What is the Y-component of a vector A, which is of magnitude
16-12 and at a 45° angle to the horizontal?

Answer:

98 J

Explanation:

Applying,

Change in kinetic energy = Final kinetic energy- initial kinetic energy

ΔK.E = mv²/2-mu²/2..............Equation 1

Where ΔK.E = Change in kinetic energy, m = mass of the box, u = initial velocity of the box, v = final velocity of the box.

From the question,

Given: m = 4.0 kg, u = 0 m/s, v = 7 ,0 m/s

Substitute these values into equation 1

ΔK.E = (4(7²)/2)-(4(0²)/2)

ΔK.E = (2×49)-0

ΔK.E = 98 J

Hence the change in kinetic energy 98 J

Answer:

False.

Explanation:

Suppose that we have an object that moves with constant acceleration A.

Then the acceleration of the object is defined by the equation:

a(t) = A

The acceleration is the rate of change of the velocity, then the velocity equation is given by the integration of the acceleration equation, we will get:

v(t) = A*t + V₀

Where V₀ is the velocity of the object at the time t = 0s.

Now, if we integrate it again, we will get the position equation:

p(t) = (1/2)*A*t^2 + V₀*t + P₀

Where P₀ is the initial position equation.

Here, we can see that the position equation is a quadratic equation (not a linear equation), then the statement is false.

Answer:

D I think I’m not for sure

Explanation:

Answer:

The right solution is "4.8° east of north".

Explanation:

Given:

Distance,

= 500 km

Speed,

[tex]\vec{v}=120 \ m/s[/tex]

Wind (towards west),

[tex]v_0=10 \ m/s[/tex]

According to the question, we get

The angle will be:

⇒ [tex]\Theta=Cos^{-1}(\frac{v_0}{v_1} )[/tex]

       [tex]=Cos^{-1}(\frac{10}{120} )[/tex]

       [tex]=85.21[/tex] (north of east)

hence,

The direction must be:

⇒ [tex]\Theta'=90-85.21[/tex]

        [tex]=4.79^{\circ}[/tex]

or,

        [tex]=4.8^{\circ}[/tex] (east of north)

Answer:

(a) Positive

(b) Electron gains energy as it moves from A to B.

Explanation:

VA = 260 V

VB = 210 V

An electron moves from lower to higher potential which is negatively charged and a positively charged particle moves  from higher to lower potential, so the charge particle is positive in nature.

(a) Positive

(b) No, electron gains energy as it moves from A to B.

When we exhale, 90% waste material is Carbon Dioxide ( CO2 ) , so, it gets exhaled out in the form of CO2 rich air and it gets removed from the body, therefore our internal body becomes more pure and helps in making our internal temperature constant at a suitable level.

The speed is in feet per seconds so change the length of the field from yards to feet.

1 yard = 3 feet

100 yards x 3 = 300 feet

The field is 300 feet long

Time = distance / speed

Time = 300 feet / 5 feet per second = 60 seconds = 1 minute

It will take 1 minute

Answer:

A person will take 60 Seconds to walk the distance of 100 yards.

Explanation:

Data Given ;

Distance ( d ) = 100 yards = 300 Ft

Speed ( v ) = 5 Ft/s

Time ( t ) = ?

The distance travelled in unit time is called speed.

formula ; [tex]v = \frac{d}{t} \\[/tex]

On putting values,

[tex]5 = \frac{300}{t}[/tex]

[tex]t =\frac{300}{5}[/tex]

[tex]t = 60 sec[/tex]

Hence the time taken by the person is 60 sec.

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Answer:

What is the answer bro idont now

Answer:

a liquid warms gas cools

b solid piece warms and liquid cools

Answer:

[tex]V_2=3.3m/s[/tex]

Explanation:

From the question we are told that:

Distance [tex]d_1=1.4m[/tex]

Tangential speed [tex]V=2.2m/s[/tex]

Distance 2 [tex]d_2=2.1m[/tex]

Generally the equation for Angular velocity is mathematically given by

 [tex]w=\frac{v}{r}[/tex]

Therefore

 [tex]\frac{v_1}{r_1}=\frac{v_2}{r_2}[/tex]

 [tex]V_2=\frac{2.2*2.1}{1.4}[/tex]

 [tex]V_2=3.3m/s[/tex]

Answer:

The moon Phobos orbits Mars (m = 6.42 x 1023 kg) at a distance of 9.38 x 106 m.

