An object whose weight is 100 lbf experiences a decrease in kinetic energy of 500 ft lbf and an increase in potential energy of 1500 ft lbf. The initial velocity and elevation of the object, each relative to the surface of the earth, are 40 ft/s and 30 ft, respectively. If g 5 32.2 ft/s2 , determine:
(a) the final velocity, in ft/s.
(b) the final elevation, in ft.

Answer:

Explanation:

The formula for angular momentum is

L = mvr where L is the angular momentum, m is the mass of the object, v is the velocity of the object, and r is the radius of the object. The problem we have that prevents us from just throwing those numbers in there is that mass has to be in kg and it's not, and radius has to be in meters and it's not.

Changing the mass to kg:

750 g = .750 kg

Changing the radius to m:

35 cm = .35 m

Now we can fill in the variables with their respective values:

L = .750(10.0)(.35) gives us

[tex]L=2.625\frac{kg*m^2}{s}[/tex]

Answer:

approaches infinity

Explanation:

There are two momentums, the classical momentum which is equal to the product of mass and velocity, and the relativistic momentum, the one we should look at when we work with high speeds, and this happens because massive objects have a speed limit, in this case, we are approaching the speed of light, so we need to work with the relativistic momentum instead of the classical momentum.

The relativistic momentum can be written as:

[tex]p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]

where

u = speed of the object relative to the observer, in this case we have that u tends to c, the speed of light.

m = mass of the object

c = speed of light.

So, as u tends to c, we will have:

[tex]\lim_{u \to c} p = \frac{1}{\sqrt{1 - \frac{u^2}{c^2} } } *m*u[/tex]

Notice that when u tends to c, the denominator on the first term tends to zero, thus, the relativistic momentum of the object will tend to infinity.

Then the correct option is infinity, as the particle speed approaches the speed of light, the relativistic momentum of the particle tends to infinity.

Answer:

Explanation:

That is an amazing fact.

The minus sign is what you have to pay attention to. The earth has a mass of 100 times that of Saturn. As someone on here once noted, Saturn has such a low density that it would float in water.

The answer is D

Answer:

independent variable

Explanation:

Answer:

The answer would be D), if the decay is beta negative.

Explanation:

Thorium-232 goes through alpha decay:

Thorium-232 --> Helium-4 + Radium-228

Radium-228 then can undergo beta positive or beta negative decay:

Beta positive = Radium-228 --> Electron + Francium-228

Beta negative = Radium-228 --> Positron + Actinium-228

Therefore, the isotope that is created is Actinium-228

Answer:

R = 0.0015Ω

Explanation:

The formula for calculating the resistivity of a material is expressed as;

ρ = RA/l

R is the resistance

ρ is the resistivity

A is the area of the wire

l is the length of the wire

Given

l = 85cm = 0.85m

A = πr²

A = 3.14*0.0018²

A = 0.0000101736m²

ρ = 1.75 × 10-8Ωm.

Substitute into the formula

1.75 × 10-8 = 0.0000101736R/0.85

1.4875× 10-8 = 0.0000101736R

R = 1.4875× 10-8/0.0000101736

R = 0.0015Ω

Answer:

the answer is 49000 joules at the maximum height

Explanation:

we know the mass (50kg)

we know the acceleration due to gravity(9.8m/s²)

we know the height too(maximum height meaning the 50th step so we multiply 50 with 20cm as each step is 20 cm and we get 1000 cm, convert to m it is 100 m

the formula is potential energy=mgh

m for mass

g for acceleration due to gravity

h for height

multiply them

50x9.8x100

we get 49000

the unit of potential energy is joules so the answer is

49000 joules

Answer:

49000 joules

Explanation:

hope it helpss

The question is incomplete. The complete question is :

Two loudspeakers are placed 1.8 m apart. They play tones of equal frequency. If you stand 3.0 m in front of the speakers, and exactly between them, you hear a minimum of intensity. As you walk parallel to the plane of the speakers, staying 3.0 m away, the sound intensity increases until reaching a maximum when you are directly in front of one of the speakers. The speed of sound in the room is 340 m/s.

