A planet has a circular orbit around a star. It is a distance of 53,000,000 km from the centre of the star.
It orbits at an average speed of 49,000 km/h.
How many Earth days does it take the planet to orbit the star?
Give your answer to 2 sf.

Answer:

a. 164°F

b. [tex]91.\overline 1 \ ^{\circ} C[/tex]

c. [tex]140.\overline 4[/tex] kJ

Explanation:

The starting temperature of the water, T₁ = 48F

The temperature at which the water boils, T₂ = 212°F

a. The difference between the initial and the boiling water temperature, ΔT = T₂ - T₁

Therefore;

ΔT = 212°F - 48°F = 164°F

The temperature by which he temperature must be raised, ΔT = 164°F

b. 48°F = ((48 - 32)×5/9)°C = (80/9)°C = [tex]8.\overline 8 \ ^{\circ} C[/tex]

212°F = ((212 - 32)×5/9)°C = 100°C

∴ ΔT = 100°C - [tex]8.\overline 8 \ ^{\circ} C[/tex] = [tex]9.\overline 1 \ ^{\circ} C[/tex]

c. The heat capacity of the water = The heat required to increase four quartz of water by 1 °C = 15.8 kJ

∴ The heat required to raise four quartz of water by [tex]9.\overline 1 \ ^{\circ} C[/tex], ΔQ = 15.8 kJ/°C × [tex]9.\overline 1 \ ^{\circ} C[/tex] = [tex]140.\overline 4[/tex] kJ.

On a hot dry day, the amount of water vapour present in atmosphere is less. Thus, water present inside the desert cooler evaporates more, thereby cooling the surroundings more. Hence, a desert cooler cools better on a hot dry day.

Answer:

[tex]\triangle V = 0.02484m^3[/tex]

Explanation:

Given

[tex]V_1 = 3.00m^3[/tex] --- initial volume

[tex]T_1 = 20.0^oC[/tex] --- initial temperature

[tex]T_2 = 60.0^oC[/tex] --- final temperature

[tex]\gamma = 207*10^{-6[/tex] ---  coefficient of thermal expansion:

Required

The change in volume

To do this, we make use of cubic expansivity formula

[tex]\triangle V = \gamma * V_2 * (T_2 - T_1)[/tex]

So, we have:

[tex]\triangle V = 207 * 10^{-6} * 3.00 * (60.0 - 20.0)[/tex]

[tex]\triangle V = 207 * 10^{-6} * 3.00 * 40.0[/tex]

[tex]\triangle V = 0.02484m^3[/tex]

The volume will expand by [tex]0.02484m^3[/tex]

Answer:

1.the friction of air, gravity2.gravity

Answer:

The softball experiences an applied force as a result of Amy’s throw. As the ball moves, it experiences drag from the air it passes through. It also experiences a downward pull because of gravity.

Explanation:

Plato

Answer:1.6875 m

Explanation:

Formula= 1/2 x at^2

Answer:

[tex]f = m \frac{v1 - v2}{t} \\ = 120 \times \frac{25 - 5}{2} \\ = 120 \times \frac{20}{2} \\ = 120 \times 10 \\ = 1200N \\ thank \: you[/tex]

Answer:

10 Joule

Explanation:

The solution and answer are well written in the Pic above.

Answer:

Because Moon and Mars has no atmosphere.

Explanation:

Moon and Mars has no atmosphere, so there is no friction on the falling object due to the atmosphere. The speed of the falling object is more at Moon and Mars.

When a small object impact on the surface of moon or Mars with high speed, the size of crater is large than the earth as out earth has atmosphere.

Answer:

1. 36.35 g = 36.35E15 micrometer.

II. 36.35 g = 363.5 millimetre.

III. 36.35 g = 0.03635 kilogram.

Explanation:

Given the following data;

To convert the mass in grams (g) to;

I. Micrometer

Conversion:

1 g = 1 exp 15 um

36.35 g = X um

Cross-multiplying, we have;

X = 36.35 * 1 exp 15 = 36.35 exp 15 um

36.35 g = 36.35E15 micrometer

II. Millimetre

Conversion:

1 g = 1 milliliter

36.35 g = X milliliter

Cross-multiplying, we have;

X = 36.35 * 1 = 36.35 milliliter

Next, we would convert milliliter to millimetre;

1 milliliter = 10 millimetre

36.35 milliliter = X millimetre

Cross-multiplying, we have;

X = 36.35 * 10 = 363.5 millimetre

36.35 g = 363.5 millimetre

III. Kilogram

Conversion:

1000 grams = 1 kilogram

36.35 g = X kilogram

Cross-multiplying, we have;

X * 1000 = 36.35 * 1

Dividing both sides by 1000, we have;

X = 36.35/1000 = 0.03635 kilogram

36.35 g = 0.03635 kilogram

Note:

Answer:

may be upside down alphabet :"T"

Explanation:

Answer:

Data given.

focal length (f)=15m÷2=7.5m

Distance of the object(U)=10m

Image distance (v)=?

