I need help on Weight vs mass.

Answer:no

Explanation:because 0.9*(30*60)=0.9*1800=1620

The turtle has already won the race

Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours

It is given

[tex]V_{t} = 0.9 \frac{m}{s}[/tex]

[tex]V_{r} = 9 \frac{m}{s}[/tex]

[tex]D=1500 m[/tex]

Time taken by turtle  

 [tex]T= \dfrac{D}{V_{t} }=\dfrac{1500}{0.9_{} }[/tex]

[tex]T=1666 minutes= 27 hours[/tex]

Time taken by  rabbit

[tex]T= \dfrac{D}{V_{r} }=\dfrac{1500}{9_{} }[/tex]

[tex]T=166 minutes[/tex]

since rabbit started 30 minutes after turtle then

[tex]T= 136+30=196 minutes[/tex]

[tex]T= 3.2 hours[/tex]

Hence Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours

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Option d (T₁ > T₂ > T₃) correctly describes the tension in the three rope system.    

Let's evaluate each tension.

Case T₃.

[tex] T_{3} - W_{3} = 0 [/tex]

For the system to be in equilibrium, the algebraic sum of the tension force (T) and the weight (W) must be equal to zero. The minus sign of W is because it is in the opposite direction of T.          

[tex] T_{3} = W_{3} [/tex]          

Since W₃ = mg, where m is for mass and g is for the acceleration due to gravity, we have:                

[tex] T_{3} = W_{3} = mg [/tex]  (1)                                                                                                     Case T₂.

[tex] T_{2} - (T_{3} + W_{2}) = 0 [/tex]    

[tex] T_{2} = T_{3} + W_{2} [/tex]   (2)

By entering W₂ = 2mg and equation (1) into eq (2) we have:

[tex] T_{2} = T_{3} + W_{2} = mg + 2mg = 3mg [/tex]

Case T₁.

[tex] T_{1} - (T_{2} + W_{1}) = 0 [/tex]  

[tex] T_{1} = T_{2} + W_{1} [/tex]    (3)

Knowing that W₁ = 3mg and T₂ = 3mg, eq (3) is:

[tex] T_{1} = 3mg + 3mg = 6mg [/tex]        

Therefore, the correct option is d: T₁ > T₂ > T₃.

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Correct answer: D. [tex]T_{1} > T_{2} > T_{3}[/tex]

First, we must construct the Equations of Equilibrium for each mass based on Newton's Laws of Motion, then we solve the resulting system for every Tension force:

Mass m:

[tex]\Sigma F = T_{3}-m\cdot g = 0[/tex] (1)

Mass 2m:

[tex]\Sigma F = T_{2}-2\cdot m \cdot g -T_{3} = 0[/tex] (2)

Mass 3m:

[tex]\Sigma F = T_{1}-3\cdot m\cdot g - T_{2} = 0[/tex] (3)

The solution of this system is: [tex]T_{3} = m\cdot g[/tex], [tex]T_{2} = 3\cdot m\cdot g[/tex] and [tex]T_{1} = 6\cdot m\cdot g[/tex], which means that [tex]T_{1} > T_{2} > T_{3}[/tex]. (Correct answer: D.)

Answer: [tex]247.12\ N[/tex]

Explanation:

Given

Magnitude of the charges

[tex]q_1=4.47\times 10^{-6}\ C[/tex]

[tex]q_2=1.86\times 10^{-6}\ C[/tex]

Distance between them [tex]d=17.4\ mm[/tex]

As both charges are of same sign, they must repel each other

Force experienced by second charge is

[tex]\Rightarrow F_{21}=\dfrac{kq_1q_2}{d^2}\\\\\Rightarrow F_{21}=\dfrac{9\times 10^9\times 4.47\times 10^{-6}\times 1.86\times 10^{-6}}{(17.4\times 10^{-3})^2}\\\\\Rightarrow F_{21}=\dfrac{74.82\times 10^{-3}}{302.76\times 10^{-6}}\\\\\Rightarrow F_{21}=0.2471\times 10^3\\\Rightarrow F_{21}=247.12\ N[/tex]

Thus, charge 2 experience a force of [tex]247.12\ N[/tex]

Answer:

The force between the two charges is 247.15 N.

Explanation:

Charge, q = 4.47 x 10^-6 C

charge, q' = 1.86 x 10^-6 C

distance, d = 17.4 mm

Let the force is F.