What is the frequency of the sound?

Solution :

Given :

The distance between the two loud speakers, [tex]d = 1.8 \ m[/tex]

The speaker are in phase and so the path difference is zero constructive interference occurs.

At the point [tex]D[/tex], the speakers are out of phase and so the path difference is [tex]$=\frac{\lambda}{2}$[/tex]

Therefore,

[tex]$AD-BD = \frac{\lambda}{2}[/tex]

[tex]$\sqrt{(1.8)^2+(3)^2-3} =\frac{\lambda}{2}$[/tex]

[tex]$\lambda = 2 \times 0.4985$[/tex]

[tex]$\lambda = 0.99714 \ m$[/tex]

Thus the frequency is :

[tex]$f=\frac{v}{\lambda}$[/tex]

[tex]$f=\frac{340}{0.99714}$[/tex]

[tex]f=340.9744[/tex] Hz

Answer:

c) 2Q

Explanation:

From the given information:

The pressure inside a pipe can be expressed by using the formula:

[tex]\Delta P = \dfrac{128 \mu L Q}{\pi D^4}[/tex]

Since the diameter in both pipes is the same, we can say:

[tex]D = D_A = D_B[/tex]

where;

length of the first pipe A [tex]L_A = L[/tex] and the length of the second pipe B [tex]L_B = 2L[/tex]

Since the difference in pressure is equivalent in both pipes:

Then:

[tex]\dfrac{128 \mu L_1Q_1}{\pi D_1^4} = \dfrac{128 \mu L_2Q_2}{\pi D_2^4}[/tex]

[tex]\dfrac{ L_1Q_1}{D_1^4} = \dfrac{ L_2Q_2}{D_2^4}[/tex]

[tex]\dfrac{ LQ_1}{D^4} = \dfrac{ 2LQ}{D^4}[/tex]

[tex]\mathbf{Q_1 = 2Q}[/tex]

The flow rate in pipe B is 2Q of the flow rate of the pipe A

The flow rate is defined as the flow of the fluid across the cross section in per unit time.

From the given information:

The pressure inside a pipe can be expressed by using the formula:

[tex]\Delta p=\dfrac{128\mu LQ}{\pi D^4}[/tex]

Since the diameter in both pipes is the same, we can say:

[tex]D=D_A=D_B[/tex]

where;

length of the first pipe A  [tex]L_A=L[/tex] and the length of the second pipe B  

[tex]L_B=2L[/tex]

Since the difference in pressure is equivalent in both pipes:

Then:

[tex]\dfrac{128\mu L_1Q_1}{\pi D_1^4}=\dfrac{128\mu L_2Q_2}{\pi D_2^4}[/tex]

[tex]\dfrac{L_1Q_1}{D_1^4}=\dfrac{L_2Q_2}{D_2^4}[/tex]

[tex]\dfrac{LQ_1}{D_1^4}=\dfrac{2LQ}{D_2^4}[/tex]

[tex]Q_1=2Q[/tex]

Hence the flow rate in pipe B is 2Q of the flow rate of the pipe A

To know more about Flow rate follow

https://brainly.com/question/26061120

Answer:

Explanation:

Newton's Second Law is pretty much the standard for all motion that involves a force. It applies to gravitational force and torque and friction and weight on an elevator. The main formula for force is

F = ma. We have to adjust that to take into account that when the elevator is moving up, that "surge" of acceleration weighs down a bit on the scale, causing it to read higher than the actual weight until the acceleration evens out and there is no acceleration at all (no acceleration simply means that the velocity is constant; acceleration by definition is a change in velocity, and if there is no change in velocity, there is 0 acceleration). The force equation then becomes

[tex]F_n-w=ma[/tex]  where [tex]F_n[/tex] is normal force. This is what the scale will read, which is what we are looking for in this problem (our unknown). Since we are looking for [tex]F_n[/tex], that is what we will solve this literal equation for:

[tex]F_n=ma+w[/tex] .  m is the mass of the boy, a is the acceleration of the elevator (which is going up so we will call that acceleration positive), and w is weight. We have everything but the unknown and the weight of the boy. We find the weight:

w = mg so

w = 28(9.8) and

w = 274.4 N BUT rounding to the correct number of significance we have that the weight is actually

w = 270 N.

