When will the Social Security fund dry up at its current level?
A.
2022
B.
2027
C.
2042
D.
2097
E.
2117

Answers

Answer 1

Answer:

C. 2042

Explanation:

trust me I know the answer

Answer 2

In 2042 the Social Security fund will dry up at its current level. The correct option is C.

The answer is not entirely clear, as projections can vary depending on various economic and demographic factors. However, based on the 2021 Social Security Trustees Report, the projected depletion date for the Social Security Trust Fund is 2034. This means that at the current rate of payouts and contributions, the Trust Fund will only be able to pay out about 78% of scheduled benefits after 2034.

Therefore, the closest option to the projected depletion date of the Social Security Trust Fund is C. 2042. However, it is important to note that this projection may change as economic and demographic factors change over time.

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Related Questions

A speeding race car primarily contains potential energy.

:True
False

Answers

True because is aid sooooo

A balloon contains 0.075 m^3 of
gas. The pressure is reduced to
100kPa and fills a box of 0.45 m^3.
What is the initial pressure inside the
balloon if the temperature remains
constant?

Answers

Answer:

600 KPa.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V1) = 0.075 m³

Final volume (V2) = 0.45 m³

Final pressure (P2) = 100 KPa

Initial pressure (P1) =?

Temperature = constant

The initial pressure can be obtained by using the Boyle's law equation as shown below:

P1V1 = P2V2

P1 × 0.075 = 100 × 0.45

P1 × 0.075 = 45

Divide both side by 0.075

P1 = 45 / 0.075

P1 = 600 KPa.

Thus, the initial pressure in the balloon is 600 KPa.

A school bus has a mass of 18,200 kg. The bus moves at 13.5 m/s. How fast must a 0.142-kg baseball move in order to have the same momentum as the bus?

Answers

Answer:

bus momentum

p_bus= m_bus x v_bus

=18,200 x 16.5

basball momentum

pball=mball x vball

=0.142 x v

p_bus = pball

18200 x 16.5 = 0.142 x v

v=(18200 x 16.5)/0.142

v is the answer for baseball

Explanation:

⚠️not my answer tryna be honest here⚠️

The momentum of the bus of 18200 kg and velocity of 13.5 m/s is 245700 Kg m/s. To have equal momentum the base ball with 145 g have to throw in a speed of 1.7 × 10 ⁶  m/s.

What is momentum?

Momentum of a moving body is the product of mass and velocity. Thus it have the unit of g m/s or Kg m/s. Momentum is a vector quantity and thus having magnitude and direction.

Given that one bus is having a mass of  18200 Kg and 13.2  m/s speed. The momentum is:

p = mv

 =18200 kg × 13.5  m/s

 = 245700 Kg m/s

To have a momentum of 245700 Kg m/s   the base ball with 0. 142 g have to have a velocity  = 245700 Kg m/s   / 0.142 g

                          =1.7 × 10 ⁶  m/s

Hence, the baseball weighs0. 142 g have to move in 1.7 × 10 ⁶  m/s

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Part A:
The primary coil of a transformer contains 100 turns; the secondary has 200 turns. The primary coil is connected to a size-AA battery that supplies a constant voltage of 1.5 volts. What voltage would be measured across the secondary coil?
Part B:
A transformer is intended to decrease the value of the alternating voltage from 500 volts to 25 volts. The primary coil contains 200 turns. Find the necessary number of turns N2 in the secondary coil.
Part C:
A transformer is intended to decrease the value of the alternating current from 500 amperes to 25 amperes. The primary coil contains 200 turns. Find the necessary number of turns N2 in the secondary coil.
Part D:
In a transformer, the primary coil contains 400 turns, and the secondary coil contains 80 turns. If the primary current is 2.5 amperes, what is the secondary current I2?
Part E:
The primary coil of a transformer has 200 turns and the secondary coil has 800 turns. The power supplied to the primary coil is 400 watts. What is the power generated in the secondary coil if it is terminated by a 20-ohm resistor?
Part F:
A transformer supplies 60 watts of power to a device that is rated at 20 volts. The primary coil is connected to a 120-volt ac source. What is the current I1 in the primary coil?
Part G:
The voltage and the current in the primary coil of a nonideal transformer are 120 volts and 2.0 amperes. The voltage and the current in the secondary coil are 19.4 volts and 11.8 amperes. What is the efficiency e of the transformer? The efficiency of a transformer is defined as the ratio of the output power to the input power, expressed as a percentage: e=100Pout/Pin.

