two forces, one of 12 N and another of 24 N, act on a body in such a way that they make an angle of 90° whith each other. find the resultant of the two forces. ​

Answer:

\(Q_net=534.67\frac{w}{m}\)

Explanation:

From the question we are told that:

Steam line diameter \(D=89mm \approx 0.089m\)

Surface emissivity \(\mu=0.8\)

Steam temp \(T_s=200\textdegree C \approx 200+273=473k\)

Surrounding temp \(T_a=20 \textdegree C \aprrox 20+273= 293k\)

Generally the equation for heat loss per unit length due to radiation \(Q_{net}\) is mathematically given by

\(Q_net=\sigma*\mu>(\pi *d)*(T_s^4-T_a^4)\)

\(Q_net=5.6*10^8*0.8*(\pi *0.089)*(473^4-293^4)\)

\(Q_net=534.67\frac{w}{m}\)

Answer:

A=1526N/m

B =40cm

Explanation:

Step one;

given

initial length=27cm= 0.27m

final length= 31.5cm= 0.315m

mass m= 7kg

Step two:

Required

A.

Force= mg

Force=weight= 7*9.81= 68.67N

extenstion e

e=31.5-27=4.5cm= 0.045m

K= spring constant

we know that F=ke

k=F/e

k=68.67/0.045

k=1526N/m

B.

New weight= 205N

F=ke

e=205/1526

e=0.13m

new length = 0.13+0.27= 0.4m

to cm=0.40*100=40cm

The magnitude of the horizontal component of the balls initial velocity if the angle between the horizontal and the direction of the 5.0 meters per second decreases from 30 degrees to 20 degrees is 4.7 m / s

cos θ = Adjacent side / Hypotenuse

θ = 20°

V = 5 m / s

Since the direction of ball makes a right angled triangle with the horizontal,

cos θ = \(V_{x}\) / V

\(V_{x}\) = cos 20° * 5

\(V_{x}\) = 0.94 * 5

\(V_{x}\) = 4.7 m / s

The formula used to solve is called as trigonometric ratios formulas. The basic formulae are:

sin θ = Perpendicular/Hypotenuse

cos θ = Base/Hypotenuse

tan θ = Perpendicular/Base

Therefore, the magnitude of the horizontal component of the balls initial velocity is 4.7 m / s

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Option C. The heat absorbed or lost by a substance while it's changing state  correctly describes latent heat

Latent heat is the heat absorbed or lost by a substance while it is changing state.

The latent heat is a type of heat that is transferred during phase change, i.e., while a substance undergoes a change of state.

For example, when ice melts into liquid water, or when liquid water evaporates into water vapor, heat is absorbed from the surroundings.

Latent heat is not associated with a temperature change; rather, it's associated with a change of state.

For instance, the temperature of water remains at 100°C while boiling.

When water is boiling, the latent heat of vaporization is absorbed and utilized to break the hydrogen bonds holding water molecules together to change water from the liquid phase to the gaseous phase.

When the water is boiling, adding more heat won't increase the water's temperature, instead, the extra heat will be absorbed to change the phase of water molecules.

Therefore, the correct answer to the given question is option C: The heat absorbed or lost by a substance while it is changing state.

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Answer:

d. 22, 700 m

Explanation:

We know that the speed of sound at 20°C is 4540 m/s and the time that the sound traveled was 5 seconds. So, we can use the following equation:

Where v is the speed, d is the distance and t is the time. Solving for d, we get:

So, replacing v = 4540 m/s and t = 5s, we get:

Therefore, the answer is

d. 22, 700 m

vf²=vi²-2ad (vf = 0, stop)

12² = 2a.3

144 = 6a

a = 24 m/s²

24² = 2 x 24 x d

576 = 48d

d = 12 m

The cost of operation for an A.C for 10 hours per day for a month will be 71,100 francs.

Power is the amount of energy transferred or converted per unit time. The unit of power is the watt, equal to one joule per second. Power is a scalar quantity.

Cost of operation for 10 hours a day;

Daily consumption = 3kW x 10 hours

Daily Consumption = 30kW

Since 1kWH = 79 francs;

Daily consumption amount = 30 x 79 francs

Daily consumption amount = 2,370 francs

Therefore, the monthly consumption (using 30days) will be;

2,370 francs x 30 = 71,100 francs

In conclusion, 71,100 francs will be spent in a month (30 days) to run the 3kW rated A.C for 10 hours a day at 1kWH.

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The density of water is approximately 1 g/cm^3. Therefore, the volume of 2800 g of water would be 2800 cm^3 because density is mass/volume, and so volume is mass/density.