Filling in the elevator equation:

[tex]F_n=28(.50)+270[/tex] and according to the rules of significant digits, we have to multiply the 28(.50) {notice that I did add a 0 there for greater significance; if not that added 0 we are only looking at 1 significant digit which is pretty much useless}, round that to 2 sig fig's, and then add to 170:

[tex]F_n=14+270[/tex] and adding, by the rules, requires that we round to the tens place to get, finally:

[tex]F_n=280N[/tex]  So you see that the surge in acceleration did in fact add a tiny bit to the weight read by the scale; conversely, if he were to have moved down at that same rate, the scale would have read a bit less than his actual weight). Isn't physics like the coolest thing ever!?

Answer:

answer

Explanation:

it is the answer which was presented in the year

Answer:

The correct answer is "[tex](0,300\times 10^{-3} \ N/C)[/tex]".

Explanation:

The given problem seems to be incomplete. Please find the attachment of the complete query.

According to the question,

At point A, we have

⇒ [tex]E_x = \frac{k q_1}{d_1^2} Cos \theta_1 - \frac{k q_2}{d_2^2} Cos \theta_2[/tex]

or,

⇒ [tex]E_x = 9\times 10^9\times [\frac{6\times 10^{-9}}{15^2}\times \frac{9}{15}-\frac{8\times 10^{-9}}{20^2}\times \frac{16}{20} ][/tex]

         [tex]=0[/tex]

and,

⇒ [tex]E_y = \frac{kq_1}{d_1^2}Sin \theta_1 +\frac{kq_2}{d_2^2}Sin \theta_2[/tex]

or,

⇒ [tex]E_y = 9\times 10^9\times [\frac{6\times 10^{-9}\times 12}{15^2\times 15}+ \frac{8\times 10^{-9}\times 12}{20^2\times 20} ][/tex]

         [tex]=0.3 \ N/C[/tex]

Solution :

Let [tex]$d_1=\frac{5.5}{2}[/tex]

          = 2.75 m

[tex]d_2 = 0.21 \ m[/tex]

And [tex]$d=|d_1-d_2|$[/tex]

       [tex]$d=(d_1+d_2) - (d_1-d_2)$[/tex]

       [tex]$d=(2.75+0.21) - (2.75-0.21)$[/tex]

       [tex]$d = 2.96-2.54$[/tex]

       [tex]d = 0.42 \ m[/tex]

a). At minimum,

[tex]$d=\frac{\lambda}{2}$[/tex]

[tex]$\lambda = 2d$[/tex]

  = 2 x 0.42

  = 0.84 m

Frequency, [tex]$\nu = \frac{v}{\lambda}$[/tex]

                      [tex]$=\frac{340}{0.84}$[/tex]

                      = 404.76 Hz

Therefore, the frequency of he sound, [tex]$\nu$[/tex] = 404.76 Hz

b). At maximum, λ = d = 0.42 m

Therefore, the frequency, [tex]$\nu = \frac{v}{\lambda}[/tex]

                                             [tex]$=\frac{350}{0.42}$[/tex]

                                             = 809.52 Hz

This question deals with projectile motion, which is a motion on both the x-axis and y-axis, simultaneously. The total time of flight of the projectile trajectory is given, while the time to reach the highest point of the projectile is required to be found.

The projectile will reach the highest point in "3 seconds".

