Newton's Law of Gravitation can be used to show that if an object weighs w pounds on the surface of the earth, then its weight at a distance x from the center of the earth is W(x)=wR2x2 (for x≥R), where R=3960 miles is the radius of the earth. Estimate the altitude at which a 120- lb pilot would weigh 119.5 lb

Answer:

θ = 41.8º

Explanation:

This is an internal total reflection exercise, the equation that describes this process is

         sin θ = n₂ / n₁

where n₂ is the index of the incident medium and n₁ the other medium must be met n₁> n₂

        θ = sin⁻¹ n₂ / n₁

let's calculate

       θ = sin⁻¹ (1.00 / 1.50)

       θ = 41.8º

Answer:

The minimum distance to the object that the radar can detect is 124.995 m

Explanation:

Here, we are to calculate the shortest distance to an object the radar in the question can detect.

Mathematically;

v = c/λ

Where v is the frequency, c is the speed of light and λ is the wave length

Thus;

v = (3 * 10^8)/2.5 = 1.2 * 10^8 Hz

Mathematically, the time period

t = 1/v = 1/(1.2 * 10^8) = 0.000000008333 = 8.333 * 10^-9 sec

From the question, we are told that the transmitting time is 100 times a single oscillation

Transmitting time = 100 * one oscillation

Hence Transmitting time = 100 * 8.33 * 10^-9 = 8.33 * 10-7

Mathematically;

Minimum distance =( Transmitting time * speed of light)/2 =

(8.33 * 10^-7 * 3 * 10^8)/2 = 124.995 m

Answer:

No. A vacuum does not exist at an altitude of 75,000 ft

Explanation:

There is naturally a decrease in air pressure as one goes higher into the atmosphere. However, vacuum conditions are not actually present within the atmosphere as there are still significant amounts of gases even at really high altitudes.

To prove this we know that planes cannot fly in a vacuum, because they need to burn air in order to gain propulsion from their engines.

One military plane - the SR-71 flew at about 90,000 feet.

This is a lot higher than 75,000 feet. For it to be able to at that altitude, it means that there was a significant quantity of air left even at 90,000 feet.

Answer:

average speed =  66.67  km/h

Explanation:

In order to find the average speed of the train, you need to calculate the total distance traveled, divided by the time it took to cover that distance. So for the total distance:

Distance= 120 km + 180 km = 300 km

For the total time we need to add three different quantities, two of which we need to derived based on the information provided:

time for first part of the trip:

[tex]time_1=\frac{D_1}{v_1} =\frac{120}{60} \,h= 2\,h[/tex]

for the time of the stop:

[tex]time_2=0.5\,\,h[/tex]

for the last part of the trip:

[tex]time_3=\frac{180}{90} \,h= 2 \,\,h[/tex]

Which gives a total of 4.5 hours

Then, the average speed is:   300/4.5   km/h = 66.67  km/h

Answer:

First, let's think in the vertical problem:

The acceleration will be the gravitational acceleration:

g = 9.8 m/s^2

a = -9.8 m/s^2

For the velocity, we integrate over time:

v(t) = (-9.8 m/s^2)*t + v0

Where v0 is the initial velocity, in this case  v0 = 30m/s.

v(t) =  (-9.8 m/s^2)*t + 30m/s

Now, for the position we integrate again over time, and get:

P(t) = (1/2)*(-9.8 m/s^2)*t^2 + 30m/s*t + p0

Where p0 is the initial position, as the ball is launched from the ground, we can use p0 = 0m

p(t) = (-4.9m/s^2)*t^2 + 30m/s*t

Now, the maximum vertical height is reached when:

v(t) = 0m/s = -9.8m/s^2*t + 30m/s

t = 30m/s/9.8m/s^2 = 3.06s

Now we can evaluate the vertical position in t = 3.06s

p(3.06s) = (-4.9m/s^2)*(3.06)^2 + 30m/s*3.06 = 62m

So, rounding down, the correct option is: C. 60 m

Explanation:

