Two blocks, one of mass 5 kg and the other of mass 2 kg, are attached to opposite ends of a light string and hung vertically from a massless, frictionless pulley. Initially the heavier block is held in place a distance 2.5 m above the floor, the lighter block is just touching the floor, and the cord is taut. Then the heavier block is released and comes crashing to the floor while the cord slackens and the lighter block continues to rise. What is the maximum height reached by the lighter block

Answer:

A vector can be written as:

(R, θ)

Where R is the magnitude, in this case, we know that the magnitude of the displacement is 5m

Then:

R = 5m

and θ defines the direction, it's an angle measured from the positive x-axis.

(In the image, θ would be the angle located at the point A)

Now, if you look at the image, you can see a triangle rectangle.

Where the adjacent cathetus has a length of 4,

the opposite cathetus has a length of 3 units

the hypotenuse has a length of 5 units.

So we can use any trigonometric rule to find the value of θ, like:

sin(θ) = (opposite cathetus)/hypotenuse

Then:

sin(θ) = 3m/5m

Now we can use the inverse sin function, Asin(x), in both sides

Asin( sin(θ)) = θ = Asin( 3/5) = 36.87°

then the vector is:

(5m, 36.87°)

Now, if we define the positive y-axis as the North, and the positive x-axis as the East.

This vector would point at 36.87° North of East.

(or almost Northeast)

Answer:

[tex]W=7.56\times 10^{-19}\ J[/tex]

Explanation:

Given that,

The work function for silver is 4.73 eV.

We need to find the value of the work function from electron volts to joules.

We know that,

[tex]1\ eV=1.6\times 10^{-19}\ J[/tex]

For 4.73 eV,

[tex]4.73\ eV=1.6\times 10^{-19}\times 4.73\\\\=7.56\times 10^{-19}\ J[/tex]

So, the work function for silver is [tex]7.56\times 10^{-19}\ J[/tex].

Answer:

Because Moon and Mars has no atmosphere.

Explanation:

Moon and Mars has no atmosphere, so there is no friction on the falling object due to the atmosphere. The speed of the falling object is more at Moon and Mars.

When a small object impact on the surface of moon or Mars with high speed, the size of crater is large than the earth as out earth has atmosphere.

Complete question is;

Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m each. If Margy is walking at 1.4 m/s and accelerates at 0.20 m/s² during one of the running portions, what is her final velocity at the end of the 100.0 m? Round your answer to the nearest tenth.

Answer:

6.5 m/s

Explanation:

We are told that she is walking at 1.4 m/s and accelerates at 0.20 m/s².

Thus;

Initial velocity; u = 1.4 m/s

Acceleration; a = 0.2 m/s²

Distance; s = 100 m

From Newton's equation of motion, we know that;

v² = u² + 2as

Where v is final velocity.

Thus;

v² = 1.4² + 2(0.2 × 100)

v² = 41.96

v = √41.96

v ≈ 6.5 m/s

Answer:

Explanation:

Mass doesn't matter here because when something is falling, gravity plays fairly; an elephant falls at the same rate of acceleration as does a feather. What DOES matter is everything pertinent to the y-dimension of free-fall:

a = -9.8 m/s/s

v₀ = 0 (since the ball was held before it was dropped)

v = ??

Δx = -8 m (negative because the ball drops this far below the point from which it was released).

Putting all this together in one equation:

v² = v₀² + 2aΔx and filling in this equation:

v² = (0)² + 2(-9.8)(-8) and

v² = 156.8 so

v = 12.5 which rounds to 13 if you're using 2 sig figs, and rounds to 10 if you're only using 1 (which you should be, according to the way the numbers have been given in this problem)

Answer 60 NEWTON

Explanation:

FORCE = MASS * acceleration

acceleration= VELOCITY / TIME

acceleration= 30 / 10 = 3   M/S2

FORCE = MASS * acceleration

FORCE = 20 *3 = 60 NEWTON

Answer:

Explanation:

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Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

[tex]F_{g0}=G \frac{mM_{E0}}{r^{2}}[/tex]

So if the mass of the earth is increased by a factor of 2, this means that:

[tex]M_{Ef}=2M_{E0}[/tex]

so:

[tex]F_{gf}=G \frac{2mM_{E0}}{r^{2}}[/tex]