Answers

Answer:

a) 0 V

b) 10 turns

c) 4000 turns

d) 12.5 A

e) 400 W

f) 0.5 A

g) 95.4%

Explanation:

A

0

B

To solve this, we would be using the simple relationship between voltage and number of turns

V1/V2 = N1/N2

500/25 = 200/N2

20 = 200/N2

N2 = 200/20

N2 = 10 turns

C

Here also, we would be using the relationship between current and the number of turns

I1/I2 = N2/N1

500/25 = N2/20

20 = N2/20

N2 = 20 * 20

N2 = 4000 turns

D

Like in the previous question, current and the number of turn relationship is used

N1/N2 = I2/I1

400/80 = I2/2.5

5 = I2/2.5

I2 = 5 * 2.5

I2 = 12.5 A

E

The power remains unchanged at 400 W

F

Power = Voltage * Current

P = VI

I = P/V

I = 60/120

I = 0.5 A

G

95.4%

The transformer is a device used to step up or step down voltage.

Part A;

Given that;

Es/Ep = Ns/Np

Es = voltage in the secondary coil

Ep = voltage in primary coil

Ns = Number of turns in secondary coil

Np = Number of coils in primary coil

Es = Ns/Np ×  Ep

Es = 200/100  × 1.5 V

Es = 3 V

Part B

Ns = Es/Ep × Np

Ns = 25/500 × 200

Ns = 10 turns

Part C

Ns/Np = Ip/Is

Ns = Ip/Is × Np

Ns = 500/25 × 200

Ns = 4000 turns

Part D

Ns/Np = Ip/Is

NsIs = NpIp

Is = NpIp/Ns

Is = 400 × 2.5/80

Is =12.5 A

Part E

The power in the primary coil is the same as the power in the secondary coil. The power in the secondary coil is 400 watts.

Part F

Power supplied = 60 watts

Voltage of primary coil = 120 V

Since;

P = IV

I = P/V = 60/120 = 0.5 A

Part G

Since;

E = 100Pout/Pin

Pin = 120 V × 2 A = 240 W

Pout =  19.4 V ×  11.8 A = 228.92 W

E = 100(228.92/240)

E = 95.4%

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If you quadrupled the mass and tripled the radius of a planet, by what factor would gg at its surface change

Answers

Answer:

4/3

Explanation:

We already know aforehand that the acceleration due to gravity on the surface of a planet is given as

GM/r²,

Now, if the mass is quadrupled, we would have

G4M/r²,

If the radius is then tripled, we would have again

G4M/3r²

And that is what we have, now, to get the factor that the acceleration changes is simply to compare it with the primary equation

HM/r² = G4M/3r²

And thus, we find out that the factor is 4/3 or 1.333

which factor does not affect the strength of an electromagnet

Answers

Factors Affecting the Strength of the Magnetic Field of an Electromagnet: Factors that affect the strength of electromagnets are the nature of the core material, strength of the current passing through the core, the number of turns of wire on the core and the shape and size of the core.

Pls give Brainiest

Answer:

the placement of the ammeter in the circuit

Explanation:

A 1 200-kg automobile moving at 25 m/s has the brakes applied with a deceleration of 8.0 m/s2. How far does the car travel before it stops?

Answers

Answer:

Δx = 39.1 m

Explanation:

Assuming that deceleration keeps constant during the braking process, we can use one of the kinematics equations, as follows:

        [tex]v_{f} ^{2} - v_{o} ^{2} = 2* a * \Delta x (1)[/tex]

        where  vf is the final velocity (0 in our case), v₀ is the initial velocity

        (25 m/s), a is the acceleration (-8.0 m/s²), and Δx is the distance

        traveled since the brakes are applied.

Solving (1) for Δx, we have:

        [tex]\Delta x = \frac{-v_{o} ^{2} }{2*a} = \frac{-(25m/s)^{2}}{2*(-8.0m/s2} = 39.1 m (2)[/tex]        

The car will travel a distance of 39.1 m before its stops.

To solve the problem above, use the equations of motion below.

Equation:

v² = u²+2as................... Equation 1

Where:

v = final velocity of the automobileu = initial velocity of the automobilea = accelerations = distance covered

From the question,

Given:

v = 0 m/s (before its stops)u = 25 m/sa = -8 m/s² (decelerating)

Substitute these values into equation 1

⇒ 0² = 25²+2(-8)(s)

Solve for s

⇒ 0²-25² = -16s⇒ -16s = -625⇒ s = -625/16⇒ s = 39.1 m

Hence, The car will travel a distance of 39.1 m before its stops.