The force is given by the Coulomb's law:

[tex]F = \frac{K q q'}{r^2}\\\\F =\frac{9\times 10^9\times 4.47\times 10^{-6}\times1.86\times 10^{-6}}{(17.4\times 10^{-3})^2}\\\\F = 247.15 N[/tex]

Answer:

Explanation:

The answer is C because the building is 80 meters high. Before the stone is dropped, it has ONLY potential energy since kinetic energy involves velocity and a still stone has no velocity. At impact, there is no potential energy because potential energy involves the height of the stone relative to the ground and a stone ON the ground has no height; here there is ONLY kinetic.

From the First Law of Thermodynamics, we know that energy cannot be created or destroyed, it can only change form. Therefore, that means that at the halfway point of 40 meters, half of the stone's potential energy has been lost, and it has been lost to kinetic energy. Here, at 40 meters, there is an equality between PE and KE. It only last for however long the stone is AT 40 meters, which is probably a millisecond of time, but that's where they are equal.

Answer:

44 N

Explanation:

Given that,

If you stand next to a wall on a frictionless skateboard and push the wall with a force of 44 N, then we need to find the force the wall push on you.

It is based on Newton's third law of motion which states that for an action there is an equal and opposite reaction. If the you push the wall with a force of 44 N, the wall push on you is 44 N also as it is based on Newton's third law of motion.

The frictional force in unit of N that the floor must exert on the ladder is approximately 232.216 N

The known values are;

The weight of the painter = 1,071.628 N

The height to which the painter needs to climb along the ladder = 1.926 m

The length of the ladder = 12.014 m

The weight of the ladder = 250 N

The points where one of the ladder's ends is resting = On the ground

The points where the other end of the ladder is resting = A perfectly smooth wall

The angle with which the ladder rises above the horizontal floor = 51.96°

The acceleration due to gravity, g ≈ 10 m/s²

The unknown values include;

The friction force that the floor must exert on the ladder

The strategy to be used;

At equilibrium, the sum of moments about a point is zero

Finding the moments about the point of contact where the ladder rests on the wall, P, is given as follows;

At equilibrium, the sum of clockwise, [tex]M_{CW}[/tex], moment about P = The sum of the counterclockwise, [tex]M_{CCW}[/tex]moment about P

[tex]\mathbf{M_{CCW}}[/tex] = (12.014 - 1.926) × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250

[tex]\mathbf{M_{CW}}[/tex] = 12.014 × cos(51.96°) × [tex]\mathbf{F_N}[/tex]

Where;

[tex]\mathbf{F_N}[/tex] = The normal reaction of the of the ground on the end of the ladder that rests on the floor

[tex]\mathbf{M_{CCW}}[/tex] = [tex]\mathbf{M_{CW}}[/tex]

∴ (12.014 - 1.926) × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250 = 12.014 × cos(51.96°) × [tex]F_N[/tex]

We get;

6,665.3068846 N·m =  7.40316448688 m × [tex]F_N[/tex]

[tex]\mathbf{F_N}[/tex] = 6,665.3068846 N·m/(7.40316448688 m) = 900.332135 N

The normal reaction of the floor on the ladder, [tex]\mathbf{F_N}[/tex] = 900.332135 N

Taking moment about the point the ladder rests on the floor, R, gives;

[tex]M_{CCW}[/tex] = 12.014 × sin(51.96°) × [tex]F_W[/tex]

Where;

[tex]\mathbf{F_W}[/tex] = The normal reaction at the wall

[tex]M_{CW}[/tex] = 1.926 × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250

At equilibrium, we have, [tex]M_{CCW}[/tex] = [tex]M_{CW}[/tex]

Therefore;

12.014 × sin(51.96°) × [tex]F_W[/tex] = 1.926 × cos(51.96°) × 1,071.628 + (12.014/2) × cos(51.96°) × 250

9.46199511627 m × [tex]F_W[/tex] = 2,197.22861125 N·m

[tex]F_W[/tex] = 2,197.22861125 N·m/(9.46199511627 m)

The reaction of the wall, [tex]\mathbf{F_W}[/tex] = 232.216206 N

We note that also at equilibrium, the sum horizontal forces = 0

The horizontal forces acting  on the ladder = The normal reaction on the, [tex]F_W[/tex] wall and the friction force on the ground, [tex]\mathbf{F_f}[/tex]

∴ At equilibrium; [tex]\mathbf{F_W}[/tex] + [tex]\mathbf{F_f}[/tex] = 0

[tex]\mathbf{F_f}[/tex] = -[tex]\mathbf{F_W}[/tex]

[tex]\mathbf{F_W}[/tex]  = 232.216206 N

Therefore;

The frictional force in unit of N that the floor must exert on the ladder, [tex]\mathbf{F_f}[/tex] = 232.216206 N ≈ 232.216 N.

