A ​12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall​ (above the fire​ truck) if the ladder makes an angle of with the horizontal

Answer:

i₀ = V / R_i

Explanation:

For this exercise we use Ohm's law

         V = i R

          i = V / R

the equivalent resistance for

         [tex]\frac{1}{R_{eq}}[/tex] =  ∑ [tex]\frac{1}{R_i}[/tex]

if all the bulbs have the same resistance, there are N bulbs

         [tex]\frac{1}{ R_{eq}} = \frac{N}{R_i}[/tex]

         R_{eq} = R_i / N

we substitute

         i = N V / Ri

where i is the total current that passes through the parallel, the current in a branch is

         i₀ = i / N

         i₀ = V / R_i

Answer:

5.04 m

Explanation:

You are told that the homeowner wants to increase their fences by 34 percent meaning Original+ 34 percent. If the original is 100 percent, then the new fence size will be 134 % of the original. You are given the original which is 3.76 meters, to find new fence size 1.34 * 3.76m to get 5.0384 meters, rounded to 5.04 m.

Answer:

5.0384m

Explanation:

% increase = 100 x (Final - Initial / | initial | )

( |~~| Bars indicate absolute value since you can't have a negative height)

Answer:

0.000064 cubic meters.

Explanation:

Given the following data;

Length of side = 4 centimeters

Conversion:

100 centimeters = 1 meters

4 cm = 4/100 = 0.04 meters

To find the volume of cuboid;

Mathematically, the volume of a cuboid is given by the formula;

Volume of cuboid = length * width * height

However, when all the sides are equal the formula is;

Volume of cuboid = L³

Volume of cuboid = 0.04³

Volume of cuboid = 0.000064 cubic meters.

Question: A commuter backs her car out of her garage with an acceleration of 1.4 m/s² (a) How long does it take her to reach a speed of 2.00 m/s

Answer:

1.43 s

Explanation:

Applying,

a = (v-u)/t........... Equation 1

Where a = acceleration, v = final velocity, u = initial velocity, t = time

make t the subject of the equation

t = (v-u)/a........... Equation 2

From the question,

Given: v = 2 m/s, u = 0m/s (from rest), a = 1.4 m/s²

Substitute into equation 2

t = (2-0)/1.4

t = 1.43 s

Answer:b

Explanation:

The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.

Answer:

Option B.

Explanation:

Remember that we can think on any movement as a sum of a movement in the y-axis, the movement in the x-axis, and the movement in the z-axis. And these are not related, this means that, for example, the movement in x does not affect the movement in y.

So, when we analyze the problem of "how long takes an object to hit the ground"

We do not care for the horizontal motion of the object, we only care for the vertical motion of the object.

So, if an object is dropped, and another has a given initial velocity in the x-axis, in both cases the initial velocity in the y-axis will zero.

And in both cases, the only vertical force acting on the balls will be the gravitational force (so both objects will have the same vertical acceleration and the same vertical initial velocity) with this, we already know that the vertical motion of both objects will be exactly the same.

So, both objects will hit the ground at the same time.

(notice that here we are ignoring things like air resistance and other complex forces)

So here the correct option is b:  Neither. Their vertical motion is the same, so they will reach the ground at the same time.

Answer:

B. Smaller than

Explanation:

This question is from the Doppler effect. As the object which is in motion goes off from the other, there's a reduction in the frequency. This is due to the fact that successive soundwave get to be longer. So that the pitch will then be lowered. When the person observing moves towards what is making the sound, each soundwave that follows gets faster than the previous.

Answer:

Yes

Explanation:

speed of truck = 20 km/h

Initially the car at rest.

maximum velocity of car = 40 km/h

When the truck and the car collide, the momentum of the truck transferred to car.

So, the car can attain the speed of 40 km/h.