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What can you infer from the fact that metals are good conductors of electricity?

Answers

Answer:

Knowing that these metals are infact good conductors of electricity we can infer that metals are able to hold and conduct certain temperatures. Another thing we can infer is that these good conductors can be used in connection to transferring energy or electricity.

Metals are an excellent conductor of electricity and heat because the atoms in the metals form a matrix through which outer electrons can move freely. Instead of orbiting their respective atoms, they form a sea of electrons that surround the positive nuclei of the interacting metal ions.

Which equation is most likely used to determine the acceleration from a velocity vs. time graph?

Answers

acceleration is the change in velocity over the change in time so divide the change in velocity by the change in time and you will get you acceleration

Answer:

a = t over delta v.

Explanation:

The speed of revolution of particle going around a circlr is doubled and its angular speed is havled. What happen to the centripetal acceleration?
a) unchanged
b) doubles
c) halves
d) becomes four times​

Answers

Answer: The correct answer is C

Explanation:

A stone dropped from a bridge strikes the water 5.6 seconds later. What is the final velocity in meters/s?
A) 179.78 meters/s
B) 5.71 meters/s
C) 1.75 meters/s
D) 54.88 meters/s

Answers

Answer: 54.88 meters/s

Explanation:

The final velocity will be calculated by using the formula:

v = u + at

where,

v = final velocity

u = initial velocity = 0

a = 9.8

t = 5.6

Therefore, we slot the value back into the formula. This will be:

v = u + at

v = 0 + (9.8 × 5.6)

v = 0 + 54.88

v = 54.88 meters per second

Therefore, the final velocity is 54.88m/s

What is the average velocity of a train moving along a straight track if its displacement is 192 m was during a time period of 8.0 s

Answers

Answer:

The average velocity of a train moving along a straight track if its displacement is 192 m was during a time period of 8.0 s is 24 [tex]\frac{m}{s}[/tex].

Explanation:

Velocity ​​is a physical quantity that expresses the relationship between the space traveled by an object and the time used for it. Then, the average velocity relates the change in position to the time taken to effect that change.

[tex]velocity=\frac{displacement}{time}[/tex]

Velocity considers the direction in which an object moves, so it is considered a vector magnitude.

In this case, the displacement is 192 m and the time period is 8 s. Replacing:

[tex]velocity=\frac{192 m}{8 s}[/tex]

Solving:

velocity= 24 [tex]\frac{m}{s}[/tex]

The average velocity of a train moving along a straight track if its displacement is 192 m was during a time period of 8.0 s is 24 [tex]\frac{m}{s}[/tex].

A 2.80 kg mass is dropped from a height of 4.50 m. Find its potential energy when it reaches the ground.

Answers

Answer:

123.48J

Explanation:

Given parameters:

Mass of the ball  = 2.8kg

Height  = 4.5m

Unknown:

Potential energy  = ?

Solution:

The potential energy is the energy due to the position of a body. It is mathematically given as;

      P.E   =  mgh

m is the mass

g is the acceleration due to gravity

h is the height

 Now insert the parameters and solve;

          P.E  = 2.8 x 4.5 x 9.8  = 123.48J

Answer:

0

Explanation:

There is 0 PE when its on the ground

A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further it will penetrate before coming to rest assuming that it faces constant resistance to motion​

Answers

[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{First \: penetrating \: length\:(s_{1}) = 3 \: cm}[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: Penetration \: length \: before \: it \: comes \: to \: rest \:( s_{2} )}[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Let \: Initial \: velocity = v\:m/s} \\\\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Left \: velocity \: after \: s_{1} \: penetration = \dfrac{v}{2} \:m/s} \\\\ [/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{s_{1} = \dfrac{3}{100} = 0.03 \: m}[/tex]

[tex]\\[/tex]

☯ As we know that,

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ \bigg(\dfrac{v}{2} \bigg)^{2} = {v}^{2} + 2a s_{1}}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} = {v}^{2} + 2 \times a \times 0.03 }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ \dfrac{ {v}^{2} }{4} - {v}^{2} = 0.06 \times a }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{\dfrac{ - 3{v}^{2} }{4} = 0.06 \times a }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{a = \dfrac{ - 3 {v}^{2} }{4 \times 0.06} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ a = \dfrac{ - 25 {v}^{2} }{2}\:m/s^{2} ......(1) }[/tex]

