which opposes the motion of the block is obtained by applying Newton's second law of motion.

F = ma

Fₓ = μF

Substitute the given parameters to calculate the frictional force on the object.

Fₓ = 0.3 x 20

Fₓ = 6 N

Thus, the force of friction opposing the block's motion is 6 N.

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When a net torque is applied to a rigid object, it always produces a change in angular velocity. (e.)

The word which best completes the given sentence, "When a net torque is applied to a rigid object, it always produces a_____" is:

According to the given question, we are asked to ask what would be produced when a net torque is applied to a rigid object based on the vector quantities involved.

As a result of this, we can see that there is a change in angular velocity when there is a net torque applied as the rigid object would change the rate at which it rotates.

Therefore, the correct answer is option E

Read more here:

https://brainly.com/question/15522344

Answer:

21,920 Pascals

Explanation:

P = ρgh

P = (1096 kg/m³) (10 m/s²) (2 m)

P = 21,920 Pa

Answer:

n = 1.6*10^9 capillaries

Explanation:

In order to calculate the number of capillaries, you take into account that the following relation must be accomplished:

[tex]A_1v_1=nA_2v_2[/tex]               (1)

A1: area of the aorta

v1: speed of the blood in the aorta = 40cm/s

n: number of capillaries = ?

A2: area of each capillary

v2: speed of the blood in each capillary

For the calculation of A1 and A2 you use the formula for the cross sectional area of a cylinder, that is, the area of a circle:

[tex]A=\pi r^2\\\\A_1=\pi r_1^2=\pi(1.0cm)^2=3.1415 cm^2\\\\A_2=\pi r_2^2=\pi (5*10^{-4}cm)^2=7.85*10^{-7}cm^2[/tex]

Where you have used the values of the radius for the aorta and the capillaries.

Next, you solve the equation (1) for n, and replace the values of all parameters:

[tex]n=\frac{A_1v_1}{A_2v_2}=\frac{(3.1415cm^2)(40cm/s)}{(7.85*10^{-7}cm^2)(0.10cm/s)}=1.6*10^9[/tex]

Then, the number of capillaries is 1.6*10^9

Answer:

The power dissipated by each bulb is  [tex]P = 10.0 \ W[/tex]

Explanation:

From the question we are told that

    The  power rating of both bulbs is  [tex]P = 40 \ W[/tex]

     The voltage  rating of both bulb is  [tex]V = 120 \ V[/tex]

     The  both bulbs are connected a voltage of  [tex]V_C = 120 V[/tex]

The amount of power rating of each  bulb is mathematically represented as

       [tex]P = \frac{V^2}{R }[/tex]

=>    [tex]R = \frac{V^2}{P}[/tex]

substituting values

       [tex]R = \frac{ (120)^2}{40}[/tex]

      [tex]R = 360 \Omega[/tex]

Now given that the bulbs are connected is  series, the equivalent resistance is  evaluated as

          [tex]R_{eq } = R +R[/tex]

substituting values

          [tex]R_{eq } = 360 + 360[/tex]

         [tex]R_{eq } =720 \ \Omega[/tex]

The  current flowing through the bulbs is mathematically evaluated as

         [tex]I =\frac{V_C}{R_{eq}}[/tex]

substituting values

      [tex]I =\frac{120}{720}[/tex]

      [tex]I = 0.1667 \ A[/tex]

Now  the power dissipated by both bulbs is mathematically represented as

           [tex]P = I ^2 * R[/tex]

substituting values      

         [tex]P = 0.1668^2 * 360[/tex]

         [tex]P = 10.0 \ W[/tex]

The power that should be dissipated by each bulb is P = 10.0 W.

Since

The power rating of both bulbs is P = 40 W.

The voltage rating of both bulbs is V = 120 V.

And, both bulks that should be connected a voltage of Vc = 120V

Now the amount of power that should be rated of each bulb should be

P = V^2/R

So, R = V^2/P

= 120^2/40

= 360Ω

The equivalent resistance should be

I = Vc/Req

= 120/720

= 0.1667 A

Now the power is = 0.1668^2 * 360

= 10.0 W

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Answer:

it is desired that the lenses stop this ray, its polarization must be vertical

Explanation:

To answer this exercise, let's analyze the rays of light reflected on a horizontal surface, when the incident light that we consider non-polarized is reflected on a surface, the electric field of light moves the electrons on the surface horizontally and this re-emits the radiation same shape, that is horizontal.

The other vertical direction the atoms have a lot of movement restricted by the attraction on the surface, so for the reflected ray this polarization is attenuated, this does not stop the transmitted ray where the two polarizations are transmitted.

Total polarizations only for one angle, but in general as we approach dominant polarization it horizontal. Specifically the angle for full polarization is

                        n = tan teaP

Now we can analyze what polarization the lenses have, if the ray that comes is polarized horizontally and it is desired that the lenses stop this ray, its polarization must be vertical

Answer:

735 N

Explanation:

If a woman has mass 75kg and u know that mass is constant everywhere then just apply the formula W=mg...as gravity on earth is 9.8 m/s2 , so her weight will be 735 N...hope it helps...