Answer:

7.9 [tex]\frac{m}{s^{2} }[/tex]

Explanation:

Take the fact that mass is inversely proportional to accelertation:

m ∝ a

Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:

[tex]\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\[/tex]

Using algebra, we can rearrange our equation to find the final acceleration, [tex]a_{2}[/tex]:

[tex]a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\[/tex]

Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:

0.85g * 9.8 [tex]\frac{m }{s^{2} }[/tex] = 8.33 [tex]\frac{m }{s^{2} }[/tex]

Plug everything in:

7.9 [tex]\frac{m }{s^{2} }[/tex] = [tex]\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}[/tex]

(1590kg the initial weight plus the weight of the added passenger)

Answer:

r = 20.22 m

Explanation:

Given that,

Charge,[tex]q=2.5\times 10^{-6}\ C[/tex]

Electric field, [tex]E=55\ N/C[/tex]

We need to find the distance. We know that, the electric field a distance r is as follows :

[tex]E=\dfrac{kq}{r^2}\\\\r=\sqrt{\dfrac{kq}{E}}\\\\r=\sqrt{\dfrac{9\times 10^9\times 2.5\times 10^{-6}}{55}}\\\\r=20.22\ m[/tex]

So, the required distance is 20.22 m.

Answer:

a.v=u+v/2

a.v=s/t

combining two equation we get,

u+v/2=s/t

(u+v)t/2=s

(u+v)t/2=s

{u+(u+at)}t/2=s

(u+u+at)t/2=s

(2u+at)t/2=s

2ut+at^2/2=s

2ut/2+at^2/2=s

UT +1/2at^2=s

proved

a=v-u/t

at=v-u

u+at=v

Answer:

Its the Federal government

The aircraft is 12 meters higher than the building so it is at 45 + 12 = 57 meters high.

For every 12 meters it travels it drops 1 m.

Divide the height by 12 to find the distance it travels:

57 / 12 = 4.75

It touches down 4.75 meters from the building.

The building is 684 meters away from the aircraft touching down on the runway.

A right-angled triangle's side ratios are the easiest way to express a function of an arc or angle, such as the sine, cosine, tangent, cotangent, secant, or cosecant. These functions are known as trigonometric functions.

As given in the problem an aircraft has a glide ratio of 12 to 1. (Glide ratio means that the plane drops 1 m in each 12 m it travels horizontally.) A building 45 m high lies directly in the glide path to the runway. If the aircraft clears the building by 12 m,

the total height of the aircraft when it clears the building = 45 +12

the total height of the aircraft when it clears the building is 57 meters

It is given that the Glide ratio is 12:1,

The distance of the building from touch down on the runway = 12 ×57

The distance of the building from the touch-down on the runway is 684 meters.

Thus, the building is 684 meters away from the aircraft touching down on the runway.

Learn more about the trigonometric functions here,

brainly.com/question/14746686

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Answer:

Action and reaction are equal in magnitude and opposite in direction but they do not balance each other because they act on different objects so they don't cancel each other out.

hope this will help you more

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

[tex]W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m[/tex]

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

[tex]W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J[/tex]

So, the new work is more than 130 J.

Answer:

1.67 N

Explanation:

Applying,

F = u(dm/dt)+m(du/dt)................ Equation 1

Where F = force, m = mass of the vehicle, u = speed.

Since u is constant,

Therefore, du/dt = 0

F = u(dm/dt)............... Equation 2

From the question,

Given: u = 10 m/s, dm/dt = 10 kg/min = (10/60) kg/s

Substitute these values into equation 2

F = 10(10/60)

F = 100/60

F = 1.67 N

Answer:

[tex]\sigma=0.014\ C/m^2[/tex]

Explanation:

Given that,

The radius of sphere, r = 5 cm = 0.05 m

Net charge carries, q = 7.5 µC = 7.5 × 10⁻⁶ C

We need to find the surface charge density on the sphere. Net charge per unit area is called the surface charge density. So,

[tex]\sigma=\dfrac{7.5\times 10^{-6}}{\dfrac{4}{3}\pi \times (0.05)^3}\\\\=0.014\ C/m^2[/tex]

So, the surface charge density on the sphere is [tex]0.014\ C/m^2[/tex].

Answer:

Both stones will land at the same time because both stones will fall with the same acceleration through the same height.

Explanation:

We are given that

Mass of stone ,m1=5 Kg

Mass of stone, m2=1 kg

We have to find which stone more faster will land and why.

[tex]h=u+\frac{1}{2}gt^2[/tex]

Initial velocity of both stones=0

[tex]h=\frac{1}{2}gt^2[/tex]

[tex]t^2=\frac{h}{g}[/tex]

[tex]t=\sqrt{\frac{h}{g}}[/tex]

[tex]t_1=t_2=\sqrt{\frac{h}{g}}[/tex]

Because both stones are thrown from the same height.