Since this volume is inside a cubic box, the length of each side of the cube (a, for instance) could be found by taking the cubic root of the volume. This is because the volume of a cube is calculated by a^3 (length of one side cubed). Hence, a = cube root of 2800 cm^3 ≈ 14.1 cm.

Converting centimeters to meters (as 1 meter is equal to 100 centimeters), we get approximately 0.141 meters.

So the filled cubic box has a side length of approximately 0.141 m.

Answer:

Refer to the step-by-step Explanation.

Step-by-step Explanation:

Simplify the equation with given substitutions,

Given Equation:

\(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2\)

Given Substitutions:

\(\omega=v/R\\\\ \omega_{_{0}}=v_{_{0}}/R\\\\\ I=(2/5)mR^2\)\(\hrulefill\)

Start by substituting in the appropriate values: \(mgh+(1/2)mv^2+(1/2)I \omega^2=(1/2)mv_{_{0}}^2+(1/2)I \omega_{_{0}}^2 \\\\\\\\\Longrightarrow mgh+(1/2)mv^2+(1/2)\bold{[(2/5)mR^2]} \bold{[v/R]}^2=(1/2)mv_{_{0}}^2+(1/2)\bold{[(2/5)mR^2]}\bold{[v_{_{0}}/R]}^2\)

Adjusting the equation so it easier to work with.\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the left-hand side of the equation:

\(mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

Simplifying the third term.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2} \Big[\dfrac{2}{5} mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{2}\cdot \dfrac{2}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v}{R} \Big]^2\)

\(\\ \boxed{\left\begin{array}{ccc}\text{\Underline{Power of a Fraction Rule:}}\\\\\Big(\dfrac{a}{b}\Big)^2=\dfrac{a^2}{b^2} \end{array}\right }\)

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2 \cdot\dfrac{v^2}{R^2} \Big]\)

"R²'s" cancel, we are left with:

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5}mv^2\)

We have like terms, combine them.

\(\Longrightarrow mgh+\dfrac{1}{2} mv^2+\dfrac{1}{5} \Big[mR^2\Big]\Big[\dfrac{v^2}{R^2} \Big]\\\\\\\\\Longrightarrow mgh+\dfrac{7}{10} mv^2\)

Each term has an "m" in common, factor it out.

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)\)

Now we have the following equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac12mv_{_{0}}^2+\dfrac12\Big[\dfrac25mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\)

\(\hrulefill\)

Simplifying the right-hand side of the equation:

\(\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac12\cdot\dfrac25\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}}{R}\Big]^2\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\Big]\Big[\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15\Big[mR^2\cdot\dfrac{v_{_{0}}^2}{R^2}\Big]\\\\\\\\\Longrightarrow \dfrac12mv_{_{0}}^2+\dfrac15mv_{_{0}}^2\Big\\\\\\\\\)

\(\Longrightarrow \dfrac{7}{10}mv_{_{0}}^2\)

Now we have the equation:

\(\Longrightarrow m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

\(\hrulefill\)

Now solving the equation for the variable "v":

\(m(gh+\dfrac{7}{10}v^2)=\dfrac{7}{10}mv_{_{0}}^2\)

Dividing each side by "m," this will cancel the "m" variable on each side.

\(\Longrightarrow gh+\dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2\)

Subtract the term "gh" from either side of the equation.

\(\Longrightarrow \dfrac{7}{10}v^2=\dfrac{7}{10}v_{_{0}}^2-gh\)

Multiply each side of the equation by "10/7."

\(\Longrightarrow v^2=\dfrac{10}{7}\cdot\dfrac{7}{10}v_{_{0}}^2-\dfrac{10}{7}gh\\\\\\\\\Longrightarrow v^2=v_{_{0}}^2-\dfrac{10}{7}gh\)

Now squaring both sides.

\(\Longrightarrow \boxed{\boxed{v=\sqrt{v_{_{0}}^2-\dfrac{10}{7}gh}}}\)

Thus, the simplified equation above matches the simplified equation that was given.  

Answer:

A. 36.6667 per second

B. 58.6667 per second

C.102.667 per second

Explanation:

u = 2.5 m/s

v = 0

t = 3sec

s = ?

s = (u+v)/t

s = (0+2.5)/3

s = 2.5/3 = 0.83 m

The minimum mass required to be placed on the 0.00 cm mark of the meter stick to maintain equilibrium is 0.120 kg.

To maintain equilibrium, the torques acting on the meter stick must balance each other. The torque is given by the formula:

τ = r * F * sin(θ)

where τ is the torque, r is the distance from the pivot point to the point where the force is applied, F is the force applied, and θ is the angle between the force vector and the lever arm.

