A 5kg mass is pushed with a force of 10N for a distance of 2.5 meters. The work done is​

Answer:

78

Explanation:

x=xi+vi(t)+1/2a(t)^2

x=10+16(4)+1/2(0.50)(4)^2

x=74+4

x=78 m

This question involves the concepts of work done and the frictional force.

a. Work done by the person is "692.82 N".

b. Work done by the frictional force is "490.5 N".

a.

Work done by the person can be given by the following formula:

[tex]W=FdCos\theta[/tex]

where,

W = work done by the person = ?

F = Force applied by the person = 80 N

d = distance traveled = 10 m

θ = angle between force and motion = 30°

Therefore,

[tex]W=(80\ N)(10\ m)Cos30^o[/tex]

W = 692.82 N

b.

Work done by the frictional force is given by the following formula:

[tex]W_f=fd\\W_f=\mu mgd[/tex]

where,

[tex]W_f[/tex] = work done by the frictional force = ?

μ = coefficient of friction = 0.5

g = acceleration due to gravity = 9.81 m/s²

Therefore,

[tex]W_f=(0.5)(10\ kg)(9.81\ m/s^2)(10\ m)[/tex]

[tex]W_f=490.5\ N[/tex]

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Answer:

negatively charged particle

valence electron  is a electron of an atom, located in the outermost shell (valence shell ) of the atom, that can be transferred to or shared with another atom.

Hi there!

The formula for velocity given acceleration:

v = at

Plug in given values:

v = 6.4(7) = 44.8 m/s

Answer:

22Ω

Explanation:

if    V ⇒ voltage

      I ⇒ current

      R ⇒ resistance

V = IR

220 = 10 x R

220 / 10 = R

22 = R

Answer:

Explanation:

0.120 rev/s(2π rad/rev) = 0.24π rad/s

At the highest point of the arc, gravity must supply the required centripetal acceleration. As the normal force is 1/4 of his normal weight, then 3/4 of gravity acceleration must be used as centripetal acceleration

0.75g = ω²R

R = 0.75(9.81) / (0.24π)²

R = 12.942198...

R = 12.9 m

The radius of the circle is equal to 12.95m which is rotating with an angular velocity of 0.120 rev/s.

A body spins in a vertical circle so that its motion at different points is different from the motion of the body is said to be vertical circular motion.

The velocity and tension vary in maximum magnitude from the lowest to the highest position because of the effect of the gravitational force of the earth.

Given, the angular velocity of the Ferris wheel, ω = 0.120 rev/s

ω = 0.120 rev/s × 2π rad/rev

ω= 0.7536 rad/s

If r is the radius of the circle and 'm' is the mass of the jack.

From newton's second law of motion, the net force will be equal to

mg - N = mrω²

mg - (mg/4) = mrω²

r = 3g/4ω²

r = 3×9.81 / (4× 0.7536)

r = 12.95 m

Therefore, the radius of the circle in which the jack travels is equal to 12.95m.

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Your question is incomplete, most probably the complete question was,

Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.120 rev/s. As Jack passes through the highest point of his circular part, the upward force that the chair exerts on him is equal to one-fourth of his weight. What is the radius of the circle in which Jack travels?

Answer:D: the velocity is zero

Explanation:

Answer:

A

Explanation:

Down the hill, the net force increases if she pushes more forward.

Answer:

its a im just did the test

Explanation:

Answer:

currents which circulate in conductors like swirling eddies in a stream in electromagnetic series....

Answer:

340

Explanation:

Sorry I don't know how to do this one yet, I just found the answer in a textbook.

The angle that vector a makes with the +x axis is closest to 23.

The direction of a vector is represented tangent of angle equal to the ratio of the y component and the x component of the vector quantity.

tangent of angle = y/x

angle = tan⁻¹ (-2.3/5.3)

angle = 23.46°

Thus, the angle that vector makes with +x is 23.

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Answer:

Explanation:

s = s₀ + v₀t + ½at²

50 = 0 + 0(4) + ½a(4²)

50 = 8a

a = 50/8 = 6.25 m/s²

Answer:

I feel like its the second one but I'm not completely sure..