(b) We know that,

1 day = 24 hours

1 hour = 3600 s

So, we found that, 1 day = 86400 s

We need to find the 360 days into seconds. So,

1 day = 86400 s

360 days = 86400×360

360 days = 31104000 seconds

(d) Weight of a body, W = 600 N

Acceleration due to gravity on mars is 3.7 m/s²

Weight, W = mg

m is mass of body

[tex]m=\dfrac{W}{g}\\\\m=\dfrac{600}{3.7}\\\\m=162.16\ kg[/tex]

(e) Mass of body, m = 100 kg

Acceleration due to gravity on the moon, 1.6 m/s²

Weight, W = 100 × 1.6

W = 160 N

Answer:

[tex]D.\ 970\ N[/tex]

Explanation:

Given:

Mass of Piano: 99kg

Required:

Calculate its weight

The weight of an object is calculate as thus;

[tex]Weight = Mass\ (m) * Acceleration\ due\ to\ gravity\ (g)[/tex]

[tex]m = 99kg[/tex] and [tex]g = 9.8m/s^2[/tex]

The formula becomes

[tex]Weight = 99kg * 9.8m/s^2[/tex]

[tex]Weight = 970.2\ kgm/s^2[/tex]

[tex]Weight = 970.2\ N[/tex]

[tex]Weight = 970\ N[/tex] Approximated

Hence, the weight of the piano is 970N

Answer:

[tex]E=6.01\times 10^{-3}\ J[/tex]

Explanation:

It is given that,

Side of a square loop, l = 24 cm = 0.24 m

Resistance of loop, R = 6.1 ohms

Initial magnetic field is 0.665 T and final magnetic field is 0 as the field is removed in 40 ms

We need to find the electrical energy dissipated in this process.

Due to change in magnetic field, the loop will induce a voltage. The induced voltage is given by :

[tex]V=-\dfrac{dB}{dt}\\\\V=\dfrac{BA}{t}[/tex]

If I is induced current then,

[tex]V=IR[/tex]

[tex]I=\dfrac{V}{R}\\\\I=\dfrac{BA}{tR}[/tex]

Power is given by voltage times current. So,

[tex]P=\dfrac{(BA)^2}{(t^2R)}[/tex]

Now, energy is given by the product of power and time. So,

[tex]E=\dfrac{(BA)^2}{(t^2R)}\times t\\\\E=\dfrac{(BA)^2}{(tR)}[/tex]

Now putting all the values in above formula. So,

[tex]E=\dfrac{(0.665\times (0.24)^2)^2}{(40\times 10^{-3}\times 6.1)}\\\\E=6.01\times 10^{-3}\ J[/tex]

So, the electrical energy of [tex]6.01\times 10^{-3}\ J[/tex] is dissipated in this process.

The electrical energy dissipated throughout this process will be "6.01 × 10⁻³ J".

According to the question,

Square loop's side, l = 24 cm or,

                                  = 0.24 m

Loop's resistance, R = 6.1 ohms

Initial magnetic field = 0.665 T

Final magnetic field = 0

We know the relation,

→ V = - [tex]\frac{dB}{dt}[/tex]

      = [tex]\frac{BA}{t}[/tex]

Also we know,

Current, V = IR

               I = [tex]\frac{V}{R}[/tex]

                 = [tex]\frac{BA}{tR}[/tex]

Now, Energy, E = [tex]\frac{(BA)^2}{t^2R}[/tex] × t or,

                          = [tex]\frac{(BA)^2}{tR}[/tex]

By substituting the values,

                          = [tex]\frac{(0.665\times (0.24)^2)^2}{40\times 10^{-3}\times 6.1}[/tex]

                          = 6.01 × 10⁻³ J

Thus the response above is correct.                                  

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Answer:

Stopping distance = 40m

Explanation:

Given the following :

Initial speed of vehicle before applying brakes = 72km/hr

Converting km/hr to m/s:

72km/hr = [(72 * 1000)m] / (60 * 60)

72km/hr = 72,000m / 3600s

72km/hr = 20m/s

Deceleration after applying brakes (-a) (negative acceleration) = - 5m/s^2

From the 3rd equation of motion:

v^2 = u^2 + 2as

Where v = final Velocity ; u= Initial Velocity ; a = acceleration and s = distance

Final velocity when the car stops will be 0

Therefore ;

v^2 = u^2 + 2as

0 = 20^2 + 2(-5)(s)

0 = 400 - 10s

10s = 400

s = 400/10

s = 40m

Therefore, the stopping distance of the car = 40 meters

Answer:

12 Ω

Explanation:

Data obtained from the question include:

Resistor 1 (R1) = 50 Ω.

