Find the weight of a man whose mass is 40 kg on earth.


(also
write complete data plus proper formula).

Answer:

    a = 5.53 g ,   a = -15g

Explanation:

This is an exercise in kinematics.

a) Let's look for the acceleration

         as part of rest v₀ = 0

          v = v₀ + a t

           a = v / t

           a = 282 / 5.2

          a = 54.23 m / s²

in relation to the acceleration of gravity

          a / g = 54.23 / 9.8

          a = 5.53 g

b) let's look at the acceleration to stop

         va = 0

         0 = v₀ -2 a y

         a = vi / y

         a = 282/2 1

         a = 141 m /s²

         a / G = 141 / 9.8

          a = -15g

Answer:

The velocity of the canoe relative to the river is 0.385 m/s, S37.26⁰W

Explanation:

Given;

velocity of the canoe relative to the earth, [tex]V_{r/e} = 0.33 \ m/s[/tex]

velocity of the river relative to the earth, [tex]V_{r/e} = 0.54 \ m/s[/tex]

The velocity of the canoe relative to the river is calculated as;

[tex]V_{(c/r)x} = V_{(c/e)x}- V_{(r/e)x} \ \ ----(1)\\\\V_{(c/r)y} = V_{(c/e)y}- V_{(r/e)y} \ \ ----(2)[/tex]

The x - component of the velocity of the canoe relative to the earth;

[tex]V_{(c/e)x} = 0.33 \times cos \ 45^0\\\\V_{(c/e)x} = 0.2333 \ m/s[/tex]

The y-component of the velocity of the canoe relative to the earth;

[tex]V_{(c/e)y} = 0.33 \times sin \ 45^0\\\\V_{(c/e)y} = 0.2333 \ m/s[/tex]

Note: velocity of the river relative to the earth has only x-component = 0.54 m/s

Apply equation (1) and (2) to calculate the velocity of the canoe relative to the river;

[tex]V_{(c/r)}x = 0.2333 - 0.54 = -0.3067 \ m/s\\\\V_{(c/r)}y = 0.2333 - 0 = 0.2333 \ m/s\\\\The \ resultant \ velocity;\\\\V_{c/r} = \sqrt{(-0.3067)^2 + (0.2333)^2} \\\\V_{c/r} = 0.385 \ ms/\\\\The \ direction:\\\\\theta = tan^{-1} (\frac{0.2333}{0.3067} ) = 37.26^0 \ south \ west \ of \ the \ river[/tex]

the answer is 396 joules :D

Answer:

As the number of free electrons increases, the resistance of this type of non-metallic material decreases with increasing temperature.

Explanation:

Answer:

Static friction

Explanation:

Answer:

Static friction

Explanation:

Answer:

the work required is 0.163 J

Explanation:

Given;

Energy applied to the spring, E = 2 J

initial length of the spring, x₀ = 32 cm

final length of the spring, x₁ = 46 cm

Extension of the spring, Δx = x₁ - x₀ = 46 cm - 32 cm = 14 cm = 0.14 m

The spring constant is calculated as follows;

E = ¹/₂kΔx²

[tex]k = \frac{2E}{(\Delta x)^2} = \frac{2\times 2}{(0.14)^2} = 204.1 \ N/m^2[/tex]

The extension of the spring when it is stretched from 37 cm + 41 cm:

Δx =  41 cm - 37 cm = 4 cm = 0.04 m

The work required:

W = ¹/₂kΔx²

W = ¹/₂ x (204.1) x (0.04)²

W = 0.163 J

Therefore, the work required is 0.163 J

Answer:Alternativa A.   Damos o nome de interferência a superposição de efeitos que ocorre ao ser produzido dois pulsos de onda, que serão propagados e acabarão inevitavelmente por se encontrar. No instante em que os pulsos se cruzarem, há então, uma superposição de efeitos individuais de cada um deles. Se durante o cruzamento, houver um reforço das ondas, estará ocorrendo a este fenômeno.