Answer:

Example:Opening of a bottle cap by tool

when we hold a tool and open the bottle cap this is because , force x tool force .

The load arm is usually shorter than the effort arm in second order levers. Moving a large weight hence requires less effort. A force multiplier lever or effort multiplier lever is the name for this kind of lever. A boat's oars, for instance, can increase the force.

Second-order levers are devices with the input force farthest from the fulcrum and the output force on the same side of the fulcrum. A wheelbarrow is an excellent illustration of a second-order lever.

A second-order lever will have an output force greater than an input force, similar to first-order levers. The output journey, however, will be shorter than the input length. Both the input and output forces in this situation will move in the same direction.

Learn more about lever here:

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Answer:

The answer is below

Explanation:

The intensity level (B) of a sound wave is given by:

B = 10log(I/I₀);

where I₀ is the threshold intensity = 1 * 10⁻¹² W/m², I is the intensity at distance 5 m, B is the intensity level = 52 dB

Substituting gives:

[tex]52=10log(\frac{I}{10^{-12}} )\\\\log(\frac{I}{10^{-12}} )=5.2\\\\I=1.58*10^{-7}\ W/m^2[/tex]

The pressure is given by:

[tex]I=\frac{p_{max}^2}{2\rho v} \\\\\rho=air\ density=1.2\ kg/m^3,v=speed\ of\ sound\ in\ air=344\ m/s,p_{max}=pressure:\\\\p_{max}=\sqrt{2\rho vI}=\sqrt{2*1.58*10^{-7}*1.2*344} =1.14*10^{-2}Pa[/tex]

Answer:

First of all the formula is F= uR,( force= static friction× reaction)

mass= 5+25=30

F= 50

R= mg(30×10)=300

u= ?

F=UR

u= F/R

u= 50/300=0.17N

Answer:

We need, mass, gravity and height.

Explanation:

When a body falls freely from a height, its initial velocity is zero, but due to the height it has some potential energy at the top and the kinetic energy is zero.

As it falls, the potential energy is gradually converted in to the kinetic energy so that the total energy of the falling body is conserved.

At the time as the body strikes the ground, the entire potential energy is converted into the kinetic energy.

Potential energy is given by

U = m g h

where, m is the mass, g is the gravity and h is the height,

So, to get the kinetic energy we require mass, gravity and height of the body.

Answer:

Option (b) is correct.

Explanation:

The amount of heat required to raise the temperature of substance of mass 1 kg by 1 degree C, is called specific heat of the substance.

The formula of the specific heat is

H = m c (T' - T)

where, m is the mass, c is the specific heat and T' - T is the change in temperature.

So, to know the rise in temperature, by adding the known amount of heat, the specific heat is required.

So, option (b) is correct.

Answer:

[b] it id heated from 4o

Explanation:

Answer:

arge number of electrons free to move between the charged ions in the lattice.

Explanation:

The metallic bond occurs when an atom with few electrons is united in its last level, therefore the best way to decrease the total energy of the system is to lose all its electrons to remain with the configuration of a noble gas. The electrons that it loses cannot be acquired by other atoms since they all have few electrons, thus leaving a large number of electrons free to move between the charged ions in the lattice.

Some important characteristics emerge from this description of the metallic bond:

* It has many free electrons therefore its electrical conductivity is high

* As the charged ions are fixed, the material can be malleable, bent without breaking since the free electrons create the bond that keeps the system stable.

* As the electrons are free when heating a part of the material, these electrons acquire energy and rapidly propagate it to the other side, giving a high thermal conductivity

* As the temperature increases, the electrons acquire more kinetic energy, which is why there are more collisions between them and consequently the resistivity of the material increases.

Answer:

[tex]F = 14.2 N[/tex]  

Explanation:

From the question we are told that:

Mass [tex]m=0.15kg[/tex]

Radius [tex]r=0.6[/tex]

Angular Velocity [tex]\omega=2rev/s[/tex]

                            [tex]\omega= =2x2 \pi rad/s=>4 \pi rad/s[/tex]

Generally the equation for Force applied is mathematically given by

 [tex]F =mrw2[/tex]

 [tex]F=0.15*0.6* (4*x3.14^)2[/tex]

 [tex]F = 14.2 N[/tex]    

Answer:

Marine geologists learn about the rocks and geologic processes of the ocean basins.