Resistor 2 (R1) = 30 Ω.

Resistor 3 (R3) = 20 Ω.

Resistor 4 (R4) = R

Galvanometer reading = Zero deflection.

The Resistor 4 (R4) in the Wheatstone bridge can be obtained as follow:

Since the galvanometer gives zero deflection, it means the bridge is balanced. Therefore,

R1/R2 = R3/R4

50/30 = 20/R

Cross multiply

50 x R = 30 x 20

Divide both side by 50

R = (30 x 20)/50

R = 12 Ω

Therefore, the value of R in the wheatstone bridge is 12 Ω.

Answer:

8.8 mT

Explanation:

Current through the conductor = 12.5 A

Magnetic force per unit length on the wire = 0.110 N/m

Recall that the magnetic force per unit length on a current carrying conductor is in a uniform magnetic field is

[tex]\frac{F}{l}= IBsin\alpha[/tex]

where B is the magnetic field magnitude

[tex]I[/tex] is the current in the conductor

α is the angle the conductor makes with the  magnetic field ( since it is perpendicular in this case, α is 90°)

imputing values into the equation, we'll have

0.11 = 12.5 x B x sin 90°

but sin 90° = 1, therefore, we have

0.11 = 12.5B

B = 0.11/12 = 8.8 x 10^-3 T

or rather = 8.8 mT

Answer:

(a) Error: Other's ; Correction: Others'

(b) Error: to ; Correction: for

(c) Error: with ; Correction: of

(d) Error: prepares ; Correction: prepare

Explanation:

a)  The error is in the word "other's" as the position of apostrophe is wrong, so the correct word will be "others'", it shows plural nouns.

b)  The error is in the word "to", so the correct word will be "for" as for is use to talk about a purpose.

c)  The error is in the word "with"  and the correct word will be "of" as of indicates relationships between other words including things that made of other things.

d)  The error is in the word "preapres" and the correct word will be "prepare".

Answer:

2 m/s²

30 m/s

Explanation:

1) F = ma

1000 N = (500 kg) a

a = 2 m/s²

2) Given:

v₀ = 0 m/s

a = 2 m/s²

t = 15 s

Find: v

v = at + v₀

v = (2 m/s²) (15 s) + 0 m/s

v = 30 m/s

Answer:

v = 46.67 km/h

Explanation:

We will use the following formula throughout this numerical:

s = vt

where,

s = distance covered

v = speed

t =  time taken

FOR FIRST 30 km:

s = 30 km

v = 30 km/h

t = t₃₀ = ?

Therefore,

30 km = (30 km/h)(t₃₀)

t₃₀ = (30 km)/(30 km/h)

t₃₀ = 1 h

FOR TOTAL 100 km:

s = 100 km

v = 40 km/h (Average Speed)

t = total time = ?

Therefore,

100 km = (40 km/h)(t)

t = (100 km)/(40 km/h)

t = 2.5 h

FOR LAST 70 km:

s = 70 km

t₇₀ = t - t₃₀ = 2.5 h - 1 h = 1.5 h

v = v₇₀ = ?

Therefore,

70 km = v(1.5 h)

v = 70 km/1.5 h

v = 46.67 km/h

Answer:

D. Mars is a long distance away

Answer:

[tex]F_{net} = 1905 N[/tex]

[tex]\boxed{a = 7.62 m/s^2}[/tex]

Explanation:

For net force:

Firstly, finding the force of friction:

[tex]F_{k} =[/tex] [tex](micro)_{k} m g cos \theta[/tex]

Where [tex](micro)_{k}[/tex] is the coefficient of friction, m =  250 kg and g = 9.8 m/s²