Explanation:

The average rate at which the neon atoms collide with one side of the container is [tex]3.31 \times 10^{26} \ atoms/s[/tex].

The given parameters;

The average kinetic energy of the gas molecules is calculated as follows;

[tex]K = \frac{3}{2} \frac{R}{N_a} T\\\\K = \frac{3 \times 8.314\times 298}{2\times 6.022 \times 10^{23}} \\\\K = 6.17\times 10^{-21} \ J/atoms[/tex]

The average speed of the gas molecules is calculated as follows;

[tex]K = \frac{1}{2}mv_{rms}^2\\\\v_{rms} = \sqrt{\frac{2K}{m} } \\\\v_{rms} = \sqrt{\frac{2\times 6.17 \times 10^{-21}}{3.35\times 10^{-26}} } \\\\v_{rms} = 607 \ m/s[/tex]

The time of collision of the gas molecules with the walls of the container is calculated as follows;

[tex]t = \frac{2d}{v} \\\\t = \frac{2\times 0.368}{607} \\\\t = 0.0012 \ s[/tex]

The average rate at which the gases collide with a single wall out of the 3 identical walls is calculated as follows;

[tex]rate =\frac{1}{3} \times \frac{n \times N_a}{t} \\\\rate = \frac{1.98 \times 6.02 \times 10^{23} \ atoms}{3 \times 0.0012 \ s} \\\\rate = 3.31 \times 10^{26} \ atoms/s[/tex]

Thus, the average rate at which the gases collide with one side of the container is [tex]3.31 \times 10^{26} \ atoms/s[/tex].

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Answer:

Explanation:

Radio waves

Radio waves have photons with the lowest energies. Microwaves have a little more energy than radio waves. Infrared has still more, followed by visible, ultraviolet, X-rays and gamma rays.

Answer:

The box arrives first.

Explanation:

Hope this helps!! :))

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According to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.

Friction may be defined as the resistance that is offered by the surfaces that are in contact when they move past each other. It is a type of force that opposes the motion of a solid object over another.

There are mainly four types of friction: static friction, sliding friction, rolling friction, and fluid friction. According to the context of this question, the sphere possesses less friction as compared to the box. This is because the box has an irregular surface that possesses high friction over the inclined surface.

Therefore, according to the information, a solid sphere is an object that arrives at the bottom first. This is because it occupies less friction as compared to the box.

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Explanation:

The range R of a projectile is given the equation

[tex]R = \dfrac{v_0^2}{g}\sin{2\theta}[/tex]

The maximum range is achieved when [tex]\theta = 45°[/tex] so our equation reduces to

[tex]R_{max} = \dfrac{v_0^2}{g}[/tex]

We can solve for the initial velocity [tex]v_0[/tex] as follows:

[tex]v_0^2 = gR_{max} \Rightarrow v_0 = \sqrt{gR_{max}}[/tex]

or

[tex]v_0 = \sqrt{(9.8\:\text{m/s}^2)(9.5×10^6\:\text{m})}[/tex]

[tex]\:\:\:\:\:\:\:=9.6×10^3\:\text{m/s}[/tex]

To find the maximum altitude H reached by the missile, we can use the equation

[tex]v_y^2 = v_{0y}^2 - 2gy = (v_0\sin{45°})^2 - 2gy[/tex]

At its maximum height H, [tex]v_y = 0[/tex] so we can write

[tex]0 = (v_0\sin{45°})^2 - 2gH[/tex]

or

[tex]H = \dfrac{(v_0\sin{45°})^2}{2g}[/tex]

[tex]\:\:\:\:\:\:= \dfrac{[(9.6×10^3\:\text{m/s})\sin{45°}]^2}{2(9.8\:\text{m/s}^2)}[/tex]

[tex]\:\:\:\:\:\:= 2.4×10^6\:\text{m}[/tex]

Answer:

everything will float up and go up to space and die

Explanation:

gravity keeps us down and once it stops everything will float up. And if it were to stop everywhere everything and everyone will die and everything will be destroyed.