Therefore:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]

When simplifying we end up with:

[tex]\frac{F_{gf}}{F_{g0}}=2[/tex]

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

[tex]M_{Ef}=3M_{E0}[/tex]

so:

[tex]F_{gf}=G \frac{3mM_{E0}}{r^{2}}[/tex]

Therefore:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]

When simplifying we end up with:

[tex]\frac{F_{gf}}{F_{g0}}=3[/tex]

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

[tex]M_{Ef}=\frac{M_{E0}}{4}[/tex]

so:

[tex]F_{gf}=G \frac{mM_{E0}}{4r^{2}}[/tex]

Therefore:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]

When simplifying we end up with:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{1}{4}[/tex]

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

Answer:

It was proposed by Isaac Newton

Explanation:

The law of universal attraction of expression

          F = [tex]G \ \frac{m_1m_2}{ r^2}[/tex]G m1m2 / r ^ 2

where G is a constant, m₁ and m₂ are the masses of the bodies and r the distance between them.

It was proposed by Isaac Newton

With this law Newton explained that the force that pulls the moon towards the earth is the same as that which attracts an apple towards the earth

Answer:

The mean is 79.[tex]\overline {54}[/tex]

The median is 80 - 100

The mode is 80 - 100

Explanation:

The given table is presented as follows;

[tex]\begin{array}{lcrc}Classes&Mid \ point &Frequency &Frequency \times Mid \ point\\0 - 20&10& 2&20\\20-40&30&2&60\\40-60&50&3&150\\60-80&70&12&840\\80-100&90&18&1620\\100-120&110&5&550\\120-140&130&2&260\end{array}[/tex]

The mean of a class of values, [tex]\overline x[/tex] = ∑(Frequency × Midpoint)/∑(Frequency)

Therefore, we get;

[tex]\overline x[/tex] = (20+60+150+840+1620+550+260)/(2+2+3+12+18+5+2) = 79.[tex]\overline {54}[/tex]

The mean, [tex]\overline x[/tex] =79.[tex]\overline {54}[/tex]

The median class = The middle value lass = The class at the 22 nd value = 80 - 100

The median = 80 - 100

The modal class = The class with the highest frequency = 80 - 100

The mode = 80 - 100

Answer:

b. 12.5 mAs, 70 kVp

Explanation:

The given parameter are;

The initial exposure factors := 10 mAs and 70 kVp

The initial Grid Ratio, G.R.₁ = 8:1

The Grid Ratio with which the radiographer desires to increase the scatter absorption, G.R.₂  = 12:1

Given that the lead content in the 12:1 grid, is higher than the lead content in 8:1 grid and that 12:1 grid needs more mAs to compensate, and provides a higher image contrast, the amount of extra mAs is given by the Grid Conversion Factors, GCF, as follows;

The GCF for G.R. 8:1 = 4

The GCF for G.R. 12:1 = 5

Therefore, given that the mAs used by the radiographer for 8:1 Grid Ratio is 10 mAs, the mAs required for a G.R. of 12:1 in order to maintain the same exposure is given as follows;

mAs for G.R. of 12:1 = 10 mAs × 5/4 = 12.5 mAs

Therefore the new exposure factors are;

12.5 mAs, 70 kVp

Explanation:

distance travelled = average speed x time

                               =30m/s*100s

                                =3000m

Answer:

3000m

Explanation:

30m/s*100s

3000m

Answer:

the mass of the body is 0.02 kg.

Explanation:

Given;

relative density of the oil, [tex]\gamma _0[/tex] = 0.875

mass of the object in oil, [tex]M_o[/tex] = 0.013 kg

mass of the object in water, [tex]M_w[/tex] = 0.012 kg

let the mass of the object in air = [tex]M_a[/tex]

weight of the oil, [tex]W_0 = M_a - 0.013[/tex]

weight of the water, [tex]W_w = M_a - 0.012[/tex]

The relative density of the oil is given as;

[tex]\gamma_0 = \frac{density \ of \ oil }{density \ of \ water} = \frac{W_0}{W_w} = \frac{M_a -0.013}{M_a -0.012} \\\\0.875 = \frac{M_a -0.013}{M_a -0.012}\\\\0.875(M_a - 0.012) = M_a - 0.013\\\\0.875M_a - 0.0105 = M_a -0.013\\\\0.875M_a - M_a = 0.0105 - 0.013\\\\-0.125 M_a = -0.0025\\\\M_a = \frac{0.0025}{0.125} \\\\M_a = 0.02 \ kg[/tex]

Therefore, the mass of the body is 0.02 kg.