Answer:

(a) Radius = 4.6 x 10^11 m

(b) speed = 16.96 km/s

Explanation:

Time period, T = 5.4 earth years

mass of sun, M = 1.989 x 10^30 kg

(a) Let the orbital radius is R.

use the formula of period

[tex]T^2 = \frac{4 \pi^2 R^3}{G M}\\\\\left ( 5.4\times 365\times 24\times 3600 \right )^2 = \frac{4\times3.14\times 3.14\times R^3}{6.67\times10^{-11}\times 1.989\times 10^{30}}\\\\R = 4.6\times 10^{11} m[/tex]

(b) Let the speed is v.

[tex]v=\frac{2 \pi\times R}{T}\\\\v=\frac{2\times 3.14\times 4.6\times 10^{11}}{5.4\times 365\times 24\times 3600}\\\\v = 16963.6 m/s =16.96 km/s[/tex]

Answer:

9.09 s

Explanation:

If the sprinter ran the 100 meters at the constant speed of 11 m/s it would take him 9.09 s to cover the full distance.

We can find this number by dividing 100 meters (the distance covered) by 11 meters per second (the speed)

[tex]\frac{100}{11} =9.09[/tex]

product of force and perpendicular distance

Answer:

F' = F/16

Explanation:

The gravitational force between masses is given by :

[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]

If the distance between the center of two objects is quadrupled, r' = 4r

New force will be :

[tex]F'=G\dfrac{m_1m_2}{r'^2}\\\\F'=G\dfrac{m_1m_2}{(4r)^2}\\\\F'=\dfrac{Gm_1m_2}{16r^2}\\\\F'=\dfrac{1}{16}\times \dfrac{Gm_1m_2}{r^2}\\\\F'=\dfrac{F}{16}[/tex]

So, the new force will change by a factor of 16.

Answer:

A

Explanation:

weight of the tree =normal force

Horizontal force =coefficient of friction x Fnormal

0.5×430=215

Answer:

The answer is given as follows,

Explanation:

Gold large-sized weight 100 g < M < 250g

50 g < Magenta small-sized weight < 100g

100g < Blue medium-sized weight < 250g

Hence,

100g < Blue medium-sized weight < 250g

50 g < Magenta small-sized weight < 100g

100 g < Gold large-sized weight < 250g.

Answer:

because it is helpful to human beings I think

Answer: static electricity

Explanation:

When the plastic ruler is rubbed, friction opposes the motion and causes the transfer of electron from one surface to another such that plastic becomes negatively charged. When ruler is brought nearer to the paper, it induces the   positive charge in the piece of paper.

Answer:

W = 20 J

Explanation:

Given that,

Charge, q = 2 C

It is moves from a point at 118 volt to a point at 128 volt.

We need to find the work done in moving the charge,

[tex]W=\Delta VQ[/tex]

Put all the values,

[tex]W=(128-118)\times 2\\\\W=20\ J[/tex]

So, the work done in moving the charge is 20 J.

Answer: 1.85 C

Explanation:

Given

charges on the conductors are [tex]-1.8\ C[/tex] and [tex]5.5\ C[/tex]

They are connected by a conducting wire for a short period of time and then disconnected. During this time charge flow from the wire and net charge becomes [tex]5.5-1.8=3.7\ C[/tex]

This charge will be equally distribute among the two conductors i.e. 1.85 C on each conductor.

Explanation:

The logarithmic damping decrement of a mathematical pendulum is DeltaT=0.5. How will the amplitude of oscillations decrease during one full oscillation of the pendulum

Answer:

0 N

Explanation:

This is a trick question, the mass of the wrench would be 0 due to it being in space and has no gravitational pull to weight it down. And since acceleration is defined as the rate and change of velocity with no respect of time and the wrench is moving at a constant velocity, that means the velocity is 0. and since F = m*a it would be F = 0 * 0 = 0 N

The mechanical advantage is the ratio of the length of the effort arm to the length of the resistance arm. Option A is correct.

Mechanical advantage is a measure of the ratio of output force to input force in a system, it is used to obtain the efficiency of the given mechanical machine.

Mechanical advantage is a measure of how much a machine multiplies the input force.

The equation for the mechanical advantage of a lever is;

MA =length of effort arm/length of resistance arm

[tex]\rm MA=\frac{L_E}{L_R}[/tex]

The mechanical advantage is the ratio of the length of the effort arm to the length of the resistance arm.