Magnification (M)=?

Solution:

From:

1/f=1/u+1/v

1/7.5=1/10+1/v=75

then v=75m

Magnification, M=u/v

=75/10=7.5

Then magnification=7.5

Answer:

v = 30 m and m = 3

Explanation:

Given that,

The radius of curvature of the mirror, R = 15 m

Focal length, f = 7.5 m

Object distance, u = -10 m

We need to find the image distance and the magnification of the object.

Using mirror's formula,

[tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{(7.5)}+\dfrac{1}{(-10)}\\\\v=30\ m[/tex]

The magnification of the object in mirror is given by :

[tex]m=\dfrac{-v}{u}\\\\m=\dfrac{-30}{-10}\\\\m=3[/tex]

So, the distance of the image from the mirror and the magnification of the object are 30 m and 3 respectively.

Answer:

(a) 1.5 nF, 1.2 nF, 1 nF

(b) 0.4 nF

Explanation:

V = 150 V

V' = 40 V, V'' = 50 V, V''' = 60 V, q = 6 x 10^-8 C

(a) C' = q/V' = 6 x 10^-8 / 40  = 1.5 x 10^-9  F

C'' = q/V'' = 6 x 10^-8 / 50 = 1.2 x 10^-9 F

C''' = q/V''' = 6 x 10^-8 / 60 = 1 x 10^-9 F

(b) The effective capacitance is

[tex]\frac{1}{C}=\frac{1}{C'}+\frac{1}{C''}+\frac{1}{C'''}\\\\\frac{1}{C}=\frac{10^9}{1.5}+\frac{10^9}{1.2}+\frac{10^9}{1}\\\\C = 0.4\times 10^{-9} F[/tex]  

Answer:

Specific heat capacity, c = 468.75 J/Kg°C

Explanation:

Given the following data;

Power = 1.5 kW to Watts = 1.5 * 1000 = 1500 Watts

Time = 5 seconds

Mass = 0.2 kg

Initial temperature = 20°C

Final temperature = 100°C

To find specific heat capacity;

First of all, we would have to determine the energy consumption of the kettle;

Energy = power * time

Energy = 1500 * 5

Energy = 7500 Joules

Next, we would calculate the specific heat capacity of water.

Heat capacity is given by the formula;

[tex] Q = mcdt[/tex]

Where;

dt = T2 - T1

dt = 100 - 20

dt = 80°C

Making c the subject of formula, we have;

[tex] c = \frac {Q}{mdt} [/tex]

Substituting into the equation, we have;

[tex] c = \frac {7500}{0.2*80} [/tex]

[tex] c = \frac {7500}{16} [/tex]

Specific heat capacity, c = 468.75 J/Kg°C

Answer:

C. 28.2 deg

Explanation:

The horizontal range of a projectile is given as:

[tex]R = \frac{v^2Sin2\theta}{g}[/tex]

where,

R = Range

v = speed

θ = angle of launch

g = acceleration due to gravity = 9.81 m/s²

First, we will find the launch speed (v) by using the initial conditions:

R = 120 m

θ = 45°

Therefore,

[tex]120\ m = \frac{v^2Sin 90^o}{9.81\ m/s^2}\\\\v = \sqrt{(120\ m)(9.81\ m/s^2)}\\\\v = 34.31\ m/s[/tex]

Now, consider the second scenario to hit the target:

R = 100 m

Therefore,

[tex]100\ m = \frac{(34.31\ m/s)^2Sin2\theta}{9.81\ m/s^2}\\\\Sin2\theta = \frac{(100\ m)(9.81\ m/s^2)}{(34.31\ m/s)^2}\\\\2\theta = Sin^{-1}(0.833)\\\\\theta = \frac{56.44^o}{2}\\\theta = 28.22^o[/tex]

Hence, the correct option is:

C. 28.2 deg

Answer:

x = 4.85 cm

Explanation:

From work energy theorem when dealing with a spring in compression, we know that total work done is;

W_t = ½kx²

Where;

k is Force constant

x is max compression

Now, we know that this is also equal to the kinetic energy.