In this case, the meter stick is in equilibrium when the torques on both sides of the pivot point cancel each other out. The torque due to the weight of the meter stick itself is acting at the center of mass of the meter stick, which is at the 50.0 cm mark.

Let's denote the mass to be placed on the 0.00 cm mark as M. The torque due to the weight of M can be calculated as:

τ_M = r_M * F_M * sin(θ)

where r_M is the distance from the pivot point to the 0.00 cm mark (which is 30.0 cm), F_M is the weight of M, and θ is the angle between the weight vector and the lever arm.

Since the system is in equilibrium, the torques on both sides of the pivot point must be equal:

τ_M = τ_stick

r_M * F_M * sin(θ) = r_stick * F_stick * sin(θ)

Substituting the given values:

30.0 cm * F_M = 20.0 cm * (0.0400 kg * 9.8 m/s^2)

Solving for F_M:

F_M = (20.0 cm / 30.0 cm) * (0.0400 kg * 9.8 m/s^2)

F_M = 0.0264 kg * 9.8 m/s^2

F_M = 0.25872 N

Finally, we can convert the force into mass using the formula:

F = m * g

0.25872 N = M * 9.8 m/s^2

M = 0.0264 kg

Therefore, the minimum mass required to be placed on the 0.00 cm mark of the meter stick to maintain equilibrium is 0.120 kg.

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Answer:

Explanation:

6) Horizontal Projectile

7) At an angle projectile

Answer:

22 m/s

Explanation:

PEf +KEf =PE0 +KE0 →PE0 −PEf =KEf

−mgΔy= 1 mv2 →v= −2gΔy = −2(9.8 m/s2)(−25 m)=22 m/s

Answer:

Explanation:

In projectile motion , formula for range is as follows

R = u² sin 2 α / g , where u is initial velocity of throw , α is angle of throw

Given R , range = 30 yards , α = 10°

30 = u² sin 20 / 9.8

u² x .342 = 294

u² = 859.65

u = 29.32 m / s

Explanation:

Work cannot be increased by using a machine of some kind.

Work cannot be increased by using a machine of some kind.

A machine is any device in which the effort applied at one end overcomes a load at the other end.

Machines are generally used to perform different tasks faster.

However, a simple machine can not be used to increase the amount of work done at any time.

Force, speed and torque can all be increased using machines.

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Answer: Delta U = 30 - 80; -50 jules

Explanation:

The system is losing 80 J, but it is also gaining 30 J because the surroundings are doing work on it. So the net change in energy is -50 J

Answer:

Second Option

Explanation:

The "hard drive" or the second option is one of the main components of storing information on a computer. You already have a hard drive built into your computer, or laptop when you buy it, and you can buy additional hard drives in the form of plugins that can store even more data if your original hard drive becomes full of data.

Hope this helps.

Answer:

B

Explanation:

Whereas memory refers to the location of short-term data, storage is the component of your computer that allows you to store and access data on a long-term basis. Usually, storage comes in the form of a solid-state drive or a hard drive.

The maximum speed that the small-mass object attains when it leaves the trebuchet horizontally is approximately 28.3 m/s.

To find the maximum speed that the small-mass object attains when it leaves the trebuchet horizontally, we can apply the principle of conservation of mechanical energy.

Initially, the trebuchet is at rest, so its total mechanical energy is zero. As the small-mass object leaves the trebuchet horizontally, it gains kinetic energy. At this point, all of the potential energy of the system is converted into kinetic energy.

The potential energy of the system can be calculated as the sum of the gravitational potential energies of the two masses:

PE = m1 * g * h1 + m2 * g * h2

Since the trebuchet is released from rest in a horizontal orientation, the initial height h1 is zero. The height h2 can be calculated as the perpendicular distance between the pivot point and the center of mass of the larger mass m2:

h2 = 13.0 cm = 0.13 m

Therefore, the potential energy simplifies to:

PE = m2 * g * h2

The kinetic energy of the small-mass object can be calculated as:

KE = (1/2) * m1 * v^2

where v is the maximum speed of the small-mass object.