[tex]F_{k} = (0.17)(25)(9.8) cos (52)\\F_{k} = (41.65)(0.616)\\F_{k} = 25.59 \ N[/tex]

Now, Finding of the pulling box:

[tex]F = mgsin\theta[/tex]

[tex]F = (250)(9.8) Sin (52)\\F = 2450 * 0.788\\F = 1930.6[/tex]

So, The net Force is

[tex]F _{net} = F - F_{k}[/tex]

[tex]F_{net} = 1930.6 -25.6\\[/tex]

[tex]F_{net} = 1905 N[/tex]

For Acceleration:

[tex]F_{net} = m a[/tex]

For a, it is:

=> [tex]a = \frac{F_{net}}{m}[/tex]

=> a = 1905 / 250

=> a = 7.62 m/s²

Answer:

A.) Find the attached file

B.) 1.536 J

C.) 123N

Explanation:

Given that the force (F) in Newtons, of his breath as a function of the distance along the straw (x) in meters, can be modeled as a linearly decreasing function for the first half of the straw,

Half of the straw = 12.5 cm

Let Force F = dependent variable

And distance x = independent variable

A.) Find the attached file for the graph.

Sinnce the force decreases by half along the first half of the straw. Assume there is negligible friction and the straw is held horizontally

B.) Given that the mass M = 12g = 12/1000 = 0.012kg

And Velocity V = 16m/s

The workdone = the kinetic energy of the split ball

WD = 1/2mv^2

Substitutes m and V into the formula

WD = 1/2 × 0.012 × 16^2

WD = 1.536 Joule

C.) The maximum force acts on the spit ball will be at maximum kinetic energy.

Since work done = force × distance

Where distance = 12.5 cm

F × S = 1/2mv^2

Substitutes all the parameters

F × 0.0125 = 1.536

Make F the subject of formula

F = 1.536/0.0125

F = 122.88 N

The maximum force F0, that acts on the spitball is therefore equal to 123 N approximately.

We have that for the Question "(a) Sketch a plot of the force of his breath as a function of position along the straw, labeling the force at x = 0 as F0.

(b) If the spitball begins from rest and leaves the straw with a speed of 16 m/s, how much work is done on the spitball?

(c) What is the maximum force F0, that acts on the spitball?"

It can be said that

From the question we are told

mass of spitball = 12g through a 25.0 cm straw

Generally the equation for workdone is mathematically given as

[tex]= \frac{1}{2}mv^2 - 0\\\\= \frac{1}{2} * 12*10^{-3} * 16^2\\\\= 1.536J[/tex]

Maximum force

[tex]w = \frac{1}{2}*\frac{x_0}{2}*\frac{F_0}{2} + x_0*\frac{F_0}{2}\\\\w = \frac{5}{8}x_0F_0[/tex]

Therefore,

[tex]F_0 = \frac{8w}{5x_0}\\\\F_0 = \frac{8*1.536}{5*25*10^[-2]}\\\\F_0 = 6.4N[/tex]

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Explanation:

Given,

I = 5 A

V = 240 V

T = 25 mins = 1500 sec

Now,

Energy dissipated = IVT= 5×240×1500 = 1800000 J

Answer:

1) a.  52.41 m/s

b. The skier will be going 15.35 m/s slower

2)  103.68 m

3) 35,127 J

4) a.  88.825 kJ

(b) 16.36 %

5) 3,071.12 J

Explanation:

1) a. The given height of the hill, h = 140.0 m

The mass of the skier at the top of the hill, m = 85.0 kg

The acceleration due to gravity, g = 9.81 m/s²

The initial potential energy, P.E of the skier = m×g×h = 85.0×140.0×9.81 = 116739 J

From the principle of conservation of energy, we have;

The potential energy, P.E. lost = The gain in kinetic energy, K.E.

m×g×h = 1/2×m×v²

116739 J = 1/2×85.0×v²

v² = 116739/(1/2*85.0)= 2746.8 m²/s²

v = √(2746.8 m²/s²) = 52.41 m/s

b. From 70 m up, we have;

The initial potential energy, P.E., of the skier is now = 85.0×70×9.81 = 58,369.5 J

The potential energy, P.E. lost = The gain in kinetic energy, K.E.