Answer:it’s C

Explanation:

by using graph of velocity vs. time for two objects, Item 4 and Item 5 statement makes an accurate comparison of the motions for objects. thus option C is correct.

velocity is the  rate of change of the position of the object with respect to  reference and it is complicated but velocity is basically speeding a particular object in a specific direction.

Velocity is a vector quantity which means both magnitude (speed) and direction are combinedly define define velocity. The SI unit of velocity is meter per second (ms-1) and the magnitude or the direction of velocity of a body changes leads to acceleration.

Speed and velocity are the two closest term but the major difference between speed and velocity is that speed gives us an idea that the object with the faster rate of movement r where as  velocity  speed up as well as  tells us the direction of the body

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The distance traveled by the body in the given time is 1,250 m.

The given parameters;

The distance traveled by the body in the given time is calculated as follows;

[tex]s =ut + \frac{1}{2} at^2\\\\s = 0 \ + \ \frac{1}{2} (4)(25^2)\\\\s =1,250 \ m[/tex]

Thus, the distance traveled by the body in the given time is 1,250 m.

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Answer:

You can jump 1.5 feet on Earth.

Explanation:

Because the moons gravity is weaker that earth so it would be easier to jump further on the moon.

Answer:

Thermal Conductivity Easily Transmits Heat Among Fine Ceramics

Answer:

D. Tungsten

Explanation:

W - Wolfram

Answer:

b Shiny

Explanation:

Trust me it's right

Answer:

Pick c is the right one

Explanation:

5.0m/s

Explanation:

The artificial satellite experiences a centripetal force [tex]F_c[/tex] as it moves around the earth and it is defined as

[tex]F_c = m\dfrac{v^2}{r} = m\left(\dfrac{2\pi r}{T}\right)^2\left(\dfrac{1}{r}\right) = \dfrac{4\pi^2mr}{T^2}[/tex]

where m is the mass of the satellite, r is its orbital radius and T is its orbital period. But we need to find the radius first.

Recall that the satellite is orbiting at a height where its acceleration due to gravity is 6.44 m/s^2. Since we know that the weight mg of the satellite is equal to the gravitational force [tex]F_G[/tex] between the earth and the satellite, we can write

[tex]mg = F_G = G\dfrac{mM}{r^2}[/tex]

[tex]\Rightarrow g = G\dfrac{M}{r^2}[/tex]

where M is the mass of the earth (=[tex]5.972×10^{24}\:\text{kg}[/tex]) and G is the universal gravitational constant (=[tex]6.674×10^{-11}\:\text{N-m}^2\text{/kg}[/tex]). Plugging in the values, we find that the radius of the satellite's orbit is

[tex]r = \sqrt{\dfrac{GM}{g}} = \sqrt{\dfrac{(6.674×10^{-11}\:\text{N-m}^2\text{/kg})(5.972×10^{24}\:\text{kg})}{6.44\:\text{m/s}^2}}[/tex]

[tex]\:\:\:\:\:= 7.87×10^6\:\text{m}[/tex]

Now that we have the value for the radius, we can now calculate the orbital period T. Recall that the centripetal force is equal to the weight of the satellite at its orbital radius. Therefore,

[tex]F_c = mg \Rightarrow \dfrac{4\pi^2mr}{T^2} = mg[/tex]

or

[tex]4\pi^2r = gT^2[/tex]

Solving for T, we get

[tex]T^2 = \dfrac{4\pi^2r}{g} \Rightarrow T = \sqrt{\dfrac{4\pi^2r}{g}}[/tex]

We can further simplify the above expression into

[tex]T = 2\pi\sqrt{\dfrac{r}{g}}[/tex]

Plugging in the values for r and g, we get

[tex]T = 2\pi\sqrt{\dfrac{(7.87×10^6\:\text{m})}{(6.44\:\text{m/s}^2)}}[/tex]

[tex]\:\:\:\:\:= 6945\:\text{s} = 1.93\:\text{hrs}[/tex]