The total time of flight of a projectile is the time during which the projectile remains in the air. For a projectile motion that ends up on the same horizontal level, from where it started, the time to reach the highest point, is equal to half of the total time of flight.

In other words, the projectile motion takes the same time, to go from the starting level to the highest point (i.e upward motion), as the time taken to reach the starting level from the highest point (i.e downward motion).

[tex]t = \frac{1}{2}T[/tex]

where,

t = time to reach the highest point = ?

T = total time of flight = 6 seconds

Therefore,

[tex]t - \frac{1}{2}(6\ seconds)[/tex]

t = 3 seconds

Learn more about the projectile motion here:

https://brainly.com/question/20689870?referrer=searchResults

Answer:

Three Types of Computer The Computer are classified into three main types:: • Analog Computers • Digital Computers • Hybrid Computers (Analog + Digital) Analog Computers:: Analog Computer measures “Physical Quantities” for example Temperature, Voltage, Pressure, and Electric Current.

Answer:

the answer should be b) 3.8 x 10-5 v

Answer:

Biosphere is the part of the earth where life exists.

Answer:

1. The image is real

2. 5.85

3. h' = 3.05 mm

4. The image is upright

Explanation:

1. Start with the first lens and apply 1/f = 1/p + 1/q

1/5.01 = 1/13.7 + 1/q

q = 7.90 cm

Since that distance is behind the first lens, and the second lens is 62.5 cm behind the first lens, that distance is 62.5 - 7.90 = 54.6 cm in front of the second lens, and becomes the object for that lens, thus,

1/25.9 = 1/54.6 + 1/q

q = 49.3 cm behind the second lens

Using that information, since q is positive, the image is real

2. Also, using that information, you have the second answer, which is 49.3 cm

The height can be found from the two magnifications.

m = -q/p

m1 = -7.9/13.7 = -.577

m2 = -49.3/54.6 = -.903

Net m = (-.577)(-.903) = .521

Then, m = h'/h

.521 = h'/5.85

3. h' = 3.05 mm

4. For the fourth answer, since the overall magnification is positive, the final image is upright

Explanation:

mass=kilogram,temperature=Klevin,power=watt,density=kilogram per cubic metre

Explanation:

the unit of mass is kg , temperature is kelvin ,power is watt and density is kilogram per cubic meter.

Answer:

a)   R₁ = 14.1 Ω,   b)  R₂ =  19.9 Ω

Explanation:

For this exercise we must use ohm's law remembering that in a series circuit the equivalent resistance is the sum of the resistances

all resistors connected

           V = i (R₁ + R₂)

with R₁ connected

           V = (i + 0.5) R₁

with R₂ connected

           V = (i + 0.25) R₂

We have a system of three equations with three unknowns for which we can solve it

We substitute the last two equations in the first

           V = i ( [tex]\frac{V}{ i+0.5} + \frac{V}{i+0.25}[/tex] )

           1 = i ( [tex]\frac{1}{i+0.5} + \frac{1}{i+0.25}[/tex] )

           1 = i ( [tex]\frac{i+0.5+i+0.25}{(i+0.5) \ ( i+0.25) }[/tex] ) =  [tex]\frac{i^2 + 0.75i}{i^2 + 0.75 i + 0.125}[/tex]

           i² + 0.75 i + 0.125 = 2i² + 0.75 i

           i² - 0.125 = 0

           i = √0.125

           i = 0.35355 A

with the second equation we look for R1

          R₁ = [tex]\frac{V}{i+0.5}[/tex]

          R₁ = 12 /( 0.35355 +0.5)

          R₁ = 14.1 Ω

with the third equation we look for R2

          R₂ = [tex]\frac{V}{i+0.25}[/tex]

          R₂ =[tex]\frac{12}{0.35355+0.25}[/tex]

          R₂ =  19.9 Ω

science is all about the world around us

Answer:

Both will have the same momentum.