Answer:

True

Explanation:

Formula for speed is;

Speed = distance/time

Now, if an object covers an equal distance in equal time intervals, it means the speed will remain the same.

For example if an object covers 3 m every 1 second it means speed will always be; 3/1 = 3 m/s.

Thus the statement is correct.

Answer:

dependent: the outcome of the experience

independent variable: everything literaly.

Independent is where you change some variables and see the result

Dependent is literaly the result, or the outcome dependent on the exprience.

Explanation:

I got u.

Answer:

  [tex]M = \left[\begin{array}{ccc}cos \ 60&0\\0&-sin \ 60\end{array}\right][/tex]

Explanation:

To find the matrix, let's decompose the vectors, the rotated angle is (-60C) for the prime system

          x ’= x cos (-60)

          y ’= y sin (-60)

we use

          cos 60 = cos (-60)

          sin 60 = - sin (-60)

we substitute

          x ’= x cos 60

          y ’= - y sin 60

the transformation system is

         [tex]\left[\begin{array}{ccc}x'\\y'\end{array}\right] = \left[\begin{array}{ccc}cos 60&0\\0&-sin60\end{array}\right] \ \left[\begin{array}{ccc}x\\y\end{array}\right][/tex]x '

the transformation matrix is

       [tex]M = \left[\begin{array}{ccc}cos \ 60&0\\0&-sin \ 60\end{array}\right][/tex]

Answer:

Acceleration is 1.2 m/s^2.

Explanation:

initial velocity, u = 0

distance, d = 60 m

time, t = 10 s  

Let the acceleration is a.

use second equation of motion

[tex]s= u t +0.5 at^2\\\\60 = 0 + 0.5 \times a \times 10\times 10\\\\a = 1.2 m/s^2[/tex]

Now according to the Newton's second law

Force = mass x acceleration

Let the mass is m.

F = m x 1.2 = 1.2 m Newton  

Scalar quantity are physical quantities that have just magnitude, not direction.

Answer:

For smooth surface:The light rays bounce off the table and all move in the same direction.

Answer:

F = 1094.4 N

Explanation:

From impulse - momentum theorem, we now that ;

Impulse = momentum

Where;

Formula for impulse = force (F) × time(t)

Momentum = mass(m) × velocity(v)

Now, we are given;

Mass of swimmer; m = 72 kg

Speed; v = 3.8 m/s

Time; t = 0.25 s

Thus;

F × t = mv

F = mv/t

F = (72 × 3.8)/0.25

F = 1094.4 N

This value of force is the magnitude of the average horizontal force by diver on the raft.

Answer:

a)   t = 20 s,  b)  x = 1000 m, As the runway is only 800 m long, the plane cannot land at this distance

Explanation:

This is a kinematics exercise

a) in minimum time to stop,

           v = vo + at

           v = 0

           t = -v0 / a

we calculate

          t = -100 / (5.00)

          t = 20 s

b) Let's find the length you need to stop

          v² = vo² + 2 a x

          x = -v0 ^ 2 / 2a

          x = - 100² / 2 (-5.00)

          x = 1000 m

As the runway is only 800 m long, the plane cannot land at this distance.

Answer:

I = 0.2 A

Explanation:

Lamp is rated at 300 mA

I_lamp = 0.3 A

Voltage is; V = 3V

Thus; Resistance is given by;

R = V/I

R = 3/0.3

R = 10 ohms

Now, since the ammeter of 5 ohms is connected in series with the lamp. Thus equivalent resistance;

R_eq = 10 + 5

R_eq = 15 ohms

Ammeter current will be;

I = V/R_eq

I = 3/15

I = 0.2 A

Answer:

b

Explanation:

bc

400 m2

Explanation:

Pressure = Force ÷ Area

5 N/m2 = 2000 N ÷ A

A = 2000 N ÷ 5

= 400 m2

If a force of 2000N is applied, the area that the force is applied to is 400 m²

The word "force" has a specific meaning in science. At this level, calling a force a push or a pull is entirely appropriate. A force is not something an object "has in it" or that it "contains."