58,369.5 J = 1/2×85.0×v²

v² = 58,369.5/(1/2*85.0) = 1373.4 m²/s²

v = 37.06 m/s

The skier will be going 52.41 - 37.06 = 15.35 m/s slower

The skier will be going 15.35 m/s slower

2) From the principle of conservation of energy, the amount of work done (energy used) = The (potential) energy gained by the load

The amount of work done by the electric hoist = 356,000 J

The mass of the load = 350.0 kg

The height to which the load is raised = h

The potential energy gained by the load = m×g×h = 350.0×9.81×h

356,000 J = 350.0×9.81×h

h = 356,000/(350.0*9.81) = 103.68 m

The height to which the load is lifted= 103.68 m

3) The initial potential energy of the roller coaster cart = 600*35.0*9.81 = 206010 J

The final potential energy = 600*28.0*9.81= 164808 J

The velocity at point 3  = 4.5 m/s

The kinetic energy at point 3 = 1/2*600*4.5^2 = 6075 J

The total energy at point 3 = 164808 + 6075 = 170,883 J

The energy loss = The initial potential energy at point 1 - Total energy at point 3

The energy loss = 206010 - 170,883 = 35,127 J

The heat energy due to friction that must have been produced between points 1 and 3 = 35,127 J

4) a. The heat energy absorbed = mass × specific heat capacity for water, [tex]C_{water}[/tex] × Temperature change

The mass of the water = 2.5×10² g = 0.25 kg

[tex]C_{water}[/tex] = 4,180 J/(kg·°C)

Initial temperature = 10.0°C

Final temperature = 95°C

The temperature change = 95.0°C - 10.0°C = 85.0°C

The heat energy absorbed = 0.25*4,180* 85 = 88,825 J = 88.825 kJ

(b) The percentage efficiency = (Heat absorbed/(Heat supplied)) × 100

The heat supplied = 543 kJ

The efficiency = (88.825/543) × 100  = 16.36 %

5) The mass of the box = 115 kg

Force acting on the rope = 255 N

The angle of inclination of the force to the horizontal = 24.5°

The distance the box is displaced = 15.0 m to the right

The work done = Force applied × distance moved in the direction of the force

The work done = Force applied × distance moved in the direction of the force

Given that the load moves a distance 15.0 m to the right,we have;

The component of the force acting in the direction of the movement of the load (to the right) is 225 × cos(24.5°) =  204.74 N

The work done = 204.7*15 = 3071.12 J

The amount of work done  = 3,071.12 J

Answer:

According to super hero logic , nothing will happen to him.

But according to science , yes he will get current shock but good news is that he wouldn't get elected until he is in contact with the wires.

He may / may not be affected but his suit will be damaged for sure as it is made of metal.

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Answer:

BETTER EXPLANATION

Explanation:

When you drink from a straw, you create a little space of low pressure inside your mouth and in the top of the straw. Then the air outside the straw pushes down on the surface of the drink and forces the liquid up through the straw and into your mouth.

Answer:

210 m

Explanation:

The speed of train 2 relative to train 1 is 15 m/s + 20 m/s = 35 m/s.

It takes 6 seconds for the train to pass, so the length of the train is:

(35 m/s) (6 s) = 210 m

Answer:

On heating gases expands than solid and liquid because the molecules gets seperated and the volume increases that causes to expand.

Explanation:

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Answer:

900 cm/s or 9 m/s.

Explanation:

Data obtained from the question include the following:

Length (L) = 30 cm

frequency (f) = 60 Hz

Velocity (v) =.?

Next, we shall determine the wavelength (λ).

This is illustrated below:

Since the wave have 4 node, the wavelength of the wave will be:

λ = 2L/4

Length (L) = 30 cm

wavelength (λ) =.?

λ = 2L/4

λ = 2×30/4

λ = 60/4

λ = 15 cm

Therefore, the wavelength (λ) is 15 cm

Now, we can obtain the speed of the wave as follow:

wavelength (λ) = 15 cm

frequency (f) = 60 Hz

Velocity (v) =.?

v = λf

v = 15 × 60

v = 900 cm/s

Thus, converting 900 cm/s to m/s

We have:

100 cm/s = 1 m/s

900 cm/s = 900/100 = 9 m/s

Therefore, the speed of the wave is 900 cm/s or 9 m/s.