(The coefficient of friction, μ = [tex]\mathbf{F_N}[/tex]/[tex]\mathbf{F_W}[/tex] = 900.332135/232.216206 ≈ 3.877).

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Answer:

i. -4m

ii. 20m

Explanation:

The car travels 8m to the east, then travels 12m to the west which is the opposite of the east. Going west, the car travels 8m back to the origin point and then another 4m due west to make 12m. The displacement from the origin point is -4 (the negative sign shows the direction because displacement is a vector quantity)

Total distance = 8m going east + 8m back to origin + 4m west = 20m

Answer:

When studying nanotechnology, scientists must be aware that their ideas may not work out. Their work could be very time consuming and cost a lot of money. Finally, scientists do not yet know all of the effects of nanotechnology on human health.

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Answer:  

t = 1.27 x 10⁹ s  

Explanation:  

First, we will find the volume of the wire:

Volume = V = AL  

where,  

A = Cross-sectional area of wire = πr² = π(1 cm)² = π(0.01 m)² = 3.14 x 10⁻⁴ m²  

L = Length of wire = 150 km = 150000 m  

Therefore,    

V = 47.12 m³

Now, we will find the number of electrons in the wire:  

No. of electrons = n = (Electrons per unit Volume)(V)  

n = (8.43 x 10²⁸ electrons/m³)(47.12 m³)  

n = 3.97 x 10³⁰ electrons  

Now, we will use the formula of current to find out the time taken by each electron to cross the wire:

[tex]I =\frac{q}{t}[/tex]  

where,  

t = time = ?  

I = current = 500 A  

q = total charge = (n)(chareg on one electron)  

q = (3.97 x 10³⁰ electrons)(1.6 x 10⁻¹⁹ C/electron)  

q = 6.36 x 10¹¹ C  

[tex]500\ A = \frac{6.36\ x\ 10^{11}\ C}{t}\\\\t = \frac{6.36\ x\ 10^{11}\ C}{500\ A}[/tex]

Therefore,

t = 1.27 x 10⁹ s

Answer:

[tex]q=41.62\ \mu C[/tex]

Explanation:

Given that,

Force between two objects, F = 11.2 N

Distance between objects, d = 1.18 m

We need to find the charge on each objects. The force between charges is as follows :

[tex]F=\dfrac{kq^2}{r^2}\\\\q=\sqrt{\dfrac{Fr^2}{k}} \\\\q=\sqrt{\dfrac{11.2\times (1.18)^2}{9\times 10^9}} \\\\q=41.62\ \mu C[/tex]

So, the charge on each sphere is [tex]41.62\ \mu C[/tex].

Answer:

Let d be the density of the water  (1000 kg / m^3   eq to 1 gm / cm^3)

P = d g h     for the pressure due to a column at the bottom of the column.

P = 1000 kg / m^3 * 10 m/s^2 * 10 m = 10^5 kg / m * s^2 = 10^5 N/m

Answer:

Minimum Area of rectangle = 24 centimeter²

Explanation:

Given:

Length of rectangle = 6 centimeter

Width of rectangle = 4 centimeter

Find:

Minimum Area of rectangle

Computation:

Minimum Area of rectangle = Length of rectangle x Width of rectangle

Minimum Area of rectangle = 6 x 4

Minimum Area of rectangle = 24 centimeter²

Answer:

28 j

Explanation:

because when you add you get 28

Answer:

v = 4.4 x 10⁷ m/s

Explanation:

The kinetic energy of the electrons will be equal to the energy supplied by the electric voltage:

Kinetic Energy = Electric Energy

[tex]\frac{1}{2}mv^2 = eV[/tex]

where,

m = mass of electron = 9.1 x 10⁻³¹ kg

v = speed of electron = ?

e = charge on electron = 1.6 x 10⁻¹⁹ C

V =Voltage = 55 kV = 55000 V

Therefore,

[tex]\frac{1}{2}(9.1\ x\ 10^{-31}\ kg)(v)^2 = (1.6\ x\ 10^{-19}\ C)(55000\ V)\\\\v^2 = \frac{(2)(8.8\ x\ 10^{-16}\ J)}{9.1\ x\ 10^{-31}\ kg}\\\\v = \sqrt{19.34\ x\ 10^{14}\ m^2/s^2}[/tex]

v = 4.4 x 10⁷ m/s

Answer:

a: true.