One thing experiences a force from another. There are both living things and non-living objects in the concept of a force.

The amount of force applied to a certain region is referred to as pressure. The force per unit area is called pressure. F in this condensed version of the equation stands in for the force, which is expressed in newtons.

Given that the pressure of 5 N/m²

Force is 2000N

Pressure = Force ÷ Area

5 N/m² = 2000 N ÷ A

A = 2000 N ÷ 5 = 400 m²

Therefore, the area that the force is applied to is 400 m².

To learn more about force, refer to the link:

https://brainly.com/question/19529052

#SPJ2

Answer:

C) Potential energy due to intermolecular forces.

Answer:

a) 98.1 Joules

b) 49.05 N × sin(θ)

c) 9.81 × sin(θ)

d) The velocity of the block at the bottom of the plane, v is approximately 6.264 m/s

e) 98.1 Joules

Explanation:

The given parameters of the block are;

The mass of the block, m = 5.0 kg

The distance down the plane the block slides, h = 2.0 m

The friction between the block and the surface = 0

Let θ represent the angle of inclination oof the plane

a) The gravitational potential energy, P.E. = m·g·h

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

∴ P.E. ≈ 5.0 kg × 9.81 m/s² × 2.0 m = 98.1 Joules

The gravitational potential energy, P.E. ≈ 98.1 Joules

b) The component of the weight of the block parallel to the plane, [tex]w_{\parallel}[/tex], is given as follows;

[tex]w_{\parallel}[/tex] = w × sin(θ) = m·g·sin(θ)

∴ [tex]w_{\parallel}[/tex] ≈ 5.0 kg × 9.81 m/s² × sin(θ) = 49.05 × sin(θ) N

The component of the weight of the block parallel to the plane, [tex]w_{\parallel}[/tex] ≈ 49.05 N × sin(θ)

c) The component of the weight along the inclined plane = The force with which the block moves along the inclined plane, therefore;

[tex]w_{\parallel}[/tex] = m·g·sin(θ) = m·a

Where a represents the acceleration of the block along the plane

Therefore, by comparison, we have;

g·sin(θ) = a

∴ a ≈ 9.81 × sin(θ)

d) Given that the motion of the block is 2.0 m downwards, we have;

The velocity of the block at the bottom of the plane, v² = 2·g·h

Therefore, v² ≈ 2 × 9.81 m/s²× 2.0 m = 39.24 m²/s²

v = √(39.24 m²/s²) ≈ 6.264 m/s

e) The kinetic energy at the bottom of the plane, K.E. = (1/2)·m·v²

∴ K.E. = (1/2) × 5.0 kg × 39.24 m²/s² = 98.1 J

In component form, the displacement vectors become

• 350 m [S]   ==>   (0, -350) m

• 400 m [E 20° N]   ==>   (400 cos(20°), 400 sin(20°)) m

(which I interpret to mean 20° north of east]

• 550 m [N 10° W]   ==>   (550 cos(100°), 550 sin(100°)) m

Then the student's total displacement is the sum of these:

(0 + 400 cos(20°) + 550 cos(100°), -350 + 400 sin(20°) + 550 sin(100°)) m

≈ (280.371, 328.452) m

which leaves the student a distance of about 431.8 m from their starting point in a direction of around arctan(328.452/280.371) ≈ 50° from the horizontal, i.e. approximately 431.8 m [E 50° N].

Answer:

The angiosperms dominate Earth's surface and vegetation in more environments, particularly terrestrial habitats, than any other group of plants. As a result, angiosperms are the most important ultimate source of food for birds and mammals, including humans.

Explanation:

plz mark brainlest

Answer:

[tex]K.E = 0.1905 J[/tex]

Explanation:

From the question we are told that:

Length [tex]L=1.4m[/tex]

Mass [tex]m=180g[/tex]

Angular Velocity [tex]\omega=1.80rads/s[/tex]

Generally the equation for Kinetic energy K.E is mathematically given by

 [tex]K.E =0.5 (1/3 ML^2 )w^2[/tex]

 [tex]K.E =0.5 ( 1/3 * 0.18 * 1.4^2 ) 1.8^2[/tex]

 [tex]K.E = 0.1905 J[/tex]