Answer:

Distance

Explanation:

Work can be defined as the energy transferred to a physical object by exertion of a force on the object to cause a displacement of the object. Thus, work is typically done when a person or simple machine move an object over a distance through the application of a force.

Mathematically, work done is given by the formula;

[tex] W = F * d[/tex]

Where,

W is the work done

F represents the force acting on a body.

d represents the distance covered by the body.

A dependent variable is the event expected to change when an independent variable is manipulated.

Hence, distance is the dependent variable because its value changes with respect to the amount of force exerted on an object.

Answer:

P = 1.4 W

Explanation:

Given that,

The work done or the energy of the battery, E = 10,000 J

Time, t = 2 h

We need to find the power consumption. Let it is P. Power is the rate of doing work. So,

[tex]P=\dfrac{W}{t}\\\\P=\dfrac{10,000}{2\times 3600}\\\\P=1.38\ W[/tex]

or

P = 1.4 W

So, the power of the battery is 1.4 W.

Answer:

The water usually rushes back too enthusiastically, causing a splash – and the bigger the rock, the bigger the splash. The splash then creates even more ripples that tend to move away from where the rock went into the water.

[tex]4.8 \times 10^8[/tex] volts

Explanation:

The electric potential due to a point charge is given by

[tex]V= \dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{Q}{r}[/tex]

where Q = charge = [tex]16 \times 10^{-9}[/tex] C

r = distance from a point = [tex]3 \times 10^{-3}[/tex] m

[tex]\varepsilon_{0}[/tex] = permitivity of free space

= 8.85×10^-12 C^2/N-m^2

Plugging in the numbers,

[tex]V = \dfrac{1}{4 \pi (8.85 \times 10^{-12})} \dfrac{16 \times 10{-9}}{3 \times 10^{-3}}[/tex]

[tex]= 4.8 \times 10^8[/tex] volts

Answer:

The speed of projection is 34 m/s.

Explanation:

Height of building, h = 51 m

horizontal distance, d = 74 m

time, t = 8 s

Let the angle is A and the speed is u.

d = u cos A x t

74 = u cos A x 8

u cos A = 9.25 .... (1)

Use second equation of motion

[tex]h = u sin A t - 0.5 gt^2\\\\-51 = u sinA \times 8 - 0.5\times 9.8\times8\times 8\\\\u sin A = 32.8 .... (2)[/tex]

Squaring and adding both the equations

[tex]u^2 = 9.25^2 + 32.8^2 \\\\u = 34 m/s[/tex]

Answer:

both swings at the same place

Explanation:

because there mother is giving same amount of force to both.

Answer:

1. 5.825 × 10¹⁷ J

2. i. $ 29.125 billion ii. I would not invest in the company

3. A nuclear reaction would occur if the payload hits an airplane flying overhead and dissipates all its kinetic energy in the collision.

4. a i. The momentum will not be relativistic

ii. This is because objects with large masses do not move at relativistic speeds

b i. 155 m/s

ii. This speed wouldn't be a problem for the ship.

Explanation:

1.  A typical payload they claim to launch will weigh 1 kilogram and be accelerated to 90% the speed of light. How much electrical energy will the rail gun require to launch the probe, assuming it is 20% efficient at converting electrical energy to projectile kinetic energy?

The kinetic energy of the payload is K = (γ - 1)mc² where m = mass of payload = 1 kg, c = speed of light = 3 × 10⁸ m/s and γ = 1/√(1 - β²) where β = 0.9 (since the payload moves at 90 % speed of light)

So, K = (γ - 1)mc²

= (1/√(1 - β²) - 1)mc²

= (1/√(1 - (0.9)²) - 1) × 1 kg × (3 × 10⁸ m/s)²

= (1/√(1 - 0.81) - 1) × 1 kg × (3 × 10⁸ m/s)²

= (1/√0.19 - 1) × 1 kg × (3 × 10⁸ m/s)²

= (1/0.436 - 1) × 1 kg × (3 × 10⁸ m/s)²

= (2.294 - 1) × 1 kg × (3 × 10⁸ m/s)²

= 1.294 × 1 kg × 9 × 10¹⁶ m²/s²

= 11.65 × 10¹⁶ kgm²/s²

= 1.165 × 10¹⁷ J

Let E be the total electrical energy of the rail gun. Since 20 % of this energy is converted to kinetic energy of the payload, we have