K.E = ½mv²

Thus;

½kx² = ½mv²

Making x the subject;

x = √(mv²/k)

We are given;

m = 5 kg

v = 2 m/s

k = 85 N/cm = 8500 N/m

Thus;

x = √(mv²/k)

x = √(5 × 2²/8500)

x = 0.0485 m

x = 4.85 cm

Answer:

If the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3, then Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4

Explanation:

In order to solve this question, we must take into account that the force of gravity is given by the following formula:

[tex]F_{g0}=G \frac{mM_{E0}}{r^{2}}[/tex]

So if the mass of the earth is increased by a factor of 2, this means that:

[tex]M_{Ef}=2M_{E0}[/tex]

so:

[tex]F_{gf}=G \frac{2mM_{E0}}{r^{2}}[/tex]

Therefore:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{2mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]

When simplifying we end up with:

[tex]\frac{F_{gf}}{F_{g0}}=2[/tex]

so if the mass of the Earth is increased by a factor of 2, then the Fgrav is increased by a factor of 2.

If the mass of the earth is increased by a factor of 3

So if the mass of the earth is increased by a factor of 2, this means that:

[tex]M_{Ef}=3M_{E0}[/tex]

so:

[tex]F_{gf}=G \frac{3mM_{E0}}{r^{2}}[/tex]

Therefore:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{3mM_{E0}}{r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]

When simplifying we end up with:

[tex]\frac{F_{gf}}{F_{g0}}=3[/tex]

so if the mass of the Earth is increased by a factor of 3, then the Fgrav is increased by a factor of 3.

If the mass of the earth is decreased by a factor of 4

So if the mass of the earth is decreased by a factor of 4, this means that:

[tex]M_{Ef}=\frac{M_{E0}}{4}[/tex]

so:

[tex]F_{gf}=G \frac{mM_{E0}}{4r^{2}}[/tex]

Therefore:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{G \frac{mM_{E0}}{4r^{2}}}{G \frac{mM_{E0}}{r^{2}}}[/tex]

When simplifying we end up with:

[tex]\frac{F_{gf}}{F_{g0}}=\frac{1}{4}[/tex]

so if the mass of the Earth is decreased by a factor of 4, then the Fgrav is decreased by a factor of 4.

Answer:

B. seafloor spreading

Explanation:

divergent motion between the Eurasian and North American, and African and South American Plates. ... In this way, as the plates move further apart new ocean lithosphere is formed at the ridge and the ocean basin gets wider

geological society

Answer:

For smooth surface:The light rays bounce off the table and all move in the same direction.

Answer:

Matter is made up of atoms

Answer:

Mater is made up of atoms.

Explanation:

Atoms come together to form molecules,which are the building blocks for all types of matter.

Answer:

i didn't understand,

Explanation:

sorry

Answer:

English only

Explanation:

When solving problems related to Electric Fields, care must be taken about symmetries. In our particular case when we take a look to at the drawings of the attached file, we realize:

1.-By symmetry each dx associated at a, has an opposite dx with point b as reference. The respective dE ( the charge is uniform ) is the same, as the charge of the wire is positive the force and the Field on a test charge (+) located at h will be upward, therefore the components dEx will cancel each other and the Electric Field becomes E = Ey = ∫ 2×dE× cosθ

The solutions:

A) Ey = 4623 N/C

B) Ey = 19.34 N/C

E = Ey = ∫ 2×dE× cosθ

Here     cosθ   = h/ d   ⇒  cosθ = h/√h² + x²      dE = K× dQ / d²

d² = h² + x²

k = 8.9 ×10⁹ Nm²C⁻²  ;   dQ = λ×dx     λ = 150×10⁻⁹ C    h = 0.08 m

Then by substitution

Ey =  2 ∫[K× λ×dx/ (h² + x²) ] × h / √h² + x²

reordering that equation:

Ey = 2×K×λ×h ∫ dx / [√ ( h² + x² ) ]³          (2)

To solve the integral we make use of a change of variables

x = h × tanα     then   x² = h² ×tan²α   and  dx = h× sec²α dα

plugging that values in equation (2)

Ey  =  2×K×λ×h ∫  h× sec²α× dα / [√ ( h² + h²tan²α)]³

Ey  = 2×K×λ×h² ∫ sec²α× dα / [ h × √ (1 + tan²α)]³            1 + tan²α = sec²α

Ey = 2×K×λ×h²× ∫ (sec²α / h³× sec³α )×dα

Ey = 2×K×λ/h × ∫ ( 1 / secα dα

Ey = 2×K×λ/h × sinα             now we αneed to come back to our original variables:

as   x = h × tanα         tanα = x/h   then x is the opposite leg in a right triangle  and h the adjacent one then the hypothenuse is √ (h² + x²)         then    sin α = x/ √ (h² + x²)      