[tex]\\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{ Initial\:velocity=v\:m/s} \\\\ [/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{ Final \: velocity = 0 \: m/s }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ {v}^{2} = {u}^{2} + 2as}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{{0}^{2} = {v}^{2} + 2 \times \dfrac{ - 25 {v}^{2} }{2} \times s }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ - {v}^{2} = - 25 {v}^{2} \times s }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s = \dfrac{ - {v}^{2} }{ - 25 {v}^{2} }}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s = \dfrac{1}{25} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s = 0.04 \: m }[/tex]

[tex]\\[/tex]

☯ For left penetration (s₂)

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{s = s_{1} + s_{2} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ 0.04 = 0.03 + s_{2}}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ s_{2} = 0.04 - 0.03 }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{s_{2} = 0.01 \: m = {\boxed{\sf{\purple{1 \: cm }}} }}[/tex]

[tex]\\[/tex]

[tex]\star\:\sf{Left \: penetration \: before \: it \: come \: to \: rest \: is \:{\bf{ 1 \: cm}}} \\ [/tex]

why do feet smell and noses run?

Answers

Answer:

Nose has mucous glands with hairs which helps the body in trapping pollutants and infectants from entering inside the body. On the other hand,our feet is composed of millions of sweat pores when dirt and other things accumulate,it smells because of sweat mixed with the dirt and other dirty things of the ground.

Explanation:

Hope this helps

Show that the pressure in liquids is given by density multiplied by gravity multiplied by height

Answers

Answer

The answer to your question is given below.

Explanation:

Pressure (P) = force (F) / Area (A)

P = F/A ........ (1)

Recall:

1. Force (F) = mass (m) × acceleration due to gravity (g)

F = mg

2. Volume (V) = Area (A) × Height (h)

V = Ah

Divide both side by h

A = V/h

Substitute the value of F and A into equation 1.

P = F/A

P = mg ÷ V/h

P = mg × h/V

P = mgh/V.... (2)

Recall:

Density (d) = mass (m) /volume (V)

d = m/V

Replace m/V in equation (2) with d.

P = mgh/V

P = dgh

Where:

P is the pressure.

d is the density.

g is acceleration due to gravity.

h is height.

Pressure = Density × gravity × height

An aluminum baking sheet with a mass of 225 g absorbs 2.4 x 104 J from an oven. If its temperature was initially 25 C, what will its new temperature be?

Answers

Answer:

The value is [tex]T_2 =416.9 \ K[/tex]

Explanation:

From the question we are told that

     The mass of the aluminum baking sheet is  [tex]m = 225 \ g = 0.225 \ kg[/tex]

      The energy absorbed is [tex]E = 2.4 *10^{4} \ J[/tex]

       The initial  temperature is  [tex]T_1 = 25 ^oC = 25 + 273 = 298 \ K[/tex]

   

Generally the heat absorbed is mathematically represented as

         [tex]Q = m * c_a * [T_2 - T_1][/tex]

Here  [tex]c_a[/tex] is the specific heat capacity of  aluminum with value  [tex]c_a = 897 \ J / kg \cdot K[/tex]

So

           [tex]2.4 *10^{4 } =0.225 * 897 * [ T_ 2- 298][/tex]

=>         [tex]T_2 - 298 = 118.915[/tex]

=>          [tex]T_2 =416.9 \ K[/tex]

What differentiates galaxy groups from clusters?
A.
Clusters are bigger than groups.
B.
Clusters are more massive than groups.
C.
Clusters contain a hot intracluster medium, whereas groups do not.
D.
Clusters are collections of galaxy groups, whereas groups are collections of galaxies.
E.
Clusters don't gravitationally bind galaxies together, while groups bind galaxies gravitationally.

Answers

Answer:

A

Explanation:

Galaxy clusters are basically very large (>50 galaxies) groups

Answer:

The correct answer would be:

C.

Clusters contain a hot intracluster medium, whereas groups do not.

#PLATOFAM

Have a nice day!