Since the total mechanical energy is conserved, we have:

PE = KE

m2 * g * h2 = (1/2) * m1 * v^2

Plugging in the given values, such as g = 9.8 m/s^2, m1 = 0.115 kg, m2 = 68.5 kg, and h2 = 0.13 m, we can solve for v:

(68.5 kg * 9.8 m/s^2 * 0.13 m) = (1/2) * 0.115 kg * v^2

Solving for v, we find:

\(v^2 = (68.5 kg * 9.8 m/s^2 * 0.13 m) / (0.115 kg)\)

\(v^2 = 800\)

v ≈ 28.3 m/s

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Yellow liquid:

Mass = 470 g

Volume = 450ml

Pink liquid

Mass= 570 g

Volume= 450ml

Volumes are equal (450ml) and masses are different (470g and 570g)

A. The liquids have the same volume and different masses.

Answer:

KE = 21.5 ft-lb

Explanation:

KE = (1/2)m*v²

KE = .5*8*(1100ft/s)²

KE = 4840000 ft-grains

KE = (4840000 ft-grains)*(1lb/7000 grains)(1 lb/32.174 slugs)

KE = 21.5 ft-lb

Answer:

Therefore it is save to carry a 62kg adult

Explanation:

From the question we are told that:

Mass \(m=380kg\)

Height of supporting Rock \(X=85cm\)

Length of Board\(L_r=4.5m\)

Mass of board \(M_b=22kg\)

Mass of adult \(M_a=62\)  

Generally the moment of balance about wedge part about  is mathematically given by

\(N -Q + R = Mg + mg\)

\(0.85*N - Mg*2.25 - mg*(2.25 + x) = 0\)

\(0.85*N = + Mg*2.25 + mg*(2.25 + x)\)

where

\(N+R=4547\)

therefore

\(N = 570.70588 + 1608.3529 + 714.823 x\)

if N=0 at fallen person

\(x=3.04m\)

Therefore it is save to carry a 62kg adult

Answer:

1800 N

Explanation:

gravitation field strength on Earth = 10N/kg

therefore force if gravity acting on paratrooper= 180kg × 10N/kg = 1800 N

The reflective surface of a concave mirror is curved inward and away from the light source. Concave mirrors focus light inward to a single focal point.

picture B correctly shows the path of the reflected light ray given an object inside the focal point.

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A procedure known as magnetic separation can be used to extract magnetic elements from coal.

This method makes use of the magnetic characteristics of some materials to distinguish them from non-magnetic materials like coal. A description of how magnetic separation can be used to remove magnetic components from coal is given below:

Putting a magnetic field around the coal and magnetic material mixture is the first step in the magnetization process. This can be achieved by creating an electromagnetic field or by putting the mixture close to a powerful magnet.

Magnetism: Magnetic materials, such as iron atoms or magnetite that are frequently found in coal, will be drawn to the magnetic field and become magnetized. They line up their magnetic moments with the magnetic field's direction.

Separation: The magnetic coal components can be physically separated from the non-magnetic coal once they have been magnetized. To create this separation, there are numerous techniques:

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Answer:

25 m

Explanation:

Let's assume that its acceleration is constant. We can determine the acceleration of the object by its definition

\(a= \frac{\Delta v}{\Delta t} = \frac{10-0(\frac ms)}{5 s} = 2 \frac m{s^2}\)

Now we can write the equation of motion

\(s(t)= s_0 + v_0t + \frac12at^2\)

where, the two terms \(s_0\ v_0\) represent the initial position and velocity respectively. Replacing the values we have ("from rest" means that initial velocity is 0)

\(s(5) = 0+0(5)+\frac12 2 (5)^2 = 25 m\)

The amount of damping force that allow shortest possible time is called​ critical damping of the system.

Critical damping is the threshold between overdamping and underdamping  at which the oscillator returns to the position of equilibrium quickly as possible.

Critical damping is frequently desired because such a system returns to and maintains equilibrium quickly. Furthermore, a constant force applied to a critically damped system moves the system to a new equilibrium position as quickly as possible without overshooting or oscillating around the new position.

Critical damping thus provides the rapid approach to zero amplitude for a damped oscillator.

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The breaking acceleration of the car must be - 44.5 m/s² so that the car stops in time and avoid hitting the Ilama.

Given in the question

The car is driving at the speed of 14 m/s and the reaction time is 0.05 seconds. So the distance traveled by car before hitting the break is,

Distance = Speed × time

Distance = 14 × 0.05

Distance = 0.7 meters

So, now the Ilama is 2.9 meters away and the distance traveled before hitting the break is 0.7 meters.

So the stopping distance = Distance of Ilama  - distance traveled before hitting the break

So stopping Distance = s = 2.9 - 0.7

stopping Distance = s = 2.2 m

Now we have

Initial Velocity = u = 14 m/s

Final Velocity = v = 0 m/s [ as after applying the break, the car comes to rest]

Stopping Distance = s = 2.2 m

Using the third equation of motion

v² - u² = 2as

Put in the value, we get

(0)² - (14)² = 2a(2.2)

4.4a =  - 196

a = -196/4.4

a = - 44.5 m/s²

Therefore, the breaking acceleration of the car must be - 44.5 m/s².

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