Ey = 2×K×λ/h × x/ √ (h² + x²) |₀⁰°⁰⁵

Ey  = 2×8.9×10⁹× 150×10⁻⁹× 5×10⁻²/8× 10⁻²× √ 10⁻² ( 8 + 5 )   N/C

Ey = 4623 N/C

To answer the second question again we will make use of symmetries if you look at drawing ( Figure 2 ) you see that again the components in direction of x-axis cancel each other and the components in y-axis direction will add. Then

Ey = ∫ dE× cosθ

following the same procedure  we will find:

Ey = ∫ [K×λ × dl/d²] × h/ d

The importan point here is that the radius of the circle is

2×π×r = 0.01      ( the length of the wire)  ⇒  r = 0.16×10⁻² m

And we need to take into account that the integration is over the circle and the length of the circle is 0.01 m or ××2×π×r. All other factors are constant. Then by substitution

Ey = [K×λ ×h×  / ( √ r² + h²)³ ] × 10⁻²    N/C

Ey = 8.9 × 10⁹ × 150× 10⁻⁹ × 6× 10⁻² × 10⁻² / √ 10⁻² ( 0.16 + 6)

Ey = 0.8 × 10² / 6

Ey = 19.34 N/C

Answer:

1

Explanation:

Specific gravity is a ratio (of 2 densities) so it has no unit.

Answer:

9m is this the way calculator its acceleration

A man is running on the straight road with the uniform velocity of 3 m/s.

Acceleration produced by the man.

Answer:

it's when heat demolished the object

Answer:

[tex]V.E=498.02m/s^2[/tex]

Explanation:

From the question we are told that:

Radius [tex]r=301Km[/tex]

Gravitational acceleration [tex]g=0.412 m/s^2[/tex]

Generally the equation for Escape velocity is mathematically given by

 [tex]V.E^2=2gr[/tex]

 [tex]V.E^2=2*0.412m/s^2*301000[/tex]

 [tex]V.E^2=248024[/tex]

 [tex]V.E=\sqrt{248024}[/tex]

 [tex]V.E=498.02m/s^2[/tex]

Answer:

Newton' second law  and third law describes the situation.

Explanation:

According to the Newton's second law, the force applied on a body is proportional to the rate of change of momentum of the body.

According to the Newton's third law, for every action there is an equal and opposite reaction.

When ice skater pushes harder means more force is applied so he moves fast and more be the action force more be the reaction force.

Thus, Newton' second law  and third law describes the situation.

Answer:

An object overall change in position

Explanation:

ans maybe corre6

Answer:

i) The pressure acting on the base of B will be half the pressure acting on the base of A

ii) The pressure acting on the base of B will be the same as the pressure acting on the base of A

iii) The pressure on the base of drum A will be slightly less than the pressure on the base of drum B

Explanation:

The pressure acting on the base of the drum, P = h·ρ·g

Where;

h = The level of the liquid in the drum

[tex]h_{max}[/tex] = The height of the drums

ρ = The density of the liquid in the drum

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

i) If A is completely filled, we have [tex]h_A[/tex] = [tex]h_{max}[/tex]

Therefore, [tex]P_A[/tex] = [tex]h_{max}[/tex]×[tex]\rho_{liquid}[/tex]×g

If B is half filled, we have, [tex]h_B[/tex] =  (1/2)·[tex]h_{max}[/tex]

[tex]P_B[/tex] = (1/2) × [tex]h_{max}[/tex]×[tex]\rho_{liquid}[/tex]×g

Therefore, [tex]P_B[/tex] = (1/2) × [tex]P_A[/tex]

The pressure acting on the base of B will be half the pressure acting on the base of A

ii) If both A and B are each filled with water (the same liquid), then the pressure on their bases will be [tex]P_A[/tex] = [tex]h_{max}[/tex]×[tex]\rho_{water}[/tex]×g = [tex]P_B[/tex], the same, given that the acceleration due to gravity, g, is constant and the same in Nepal and India

iii) If A is filled with water, and B is filled with salty water, we have that, the density of salty water is slightly higher than water, therefore, we get;

[tex]P_A[/tex] = [tex]h_{max}[/tex]×[tex]\rho_{water}[/tex]×g <  [tex]P_B[/tex] =

The pressure on the base of drum A will be less than the pressure on the base of drum B.

Explanation:

If the particles that make up an object begin to move quickly, their average kinetic energy increases the object's temperature rises. Group of answer choices

Answer:Yes, A body can have constant speed but still accelerate as in case of uniform circular motion. In uniform circular motion speed remains constant and direction of velocity changes with every point in the direction of tangent drawn from that point.

Explaination:

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