A ball is throw at an angle of 30 degrees off the horizontal, with an initial velocity of 28 m/s. what is the maximum height the ball will reach?​

Answers

[tex]{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Angle \ of \ projection = 30^{\circ} }[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ projectile = 28 \: m/s^{-1} }[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\[/tex]

[tex]\:\:\:\:\bullet\:\:\:\sf{Height_{\:(max)}\: reached\: by \:the \:projectile }[/tex]

[tex]\\[/tex]

[tex]{\mathfrak{\underline{\purple{\:\:\: Calculation:-\:\:\:}}}} \\ \\[/tex]

As we know that,

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ H = \dfrac{u^2\;sin^2\theta}{2\;g} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{H = \dfrac{(28)^2\;sin^2 30^{\circ}}{2\;(9.8)} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{H = \dfrac{784 \times \;sin^230^{\circ}}{19.6} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times sin^2 30^{\circ}}[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \bigg(\dfrac{1}{2}\bigg)^2 }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: \sf{ H = \dfrac{784}{19.6}\times \dfrac{1}{4} }[/tex]

[tex]\\[/tex]

[tex]\dashrightarrow\:\: {\boxed{\sf{H=10\:m }}}[/tex]

A 950 kg car rounds an unbanked curve at a speed of 25 m/s. If the radius of the curve is 72 m, what is the minimum coefficient of friction between the car and the road required so that the car does not skid?

Answers

Compute the car's weight:

W = m g = (950 kg) (9.8 m/s²) = 9310 N

The net vertical force on the car is

F = N - W = 0

so the normal force has magnitude

N = W = 9310 N

Then the friction force that keeps the car from skidding has magnitude F = µ N, where µ is the coefficient of friction, and it's friction that makes up the net horizontal force on the car. By Newton's second law, we have

F = m a

µ N = m v ² / R

µ (9310 N) = (950 kg) (25 m/s)² / (72 m)

µ ≈ 0.89

A pmdc has a stall torque of 10 and maximum mechanical power of 200. What is the maximum angular velocity?

Answers

Answer:

The maximum angular velocity is 20 rad/s

Explanation:

Given;

torque, τ = 10 N

maximum mechanical power, P = 200 J/s

The output power of the pmdc is given as;

P = τω

where;

P is the maximum mechanical power

ω is the maximum angular velocity

ω = P / τ

ω = (200) / (10)

ω = 20 rad/s

Therefore, the maximum angular velocity is 20 rad/s

g You heard the sound of a distant explosion (3.50 A/10) seconds after you saw it happen. If the temperature of the air is (15.0 B) oC, how far were you from the site of the explosion

Answers

Answer:

The answer is "1557 meters".

Explanation:

speed of sound in ([tex]\frac{m}{s}[/tex]) [tex]= 331.5 + 0.60 \ T^{\circ}\ C\\\\[/tex]

[tex]\to V = 331.5 + 0.6 \times 24 = 346 \frac{m}{s}\\\\\to t = 4.5 \ seconds \\\\\to S = vt = 346 \times 4.5 = 1557 \ meters[/tex]


PLEASEEEE

Calculate the mechanical advantage of a ramp if the box you are trying to move has a mass of 10 kilograms, the
board is 15 feet long and the height of the ramp is 5 feet.

Answers

Answer:

add answer +5 so so so so so

Answer:

3 trust me

Explanation:

A deuterium atom is a hydrogen atom with a neutron added to its nucleus. Approximate the binding energy of this nucleus, given that the mass of the deuterium atom is 2.014102 u and the masses of a hydrogen atom and a neutron are 1.007825 u and 1.008665 u, respectively.
a. 2 GeV.
b. 2 keV.
c. 2 MeV.
d. 2 eV.

Answers

Answer:

c. 2 MeV.

Explanation:

The computation of the binding energy is shown below

[tex]= [Zm_p + (A - Z)m_n - N]c^2\\\\=[(1) (1.007825u) + (2 - 1 ) ( 1.008665 u) - 2.014102 u]c^2\\\\= (0.002388u)c^2\\\\= (.002388) (931.5 MeV)\\\\=2.22 MeV[/tex]

= 2 MeV

As 1 MeV = (1 u) c^2

hence, the binding energy is 2 MeV

Therefore the correct option is c.

We simply applied the above formula so that the correct binding energy could come

And, the same is to be considered

to determine the height of a steep cliff an experimenter stations a sensor on the top of the cliff then fires a pellet vertically upward with an initial velocity of 80 m/s . the sensor reports that the pellet reached a maximum height 3 meters above the edge of the cliff. how high is the cliff?​
a. 77 m
b. 237 m
c. 317 m
d. 637 m
e. 797 m​

Answers

Answer:

c. 317 m

Explanation:

Vertical Launch Upwards

It occurs when an object is launched vertically up without taking into consideration any kind of friction with the air.