20 % of E = K

0.2E = K

E = K/0.2

= 1.165 × 10¹⁷ J/0.2

= 5.825 × 10¹⁷ J

2 Given that typical electrical costs are about 5 cents/MJ, how much would this launch cost? Would you invest in this company?

i. how much would this launch cost?

Since the total energy required is E = 5.825 × 10¹⁷ J = 5.825 × 10¹¹ MJ and it costs 5 cent/MJ. So the total cost of energy will be total energy rate = 5.825 × 10¹¹ MJ × 5 cent/MJ = 29.125 × 10¹¹ = 2.9125 × 10¹² cents. Converting this to dollars, we have 2.9125 × 10¹² cents/100 cents/dollar = 2.9125 × 10¹⁰ dollars = 29.125 × 10⁹ dollars = 29.125 billion dollars = $ 29.125 billion

ii. Would you invest in this company?

I would not invest in the company

3. You are also concerned about safety. What happens if this projectile were to hit an airplane that is flying overhead and dissipate all of its kinetic energy in the collision? To give you a sense of scale, a large nuclear explosion generates about 1015 J of energy.

Since the kinetic energy of the payload is 1.165 × 10¹⁷ J and a nuclear explosion generates about 10¹⁵ J of energy,  then a nuclear reaction would occur if the payload hits an airplane flying overhead and dissipates all its kinetic energy in the collision.

4. The system must be able to launch probes to all parts in the sky and must be transportable on a ship. Assume that the rail gun is mounted on a frigate-class navy ship (weight = 4,000 metric tons).

a. Will the recoil momentum of the ship be relativistic? Justify your argument.

i. Will the recoil momentum of the ship be relativistic?

The momentum will not be relativistic.

ii. Justify your argument.

This is because objects with large masses do not move at relativistic speeds. Since the speed cannot be relativistic, its momentum which is the product of mass and speed is non-relativistic

b. At what speed will the ship recoil after it launches a probe? Do you think that this is a problem for the ship?

i. At what speed will the ship recoil after it launches a probe?

Since the total energy of the payload E' = K + mc² = 1.165 × 10¹⁷ J + 1 kg × (3 × 10⁸ m/s)² = 1.165 × 10¹⁷ J + 1 kg × 9 × 10¹⁶ m²/s² = 11.65 × 10¹⁶ J + 9 × 10¹⁶ J = 20.65 × 10¹⁶ J

Also, E'² = (pc)² + (mc²)² where p = momentum of payload

So, making p subject of the formula, we have

(pc)² = E'² - (mc²)²

pc = √[E'² - (mc²)²]

p = √[E'² - (mc²)²]/c

substituting the values of the variables into the equation, we have

p = √[E'² - (mc²)²]/c

p = √[(20.65 × 10¹⁶ J)² - 1kg × (3 × 10⁸ m/s²)²]/3 × 10⁸ m/s

p = √[(20.65 × 10¹⁶ J)² - (1kg × 9 × 10⁸ m²/s²)²]/3 × 10⁸ m/s

p = √[426.4225 × 10³² J² - 81 × 10³² J²]/3 × 10⁸ m/s

p = √[345.4225 × 10³² J²]/3 × 10⁸ m/s

p = 18.59 × 10¹⁶/3 × 10⁸ m/s

p = 6.20 × 10⁸ kgm/s

From the law of conservation, this momentum of the payload equals the momentum of recoil of the ship.

So, p = m'v where m' = mass of navy ship = 4,000 metric tons = 4,000 × 1000 kg = 4 × 10⁶ kg and v = speed of navy ship

So, v = p/m'

= 6.20 × 10⁸ kgm/s ÷ 4 × 10⁶ kg

= 1.55 × 10² m/s

= 155 m/s

ii. Do you think that this is a problem for the ship?

Since the ship's speed is 155 m/s, which is small for an object with such a large mass, this speed wouldn't be a problem for the ship.