Explanation:

We can define an equipotential surface as a surface where the potential at any point of the surface is constant.

For example, for a punctual charge, the equipotential surfaces are spheres centered at the punctual charge.

Or in the case of an infinite plane of charge, the equipotential surfaces will be planes parallel to our plane of charge.

Now we want to see if the electric field is always perpendicular to these equipotential surfaces.

You can see that in the two previous examples this is true, but let's see for a general case.

Now suppose that you have a given field, and you have a test charge in one equipotential surface.

So, now we can move the charge along the equipotential surface because the potential in the surface is constant, then the potential energy of the charge does not change. And because there is no potential change, then there is no work done by the electric field as the charge moves along the equipotential surface.

But the particle is moving and the electric field is acting on the particle, so the only way that the work can be zero is if the force (the one generated by the electric field, which is parallel to the electric field) and the direction of motion are perpendiculars.

Then we can conclude that the electric field will be always perpendicular to the equipotential surfaces.

The correct option is a.

Answer:

(a) The ratio of turns in the primary and secondary coils of her transformer is 1.833

(b) The ratio of input to output current is 0.55

(c) To increase the output voltage, you can either increase the number of turns in the secondary coil (step-up) or increase the input current. Therefore, the Chinese person has to increase the input current of the transformer to achieve an increased output voltage that can power her 220 V appliances.

Explanation:

Given;

input voltage, [tex]V_p[/tex] = 220 V

output voltage, [tex]V_s[/tex] = 120 V

General transformer equation is given as;

[tex]\frac{V_p}{V_s} = \frac{N_p}{N_s} = \frac{I_s}{I_p}[/tex]

where;

Np is number of turns in the primary coil

Ns is number of turns in the secondary coil

Is - is the secondary current or output current

Ip - is the primary current or input current

(a) The ratio of turns in the primary and secondary coils of her transformer;

[tex]\frac{N_p}{N_s} = \frac{V_p}{V_s} \\\\\frac{N_p}{N_s} = \frac{220}{120} = 1.833[/tex]

(b) The ratio of input to output current;

[tex]\frac{I_p}{I_s} = \frac{V_s}{V_p} \\\\\frac{I_p}{I_s} = \frac{120}{220} \\\\\frac{I_p}{I_s} = 0.55[/tex]

(c) To increase the output voltage, you can either increase the number of turns in the secondary coil (step-up) or increase the input current. Therefore, the Chinese person has to increase the input current of the transformer to achieve an increased output voltage that can power her 220 V appliances.

Answer:

The required fraction is 0.023.

Explanation:

Given that

Mass of a car, m = 1030 kg

Mass of 4 wheels = 12 kg

We need to find the fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles.

The rotational kinetic energy due to four wheel is

[tex]=4\times \dfrac{1}{2}I\omega^2\\\\=4\times \dfrac{1}{2}\times \dfrac{1}{2}mR^2(\dfrac{v}{R})^2\\\\=mv^2[/tex]

Linear kinetic Energy of the car is:

[tex]=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times Mv^2[/tex]

Fraction,

[tex]f=\dfrac{mv^2}{\dfrac{1}{2}Mv^2}\\\\f=\dfrac{m}{\dfrac{1}{2}M}\\\\f=\dfrac{12}{\dfrac{1}{2}\times 1030}\\\\=0.023[/tex]

So, the required fraction is 0.023.

Answer:

a)  v = 8,878 km / h, θ’= 238.8º,  b) t = 1890.9 s

Explanation:

a) In this exercise we must find the resulting speed of the boat.

