Which refers to an object’s resistance to any change in its motion? force acceleration gravity inertia

Answer:

(a) 2.33 A

(b) 15.075 V

Explanation:

From the question,

The total resistance (Rt) = R1+R2 = 3.85+6.47

R(t) = 10.32 ohms.

Applying ohm's law,

V = IR(t)..........equation 1

Where V = Emf of the battery, I = current flowing through the circuit, R(t) = combined resistance of both resistors.

Note: Since both resistors are connected in series, the current flowing through them is the same.

Therefore,

I = V/R(t)............. Equation 2

Given: V = 24 V, R(t) = 10.32 ohms

Substitute these values into equation 2

I = 24/10.32

I = 2.33 A.

Hence the current through R1 = 2.33 A.

V2 = IR2.............. Equation 3

V2 = 2.33(6.47)

V2 = 15.075 V

The current is 2.33 A

And, the potential difference is 15.075 V

Since Two resistors, R1=3.85 Ω and R2=6.47 Ω

So,

The total resistance (Rt) is

= R1+R2

= 3.85+6.47

= 10.32 ohms.

Now here we applied ohm law

V = IR(t)..........equation 1

Here V = Emf of the battery,

I = current

R(t) = combined resistance

Now

I = V/R(t)............. Equation 2

Here

Given: V = 24 V, R(t) = 10.32 ohms

So, I should be

I = 24/10.32

I = 2.33 A.

Now the potential difference is

V2 = IR2.............. Equation 3

V2 = 2.33(6.47)

V2 = 15.075 V

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Answer:

please check attachment and follow the steps.

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:

Gabe is wrong in concluding that the polygraph tests are infallible. Even though the polygraph machine is a man-made machine, there is a small margin of error factored into it.

This error could be caused as a result of the anxiety of the test taker, the heart beat rate, the blood pressure etc which would lead to the answers provided by the person to be flagged as false despite being true.

In most cases, the test result will be declared inconclusive if there was an error in the readings.

Explanation:

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:

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:

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: 2.13 × 10^-4 A

Explanation:

Given that the RLC circuit has a resistance R = 200 Ω and an inductance L = 15 mH.

Its oscillation frequency F = 7000 Hz

The initial current I = 25 mA = 25/1000 or 25 × 10^-3 A

Since there is no charge on the capacitor, the current after complete 5 circle will be achieved by using the formula in the attached file.

Please find the attached file for the remaining explanation for the solution.

The current in the RLC circuit after five (5) complete cycles is equal to [tex]2.15 \times 10^{-4} \; Amperes[/tex]

Given the following data:

To determine the current in the RLC circuit after five (5) complete cycles, we would use the following formula:

Note: There is no electrical charge on the capacitor.

After five (5) complete cycles, the formula becomes:

Substituting the given parameters into the formula, we have;

[tex]I_5 = 0.025 e^{\frac{-200}{2\; \times \;0.015} \times \frac{5}{7000} }\\\\I_5 = 0.025 e^{\frac{-200}{0.03} \times \frac{5}{7000} }\\\\I_5 = 0.025 e^{\frac{-1000}{210} }\\\\I_5 = 0.025 e^{-4.7619}\\\\I_5 = 0.025 \times 0.0086\\\\I_5 = 0.00215 \\\\I_5 = 2.15 \times 10^{-3} \; Amps[/tex]

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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:

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:

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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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

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: 529.9 Hz

Explanation:

Here we need to use the Doppler equation, so we have:

f' = f*(v + v0)/(v - vs)

Here, f is the frequency = 500Hz

v is the velocity of the wave, = 334m/s

v0 is the velocity of the observer = 20m/s

vs is the velocity of the source = 0m/s

Then we have:

f' = 500Hz*(334m/s + 20m/s)/(334m/s) = 529.9 Hz

Answer:

Taurus constellation

Explanation:

Because Taurus constellation is mostly seen In the Northern Hemisphere, the bull passes through the sky from November to March, but the constellation's at its most visible in January. Taurus covers 797 square degrees.

Right ascension: 4 hours

Declination: 15 degrees

Best visible between latitudes 90 degrees and minus 65 degrees

Answer:

Do u have a picture of the graph?

Explanation:

I can solve it with refraction

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:

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:

[tex]F = 4.3671 * 10^{-8}\ Newtons[/tex]

Explanation:

The gravitational force between two corpses is given by the following equation:

[tex]F = GMm/d^2[/tex]

Where F is the force, G is the gravitational constant

([tex]G = 6.67408*10^{-11}\ m^3kg^{-1}s^{-2}[/tex]), M and m are the masses of the corpses and d is the distance between them.

So we have that:

[tex]F = 6.67408*10^{-11} * 7.1*10^4 * 73/89^2[/tex]

[tex]F = 4.3671 * 10^{-8}\ Newtons[/tex]

Answer:

The length of the rod for the condition on the question to be met is [tex]L = 1.5077 \ m[/tex]

Explanation:

The  Diagram for this  question is  gotten from the first uploaded image  

From the question we are told that

          The mass of the rod is [tex]M = 3.41 \ kg[/tex]

           The mass of each small bodies is  [tex]m = 0.249 \ kg[/tex]

           The moment of inertia of the three-body system with respect to the described axis is   [tex]I = 0.929 \ kg \cdot m^2[/tex]

             The length of the rod is  L  

     Generally the moment of inertia of this three-body system with respect to the described axis can be mathematically represented as

        [tex]I = I_r + 2 I_m[/tex]

Where  [tex]I_r[/tex] is the moment of inertia of the rod about the describe axis which is mathematically represented as  

        [tex]I_r = \frac{ML^2 }{12}[/tex]

And   [tex]I_m[/tex] the  moment of inertia of the two small bodies which (from the diagram can be assumed as two small spheres) can be mathematically represented  as

           [tex]I_m = m * [\frac{L} {2} ]^2 = m* \frac{L^2}{4}[/tex]

Thus  [tex]2 * I_m = 2 * m \frac{L^2}{4} = m * \frac{L^2}{2}[/tex]

Hence

       [tex]I = M * \frac{L^2}{12} + m * \frac{L^2}{2}[/tex]

=>   [tex]I = [\frac{M}{12} + \frac{m}{2}] L^2[/tex]

substituting vales  we have  

        [tex]0.929 = [\frac{3.41}{12} + \frac{0.249}{2}] L^2[/tex]

       [tex]L = \sqrt{\frac{0.929}{0.40867} }[/tex]

      [tex]L = 1.5077 \ m[/tex]

Answer:

21,920 Pascals

Explanation:

P = ρgh

P = (1096 kg/m³) (10 m/s²) (2 m)

P = 21,920 Pa

Answer:

It is likely that vertically developed cumuliform clouds would begin to form

Explanation:

This is because since the air is conditionally unstable below 3000mand air were to be forced to rise to a point of saturation within this particular layer of the atmosphere

Explanation:

A metal such as copper is a conductor because it provides a pathway for electric charges to move easily. A material such as rubber is an insulator because it resists the flow of electric charges. A material that partially conducts electric current is a semiconductor. These materials include group 3 and group 5 elements.

Answer:

conductor

insulator

resists

semiconductor

group 3 and group 5

Explanation:

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: 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.

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:

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 ;)

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:

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