P = M v     momentum

v = a t   for uniform acceleration

P = M a t

But a = F / M

P = M (F / M) t = F t    so both have the same momentum

Answer:

The elastic energy is 245 J.

Explanation:

Length, L = 5 m

Diameter, D = 10^-3 m

Stretch, l = 3 x 10^-4 m

Load, F = 10 x 9.8 = 98 N

Let the elastic energy is U.

[tex]U = \frac{1}{2}\times stress\times strain\times volume\\\\U = 0.5\times \frac{Force}{area}\times \frac{l}{L}\times Area\times L\\\\U = 0.5 \times F\times l\\\\U = 0.5\times 98\times 5\\\\U = 245 J[/tex]

Answer:

the initial velocity of the car is 12.04 m/s

Explanation:

Given;

force applied by the break, f = 1,398 N

distance moved by the car before stopping, d = 25 m

weight of the car, W = 4,729 N

The mass of the car is calculated as;

W = mg

m = W/g

m = (4,729) / (9.81)

m = 482.06 kg

The deceleration of the car when the force was applied;

-F = ma

a = -F/m

a = -1,398 / 482.06

a = -2.9 m/s²

The initial velocity of the car is calculated as;

v² = u² + 2ad

where;

v is the final velocity of the car at the point it stops = 0

u is the initial velocity of the car before the break was applied

0 = u² + 2(-a)d

0 = u² - 2ad

u² = 2ad

u = √2ad

u = √(2 x 2.9 x 25)

u =√(145)

u = 12.04 m/s

Therefore, the initial velocity of the car is 12.04 m/s

Answer:

d

Explanation:

i think it is rotational inertia

because analogue of mass in rotational motion is moment of inertia. It plays the same role as mass plays in transnational motion.  

hope it's right & helps !!!!!!!!!

Answer:

 emf = 312 V

Explanation:

In this exercise the electromotive force is asked, for which we must use Faraday's law

           emf =  [tex]- N \frac{d \Phi }{dt}[/tex]- N dfi / dt

           Ф = B. A = B A cos θ

bold type indicates vectors.

They indicate that the magnetic field is constant, the angle between the normal to the area and the magnetic field is parallel by local cosine values ​​1

It also indicates that the area is reduced from  a₀ = 0.325 me² to a_f = 0 in a time interval of ΔT = 0.100 s, suppose that this reduction is linear

            emf = -N B [tex]\frac{dA}{dT}[/tex]

            emf = - N B (A_f - A₀) / Dt

we calculate

           emf = - 60 1.60 (0 - 0.325) /0.100

           emf = 312 V

The direction of this voltage is exiting the page

Answer:

B) The same as the momentum change of the heavier fragment.

Explanation:

Since the initial momentum of the system is zero, we have

0 = p + p' where p = momentum of lighter fragment = mv where m = mass of lighter fragment, v = velocity of lighter fragment, and p' = momentum of heavier fragment = m'v' where m = mass of heavier fragment = 25m and v = velocity of heavier fragment.

0 = p + p'

p = -p'

Since the initial momentum of each fragment is zero, the momentum change of lighter fragment Δp = final momentum - initial momentum = p - 0  = p

The momentum change of heavier fragment Δp' = final momentum - initial momentum = p' - 0 = p' - 0 = p'

Since p = -p' and Δp = p and Δp' = -p = p ⇒ Δp = Δp'

So, the magnitude of the momentum change of the lighter fragment is the same as that of the heavier fragment.  

So, option B is the answer

Answer:

Explanation:

a)     Fx = F cos (θ)

           = (200) cos(60)

           = 100 N

b)     FR = ma

       Fx + Ff = ma

      100 + Ff = (50)(1,5)

       Ff     = 75 - 100

               =  -25 N

c)    Fy = F sin θ

           = (200) sin(60)

           = 173,2 N

Answer:

B. A car sits at rest at a stop sign.