If a woman has a mass of 75 kilograms then her weight on the earth would be 735.75 Newtons, because the weight of the woman is the multiplication of the mass of the woman and the acceleration due to the gravity of the earth.

It can be defined as the force by which a body attracts another body toward its center as the result of the gravitational pull of one body and another,

As given in the problem if a woman has a mass of 75 kilograms we have to find the weight of the woman,

The weight of the woman = mass of the woman × acceleration due to the gravity of the earth

                                           = 75 kilograms × 9.81

                                           =735.75 Newtons

Thus, the weight of a woman who has a mass of 75 kilograms would be 735.75 Newtons

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#SPJ5

Answer:

 f = 6.66 cm

Explanation:

For this exercise we will use the constructor equation

         1 / f = 1 / p + 1 / q

where f is the focal length, p is the distance to the object and q is the distance to the image

the expression for magnification is

          m = h '/ h = - q / p

with this we have a system of two equations with two unknowns, in the problem they give us the distance to the image q = 10 cm and a magnification of m = -0.5

        -0.5 = - q / p

         p = - q / 0.5

         p = - 10 / 0.5

         p = 20 cm

now we can with the other equation look for the focal length

         1 / f = 1/20 + 1/10

         1 / f = 0.15

          f = 6.66 cm

Answer:

B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave

Answer:

B) The wavelength of both transverse and longitudinal waves is measured parallel to the direction of the travel of the wave.

Explanation: hope this helps ;)

Find the acceleration, a .

Force = Mass × Acceleration

We know that ,

Force = Mass × Acceleration

★ Substituting the values in the above formula,we get:

⇒ 480 = 40 × Acceleration

⇒ Acceleration, a = 480/40

⇒ Acceleration,a = 12 m/s

Thus,the acceleration of a body is 12 metres per seconds.

Answer:

the internal energy of the mixture at final state = 238kJ/kg

Explanation:

Given

V= 0.6m³

m=5kg

R=0.287kJ/kg.K

T=320 K

from ideal gas equation

PV = nRT

where P is pressure, V is volume, n is number of mole, R is ideal gas constant , T is the temperature.

Recall, mole = mass/molar mass

attached is calculation of the question.

Answer:

Object 2 has the larger drag coefficient

Explanation:

The drag force, D, is given by the equation:

[tex]D = 0.5 c \rho A v^2[/tex]

Object 1 has twice the diameter of object 2.

If [tex]d_2 = d[/tex]

[tex]d_1 = 2d[/tex]

Area of object 2, [tex]A_2 = \frac{\pi d^2 }{4}[/tex]

Area of object 1:

[tex]A_1 = \frac{\pi (2d)^2 }{4}\\A_1 = \pi d^2[/tex]

Since all other parameters are still the same except the drag coefficient:

For object 1:

[tex]D = 0.5 c_1 \rho A_1 v^2\\D = 0.5 c_1 \rho (\pi d^2) v^2[/tex]

For object 2:

[tex]D = 0.5 c_2 \rho A_2 v^2\\D = 0.5 c_2 \rho (\pi d^2/4) v^2[/tex]

Since the drag force for the two objects are the same:

[tex]0.5 c_1 \rho (\pi d^2) v^2 = 0.5 c_2 \rho (\pi d^2/4) v^2\\4c_1 = c_2[/tex]

Obviously from the equation above, c₂ is larger than c₁, this means that object 2 has the larger drag coefficient

Answer:

(a) The power wasted for 0.289 cm wire diameter is 15.93 W

(b) The power wasted for 0.417 cm wire diameter is 7.61 W

Explanation:

Given;

diameter of the wire, d = 0.289 cm = 0.00289 m

voltage of the wire, V = 120 V

Power drawn, P = 1850 W

The resistivity of the wire, ρ = 1.68 x 10⁻⁸ Ω⋅m

Area of the wire;

A = πd²/4

A = (π x 0.00289²) / 4

A = 6.561 x 10⁻⁶ m²

(a) At 26 m of this wire, the resistance of the is

R = ρL / A

R = (1.68 x 10⁻⁸  x 26) / 6.561 x 10⁻⁶

R = 0.067 Ω

Current in the wire is calculated as;

P = IV

I = P / V

I = 1850 / 120

I = 15.417 A

Power wasted = I²R

Power wasted = (15.417²)(0.067)

Power wasted = 15.93 W

(b) when a diameter of 0.417 cm is used instead;

d = 0.417 cm = 0.00417 m

A = πd²/4

A = (π x 0.00417²) / 4

A = 1.366 x 10⁻⁵ m²

Resistance of the wire at 26 m length of wire and  1.366 x 10⁻⁵ m² area;

R = ρL / A

R = (1.68 x 10⁻⁸  x 26) / 1.366 x 10⁻⁵

R = 0.032 Ω

Power wasted = I²R

Power wasted = (15.417²)(0.032)

Power wasted = 7.61 W