If vo is the initial speed and g is the acceleration of gravity, the maximum height reached by the object is given by:

[tex]\displaystyle h_m=\frac{v_o^2}{2g}[/tex]

We'll assume the acceleration of gravity as [tex]g=10\ m/s^2[/tex].The pellet is vertically upward launched with vo=80 m/s. The maximum height is:

[tex]\displaystyle h_m=\frac{80^2}{2*10}=320[/tex]

[tex]h_m = 320\ m[/tex]

That height is 3 meters above the edge of the cliff, thus the cliff is 320-3=317 m hight

c. 317 m

Two cars each have a mass of 1050 kg. If the gravitational force between
them is 2.27 x 10-7N, how far apart are they? G = 6.67 10-11 N:(m/kg)2
A. 21 m
B. 5.6 m
C. 33 m
D. 18 m
DUBMIT

Answers

Answer: D.

Explanation:

d = ± √ G m 1 m 2 /F

 Because distance cannot be negative there it goes:

d = √ G m 1 m 2 /F

d = √ 6.67 ⋅ 10 − 11 ⋅ 1050 ⋅ 1050 2.27 ⋅ 10 − 7

d = √ 0.0000735 2.27 ⋅ 10 − 7  

d = √ 3239504.405

d ≈ 1799.86 m = 18m

WHAT DOES DENSITY HAVE TO DO WITH PLATE TECTONICS?
Explain

Answers

Answer: The reason for the differences in density is the composition of rock in the plates. When two plates come in contact with each other through plate tectonics, scientists can use the density of the plates to predict what will happen. Whichever plate is more dense will sink, and the less dense plate will float over it.

Explanation:

Hope this helps ( not copied and pasted, this answer was done by me so I don't know if it's good or not)

A 5 kg block rests on an inclined plane with a coefficient of static friction equal to 0.30. What is the minimum angle at which the block will begin to slide

Answers

Answer:

[tex]\theta = 16.70 ^{\circ}[/tex]

Explanation:

The coefficient of static friction is equal to the tangent of the minimum angle at which an object will begin to start sliding down a ramp.  

[tex]\displaystyle u_s=\frac{F_f}{F_N} = \frac{F_g\ \text{sin}\theta}{F_g\ \text{cos} \theta} = \text{tan} \theta[/tex]

Since we are given the coefficient of static friction we can solve for the minimum angle that the block will begin to slide.

Let's solve for the force of gravity that is acting on the block. The force of gravity is also known as the weight force, which can be calculated by using w = mg.

[tex]w=mg[/tex]

We are given the mass of the block (kg) and we know that g = 9.8 m/s².

[tex]w=(5)(9.8) = 49 \ \text{N}[/tex]

Now we can use this force in the equation:

[tex]\displaystyle u_s = \frac{F_g \ \text{sin} \theta }{F_g \ \text{cos} \theta}[/tex]

Plug [tex]\displaystyle u_s = 0.30[/tex] and 49 N into the equation.

[tex]\displaystyle 0.30 = \frac{(49) \ \text{sin} \theta }{(49) \ \text{cos} \theta}[/tex]  [tex]0.30=\text{tan} \theta[/tex]

Notice that the gravitational force cancels out in the end, so we can actually start with [tex]0.30=\text{tan} \theta[/tex].

Evaluate this equation by taking the inverse tangent of both sides of the equation.

[tex]\text{tan}^-^1 (0.30) = \text{tan}^-^1 (\text{tan}\theta)[/tex] [tex]\text{tan}^-^1 (0.30) =\theta[/tex] [tex]\theta = 16.69924423[/tex]

The minimum angle at which the block will begin to slide is about 16.70 degrees.

10. A boy weighs 475 N. What is his mass? (acceleration due to gravity on Earth is 9.8m/s2 = g)


Answers

Answer: mass = 48.47 kg.

Explanation:

Formula : Weight = mg  , where m = mass of body , g= acceleration due to gravity .

Given: Weight  = 475 N

[tex]g= 9.8\ m/s^2[/tex]

Substitute all values in formula ,  we get

[tex]475= m \times9.8\\\\\Rightarrow\ m = \dfrac{475}{9.8}\\\\\Rightarrow\ m \approx 48.47\ kg[/tex]

Hence, his mass = 48.47 kg.

20 pts

Which of the following statements is true?
1..LIghtning is a form of statlc energy.

2..Water Is a conductor of electricIty.

3Electricity must flow in a complete circuit.

4 all of the above

Answers

Answer: 2

Explanation:

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