Let's use trigonometry to break down the speed of the boat (v1)

            cos 225 = v₁ₓ / v₁

            sin 225 = v_{1y} / v₁

            v₁ₓ = v₁ soc 225

            v_{1y} = v₁ sin 225

            v₁ₓ = 6.5 cos 225 = -4.596 km / h

            v_{1y} = 6.5 sin 225 = -4.596 km / h

to find the velocity we add each component

           vₓ = v₁ₓ

           vₓ = - 4,596 km / h

           v_y = v_{1y} + v₂

           v_y = -4.596 - 3

           v_y = - 7,596 km / h

Now let's compose the speed

Let's use the Pythagorean theorem for the module

           v = [tex]\sqrt{v_x^2 + v_y^2 }[/tex]

           v = Ra 4.596² + 7.596²

           v = 8,878 km / h

Let's use trigonometry for the direction

          tan θ = v_y / vₓ

          θ = tan⁻¹ v_y / vₓ

          θ = tan⁻¹  ( [tex]\frac{-7.596}{ -4.596}[/tex] )

          θ = 58.8º

measured from the positive side of the x-axis

          θ'= 180 + 58.8

          θ’= 238.8º

b) Let's reduce the river width to the SI system

          x = 1.5 miles (1,609 km / 1 mile) = 2,414 km

to cross the river the speed is on the x axis which is the width of the river

         v = x / t

         t = x / v

         t = 2.414 /4.596

         t = 0.525 h

let's reduce to the SI system

         t = 0.525 h (3600 s / 1h)

         t = 1890.9 s

Explanation:

Inertia of motion

It is also known as Newton's first law of motion.

It states that,

An object remains in a state of rest or of uniform motion in a straight line unless compelled to change its state by an applied external force.

Answer:

height is 78.4m

Explanation:

h=u.t + 0.5.g.t^2

= 0 + 0.5x9.8x4^2

= 78.4m

Answer:

An axis is an invisible line around which an object rotates, or spins. The points where an axis intersects with an object's surface are the object's North and South Poles.

Explanation:

The Earth's axis is represented by the red line. The white circle represents axial precission, the slow "wobble" of the axis.

Answer: [tex]283.2\times 10^{-9}\ nC[/tex]

Explanation:

Given

Cross-sectional area [tex]A=0.4\ cm^2[/tex]

Dielectric constant [tex]k=4[/tex]

Dielectric strength [tex]E=2\times 10^8\ V/m[/tex]

Distance between capacitors [tex]d=5\ mm[/tex]

Maximum charge that can be stored before dielectric breakdown is given by

[tex]\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC[/tex]

Answer:

The maximum charge is 7.08 x 10^-8 C.

Explanation:

Area, A = 0.4 cm^2

K = 4

Electric field, E = 2 x 10^8 V/m

separation, d = 5 mm = 0.005 m

Let the capacitance is C and the charge is q.

[tex]q = CV\\\\q=\frac{\varepsilon o A}{d}\times E d\\\\q = \varepsilon o A E\\\\q = 8.85\times 10^{-12}\times0.4\times 10^{-4}\times 2\times 10^8\\\\q = 7.08\times 10^{-8}C[/tex]

Answer:

Between F2 and 2F2

Explanation:

Diagram attached from Teachoo.

Link to website if you need to refer

https://www.teachoo.com/10838/3118/Convex-Lens---Ray-diagram/category/Concepts/

Answer:

incorporators and it is the one you for the delay to get it for now that the new to me to the same as last week to week in my opinion of your

Answer:

oh my God I got really confused right now

Answer:

Explanation:

a)

Final velocity v = 0 ; initial velocity u = 30 m/s , time t = .0015 s

v = u + a t

0 = 30 m/s + a x .0015 s

a = - 30 / .0015

= - 20000 m / s²

b )

Magnitude of force = m x a

= .140 kg x 20,000 m / s²

= 2800 N = 2.8 kN.

c )

The force applied by baseball = 2.8 kN .

d )

Since ball can withstand a force of 1.3 kN so it will break if 2.8 kN force acts on it . SO, head will fracture.

Answer:

Option (D) is correct.

Explanation:

Let the speed is v.

[tex]\Delta t = \gamma \Delta t'\\\\\Delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\times \frac{\Delta t}{2}\\\\\sqrt{1-\frac{v^2}{c^2}} =\frac{1}{2}\\\\1-\frac{v^2}{c^2}=\frac{1}{4}\\\\\frac{3}{4}c^2 = v^2\\\\v = 0.87 c[/tex]

Option (D) is correct.

Answer:

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