When a rope is wrapped around a wheel (or pulley) and pulled, the lever arm is the ________ Blank 1 of the wheel and the angle between this and the force is ALWAYS ________ degrees. Blank 2

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:

may be upside down alphabet :"T"

Explanation:

Answer:

Parte A

El ángulo con respecto al horizonte, de la segunda cuerda es de aproximadamente 19,47°

Parte B

La masa de la caja que se va a sostener es de aproximadamente 0,7808 kg.

Explanation:

Parte A

Los parámetros dados son;

La tensión en la cuerda, T₁ = 8 N

La tensión en la cuerda, T₂ = 6 N

El ángulo de inclinación de la primera cuerda con la horizontal, θ₁ = 45°

Sea θ₂ el ángulo de inclinación de la segunda cuerda, obtenemos;

T₁·cos (θ₁) = T₂·cos (θ₂)

∴ 8 N × cos (45°) = 6 N × cos (θ₂)

cos (θ₂) = 8 N × cos (45°) / (6 N) = (√2)/2 × (4/3) = (2·√2)/3

θ₂ = arcos ((2·√2) / 3) ≈ 19,47°

El ángulo con respecto al horizonte, de la segunda cuerda, θ₂ ≈ 19,47°

Parte B

El peso de la caja, W = T₁·sin (θ₁) + T₂·sin (θ₂)

∴ W = 8 N × sen (45 °) + 6 N × sen (19,47 °) ≈ 7,66 N

El peso de la caja que se va a sostener, W ≈ 7,66 N

La masa de la caja que se va a sostener, m ≈ 7,66 N / (9,81 m/s²) ≈ 0,7808 kg

Answer:1.6875 m

Explanation:

Formula= 1/2 x at^2

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:

Determine the direction that the pilot is aiming the plane.

Explanation:

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:

(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:

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:

i didn't understand,

Explanation:

sorry

For a ball that is launched upward at an angle from the ground, the direction of its acceleration at the top of the projectile curve is downwards.

As the ball is projected upward its acceleration points upwards until the ball reaches the maximum height or top of the projectile path.

At the top of the projectile path, its acceleration start pointing downwards.

Thus, for a ball that is launched upward at an angle from the ground, the direction of its acceleration at the top of the projectile curve is downwards.

Learn more about downward acceleration here: https://brainly.com/question/22048837

#SPJ1

Answer:

vertical

Explanation:

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:

[tex]W=7.56\times 10^{-19}\ J[/tex]

Explanation:

Given that,

The work function for silver is 4.73 eV.

We need to find the value of the work function from electron volts to joules.

We know that,

[tex]1\ eV=1.6\times 10^{-19}\ J[/tex]

For 4.73 eV,

[tex]4.73\ eV=1.6\times 10^{-19}\times 4.73\\\\=7.56\times 10^{-19}\ J[/tex]

So, the work function for silver is [tex]7.56\times 10^{-19}\ J[/tex].

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.

Solution :

When the spacecraft is at halfway point, the distance from the Earth as well as Mars are same. We have to account the masses of the planets. The gravitational force that is exerted by the Earth is greater because of its combined mass with the space probe.

The mass of Earth is greater than the mass of Mars. Therefore, the force of Earth is more than Mars.

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:

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:

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:

D

Explanation:

The tectonic plates move on the mantle, sort of floating on it as they are part of the crust. When they collide things like mountain ranges form, and big earthquakes happen.

Answer:

(a) The optimum banking Decreases

(b) The optimum banking Increases

(c) The optimum banking is approximately 86.88°

Explanation:

(a) The equation of motion on a banked road is given as follows;

[tex]v = \sqrt{R \cdot g \times \left(\dfrac{tan (\theta) + \mu}{1 - \mu \cdot tan (\theta) }\right) }[/tex]

For no friction, we have;

v = √(R·g·tan(θ))

Where;

R₁ = The radius of the road

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

θ₁ = The bank angle

μ = The coefficient pf friction = Constant

v = The vehicle's speed

If the radius doubles, for no friction, we have;

v² = R·g·(tan(θ))

tan(θ) = v²/(R·g)

Therefore, when the radius doubles, tan(θ) becomes smaller and therefore, the optimum banking angle θ decreases (becomes smaller)

(b) When the speed doubles, we have;

v₁ = 2·v

∴ tan(θ₁) = (v₁)²/(R·g) = 4·(v)²/(R·g) = 4·tan(θ)

When the speed doubles, tan(θ) increases and therefore, the optimum banking angle θ increases increases

(c) The radius negotiated by the car, R = 80 cm = 0.8 m

The speed of the car, v = 12 m/s

From tan(θ) = v²/(R·g), we have;

tan(θ) = 12²/(0.8 × 9.81) ≈ 18.349

θ ≈ arctan(18.349°) ≈ 86.88°

Answer:

2r or diameter

Explanation:

After 3 1/2 rounds it will end up on the other side of the circle and displacement will be 2 x the radius = d

Distance = 7 π R

Displacement = 2 R from the starting point directed through the center of the circle

Answer: The HUMAN EYE

Explanation:

The human eye is made up of different parts which ranges from controlling the amount of light that enters the eye to the focusing of the image that is formed. The camera is a device which is both mechanically and electronically operated which shares a number of similarities with the eye.

In the human eye, the IRIS helps to regulate the amount of rays passing through the pupil to the lens by either contracting or dilating in light or dark environment respectively. While in the camera, the DIAPHRAGM controls the amount of light entering the camera.

The PUPIL serves as the passage for light into the eye while in the camera, the APERTURE does the same.

The photosensitive surface in the eye is the YELLOW SPOT while in the camera, the photosensitive surface is the PHOTOGRAPHIC FILM.

Answer:

The angiosperms dominate Earth's surface and vegetation in more environments, particularly terrestrial habitats, than any other group of plants. As a result, angiosperms are the most important ultimate source of food for birds and mammals, including humans.

Explanation:

plz mark brainlest

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:

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.

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:

1

Explanation:

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

Answer:

A. There are 6.02 x 1023 items in a mole, which equals Avogadro's

number

Explanation:

The mole of a substance is