Both stones will land at the same time because both stones will fall with the same acceleration through the same height  and the acceleration  does not depend of its mass.

Answer:

Physical properties are properties that can be measured or observed without changing the chemical nature of the substance. Some examples of physical properties are:

color (intensive)

density (intensive)

volume (extensive)

mass (extensive)

boiling point (intensive): the temperature at which a substance boils

melting point (intensive): the temperature at which a substance melts

Explanation:

Explanation:

Let x = distance of [tex]F_1[/tex] from the fulcrum and let's assume that the counterclockwise direction is positive. In order to attain equilibrium, the net torque [tex]\tau_{net}[/tex] about the fulcrum is zero:

[tex]\tau_{net} = -F_1x + F_2d_2 = 0[/tex]

[tex] -m_1gx + m_2gd_2 = 0[/tex]

[tex]m_1x = m_2d_2[/tex]

Solving for x,

[tex]x = \dfrac{m_2}{m_1}d_2[/tex]

[tex]\:\:\:\:=\left(\dfrac{105.7\:\text{g}}{65.7\:\text{g}} \right)(13.8\:\text{cm}) = 22.2\:\text{cm}[/tex]

The question is incomplete. The complete question is :

High-speed stroboscopic photographs show that the head of a 200 g golf club is traveling at 60 m/s just before it strikes a  50 g golf ball at rest on a tee. After the collision, the club head travels (in the same direction) at 40 m/s. Find the speed of the golf ball just after impact.

Solution :

We know that momentum = mass x velocity

The momentum of the golf club before impact = 0.200 x 60

                                                                             = 12 kg m/s

The momentum of the ball before impact is zero. So the total momentum before he impact is 12 kg m/s. Therefore, due to the conservation of momentum of the two bodies after the impact is 12 kg m/s.

Now the momentum of the club after the impact is = 0.2 x 40

                                                                                    = 8 kg m/s

Therefore the momentum of the ball is = 12 - 8

                                                                = 4 kg m/s

We know momentum of the ball, p = mass x velocity

                                                     4 = 0.050 x velocity

∴ Velocity =  [tex]$\frac{4}{0.050}$[/tex]

                 = 80 m/s

Hence the speed of the golf ball after the impact is 80 m/s.

Answer:

Look at work

Explanation:

a) I am not sure if you want tangential or centripetal but I will give both

Centripetal acceleration = r*α

Since ω is constant, α is 0 so centripetal acceleration is 0m/s^2

Tangential acceleration = ω^2*r

convert 200rev/min into rev/s

200/60= 10/3 rev/s

a= 100/9*1.2= 120/9= 40/3 m/s^2

b) Rotational Kinetic Energy = 1/2Iω^2

I= mr^2

Plug in givens

I= 43.2kgm^2

K= 1/2*43.2*100/9=2160/9=240J

Answer:

The speed of the car, v = 21.69 m/s

Explanation:

The diameter is  = 48.01 m

Therefore, the radius of the loop R = 24.005 m

Weight at the top is n = mv^2/R - mg

Since the apparent weight is equal to the real weight.

So, mv^2/R - mg = mg

v = √(2Rg)

v = √[2(24.005 m)(9.8 m/s^2)]

The speed of the car, v = 21.69 m/s

Answer:

The speed is 15.34 m/s.

Explanation:

Diameter, d = 48.01 m

Radius, R = 24.005 m

Let the speed is v and the mass is m.

Here, the weight of the car is balanced by the centripetal force.

According to the question

[tex]m g = \frac{mv^2}{R}\\\\v =\sqrt{24.005\times9.8}\\\\v = 15.34 m/s[/tex]

Answer:

A gold medal has the (minimum) dimensions of:

diameter = 60mm

thickness = 3mm

So we will work with those dimensions.