Explanation:

A theory is a way of explaining an object or event using a set of facts.

Answer:

The teacher should swim in a direction 29.24° North of East

Explanation:

Given that the there is a water  current across the lake, and the physics teacher intends to swim directly across the lake, the direction the physics teacher will have to swim is found as follows;

The speed of the water current is given by the speed of the floating leaf traveling with the water current  

Distance traveled by the leaf = 4.2 m South

Time of travel of the leaf = 5.0 s

Speed of leaf = 4.2/5 = 0.84 m/s = Speed of the water current

Swimming peed of the teacher, v = 1.9 m/s

To swim directly across the lake, the teacher has to swim slightly in the opposite direction of the water current, the y-component of the teacher's swimming speed should be equal to and opposite that of the speed of the water current.

Y-component of v = v×sin(θ), where θ is the angle of the direction, the teacher should swim

Therefore;

1.9 × sin(θ) = 0.84

sin(θ) = 0.84/1.9 = 0.44

θ = 26.24°

That is the teacher should swim in a direction 29.24° North of East.

To cross the lake the teacher has to swim in a direction 29.24° North of the East

The speed of the water current can be derived from the speed of the floating leaf :

The distance traveled by the leaf L = 4.2 m South

Time taken T = 5s

So, the speed of the leaf is:

u = 4.2/5

u = 0.84 m/s South

So, the speed of the current is 0.84 m/s South

Now, it is given that the speed of the teacher is, v = 1.9 m/s East

To cross the lake the speed of the teacher must be in a Northeast direction so that the North component of the speed of the teacher cancels out the speed of the current which is directed towards the South.

Let, the speed of the teacher makes an angle of θ from the EAST.

So, the North component is given by:

v(north) = vsinθ

it must be equal to the speed of the current:

vsinθ = u

1.9 × sinθ = 0.84

sinθ = 0.84/1.9

sinθ = 0.44

θ = 26.24°

The teacher should swim in a direction 29.24° North of East.

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Answer:

He should aim it at the monkey

Explanation:

The question is related to projectile motion of an arrow from a bow such that the arrow is able to reach a specific target

Therefore, since the monkey is going to drop from the tree the instant the person launches the food, then for the monkey to be able reach the food, we have;

Final level position of the launched food should be ≥ The level of the monkey

As the monkey is dropping from the tree the destination of the launched food should be at the monkey so that before the food gets to the position where the monkey was when launching the food with the arrow, the monkey will be slightly lower so as to be able to catch the food or return up the tree to get the food rather than the food falling to the ground.

Answer:

(a) 1900 kg m²

(b) 8950 Nm

Explanation:

(a) The moment of inertia of a rod about its end is I = ⅓mL².

For 3 rods of mass m = 135 kg and length L = 3.75 m, the total moment of inertia is:

I = 3 (⅓ (135 kg) (3.75 m)²)

I = 1900 kg m²

(b) Net torque = moment of inertia × angular acceleration

∑τ = Iα

First, find the angular acceleration.

ω₀ = 0 rad/s

ω = 6.0 rev/s (2π rad/rev) = 37.7 rad/s

t = 8.0 s

α = (37.7 rad/s − 0 rad/s) / 8.0 rad/s = 4.71 rad/s²

∑τ = Iα

∑τ = (1900 kg m²) (4.71 rad/s²)

∑τ = 8950 kg m² / s²

∑τ = 8950 Nm

Answer:

Option D. 4.4 m/s²

Explanation:

The following data were obtained from the question:

Velocity (v) = 21 m/s

Radius (r) = 100 m

Centripetal acceleration (a) =.?

The centripetal acceleration of the car can be obtained as follow:

Centripetal acceleration (a) = Velocity square (v²) / radius (r)

a = v²/r

a = 21²/100

a = 441/100

a = 4.41 ≈ 4.4 m/s²

Therefore, the centripetal acceleration of the car is 4.4 m/s².