Hence, option A is correct.

To learn more about the mechanical advantage, refer to the link;

https://brainly.com/question/7638820

#SPJ1

The mechanical advantage is the ratio of the length of the effort arm to the length of the resistance arm. Option A is correct.

What is the mechanical advantage?

Mechanical advantage is a measure of the ratio of output force to input force in a system, it is used to obtain the efficiency of the given mechanical machine.

Mechanical advantage is a measure of how much a machine multiplies the input force.

The equation for the mechanical advantage of a lever is;

MA =length of effort arm/length of resistance arm

Answer:

b)      v_y = 4.57 m / s

a)      vₓ = 4.43 m / s

Explanation:

This is an exercise in kinematics, where we assume that the acceleration is in the direction of the force and the initial body with zero velocity

          v = v₀ + a t

          v = 0 + a t

          v = 186  0.0342

          v = 6.36 m / s

let's use trigonometry to decompose this velocity

           sin 45.9 = v_y / v

           cos 45.9 = vₓ / v

           v_y = v sin 45.9

           vₓ = v cos 45.9

           v_y = 6.36 sin 45.9

           vₓ = 6.36 cos 45.9

           v_y = 4.57 m / s

           vₓ = 4.43 m / s

ₓ

Answer:

[tex]V_{a/c} = V_{a/b} + V_{b/c}[/tex]

Explanation:

The relative velocity of an object is the velocity of the object relative to the observer or frame of reference.

The velocity of particle "A" with respect to particle "B" is written as [tex]V_{A/B} = V_{A} - V_{B}[/tex]

From the given options, the second option is the correct answer.

[tex]V_{a/c} = V_{a/b} + V_{b/c}\\\\Re-arrange \ the \ above \ equation;\\\\V_{a/c} - V_{b/c}= V_{a/b}\\\\or\\\\V_{a/b}= V_{a/c} - V_{b/c}[/tex]

Answer:

25

Explanation:

5 theroical name indvivdual perspective asssssoitive each persp

True

Hope this helps! :)

Answer:

true because when trow the frisbee gives u level

Answer:

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Answer:

The number of photons in the pulse is 1.04 x 10²⁴

Explanation:

Given;

frequency of the emitted photons, f = 2.83 x 10¹¹ Hz

duration of the incident light, t = 3 ms = 3 x 10⁻³ s

power of the incident light, P = 65,000 W

The energy of each photon emitted is calculated as;

E = hf

where;

h is Planck's constant, = 6.626 x 10⁻³⁴ Js

E =  6.626 x 10⁻³⁴ x  2.83 x 10¹¹

E = 1.875 x 10⁻²² J

let the number of photons in the pulse = n

n(E)= Power x time

[tex]n = \frac{Pt}{E} \\\\n = \frac{65,000 \times 3\times 10^{-3}}{1.875 \times 10^{-22}} \\\\n = 1.04 \times 10^{24} \ photons[/tex]

Answer:

Answer: Speed is not a vector quantity. It has only magnitude and no direction and hence it is a scalar quantity.

Answer:

22.5 m

Explanation:

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

Initial velocity (u) = 30 m/s

Time (t) = 1.5 s

Final velocity (v) = 0 m/s

Distance (s) =?

The distance to which the car move before stopping from the time the driver applied the brake can be obtained as follow:

s = (u + v)t/2

s = (30 + 0)1.5 / 2

s = (30 × 1.5) / 2

s = 45 / 2

s = 22.5 m

Thus, the car will move to a distance of 22.5 m before stopping from the time the driver applied the brake.

Answer:

0.0928J

Explanation:

the pulling force of spring F=-kx

where x is the displacement from equilibrium position.

energy stored:

[tex]\int\limits^x_0 {-F} \, dx \\=\int\limits^x_0 {kx} \, dx \\\\=\frac{kx^{2} }{2}[/tex]

*** Its fine if you know nothing about calculus. Just apply the equation

    [tex]U=\frac{kx^{2} }{2}[/tex]

  where U is the potential energy of the spring***

  put x=0.150, [tex]U=\frac{8.25}{2}[/tex]×[tex]0.150^{2}[/tex] = 0.0928J (corr. to 3 sig. fig.)