The medal is then a cyinder of diameter

D = 60mm = 6cm

and height:

H = 3mm = 0.3cm

Remember that the volume of a cylinder is:

V = pi*(D/2)^2*H

where pi = 3.14

Then the volume of a medal is:

V = 3.14*(6cm/3)^2*0.3cm = 3.768 cm^3

The density of the gold in g/cm^3 is:

d = 19.3 g/cm^3

And remember that:

density = mass/volume

So, if the volume is 3.768 cm^3

Then the mass will be:

mass = density*volume =  19.3 g/cm^3*3.768 cm^3 = 72.7 g

So, a single gold medal would weight 72.7 grams

And each gram of gold costs $40

Then the total cost of the gold medal would be:

value = $40*72.7 = $2,908

Now, if yo think that in the Olympics there are 35 sports (a lot with a large number of players) and near 50 disciplines, they need a lot of gold medals.

And each gold medal costs $2,908

So the total cost (only for the gold medals, ignoring the others) would be to high.

This is why the gold medals are made mostly of silver.

Answer:

Explanation:

First last of thermodynamics, just discusses the changes that a system is undergoing and the processes involved in it. It explains conservation of energy for a system undergoing changes or processes.

Second law of thermodynamics helps in defining the process and also the direction of the processes. It tells about the possibility of a process or the restriction of a process. It states that the entropy of a system always increases.

For this to occur the energy contained by a body has to diminish without converting to work or internal energy. So imagine a machine which works with less than efficiency, this means there are losses but they don’t show up anywhere. But the energy is obtained from a higher energy source to lower.

The easy way to do this is with an imaginary device that extracts zero-point energy to heat a quantity of gas. Energy is being created, so the first law is violated, and the entropy of the system is increasing as the gas heats up.

First law is violated since the energy conversion don't apply but the direction of work is applied so second law is satisfied.

The complete question is as follows: At 700 K, [tex]CCl_{4}[/tex] decomposes to carbon and chlorine. The Kp for the decomposition is 0.76.

Find the starting pressure of [tex]CCl_{4}[/tex] at this temperature that will produce a total pressure of 1.1 atm at equilibrium.

Answer: The starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.

Explanation:

The equation for decomposition of [tex]CCl_{4}[/tex] is as follows.

[tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]

Let us assume that initial concentration of [tex]CCl_{4}[/tex] is 'a'. Hence, the initial and equilibrium concentrations will be as follows.

                   [tex]CCl_{4}(g) \rightleftharpoons C(s) + 2Cl_{2}(g)[/tex]

Initial:            a                0          0

Equilibrium:  (a - x)          0          2x

Total pressure = (a - x) + 2x = a + x

As it is given that the total pressure is 1.1 atm.

So, a + x = 1.1

a = 1.1 - x

Now, expression for equilibrium constant for this equation is as follows.

[tex]K_{p} = \frac{P^{2}_{Cl_{2}}}{P_{CCl_{4}}}\\0.76 = \frac{(2x)^{2}}{(a - x)}\\0.76 = \frac{4x^{2}}{1.1 - x - x}\\0.76 = \frac{4x^{2}}{1.1 - 2x}\\x = 0.31 atm[/tex]

Hence, the value of 'a' is calculated as follows.

a + x = 1.1 atm

a = 1.1 atm - x

  = 1.1 atm - 0.31 atm

  = 0.79 atm

Thus, we can conclude that starting pressure of [tex]CCl_{4}[/tex] is 0.79 atm.

Answer:

The drop time ball 1 is less than the drop time of ball 2. A further explanation is provided below.

Explanation:

The net force acting on the ball will be:

⇒ [tex]F_{net}=mg-F_r[/tex]

Here,

F = Force

m = mass

g = acceleration

Now,

According to the Newton's 2nd law of motion, we get

⇒ [tex]F_{net} = ma[/tex]

To find the value of "a", we have to substitute "[tex]F_{net}=ma[/tex]" in the above equation,

⇒ [tex]ma=mg-F_r[/tex]

⇒    [tex]a=g-\frac{F_r}{m}[/tex]

We can see that, the acceleration is greater for the greater mass of less for the lesser mass. Thus the above is the appropriate solution.

Answer:

Both the ball takes equal time to reach to the ground.  

Explanation:

Two balls of same diameter

Let the height is h.

Mass of ball 1 is more than the mass of ball 2.

The second equation of motion is

[tex]h = u t +0.5 gt^2[/tex]

Here, the buoyant force due to air is same. So, the time of fall is independent of the mass.

So, both the ball takes equal time to reach to the ground.  

I see six (6) loops.

I attached a drawing to show where I get six loops from.