At high noon, the sun is almost directly above (about 2.0 degrees from the vertical) and a tall redwood tree casts a shadow that is 10m long. How tall is the redwood tree?

Answer:

D. Height of the ramp.

Explanation:

The solid spherical ball is expected to have more mass than that of the hollow spherical ball. And the speed of both balls would be influenced by the gravitational force as they roll down the ramp. Thus, the masses would move at different speed.

At the bottom of the ramp, the speed of the balls can be varied by varying the height of the ramp. So that the speed of both balls depend on the height of the ramp. As the height of the ramp increases, consequently, the speed of the balls increases. And if the height of the ramp decreases, the speed of the balls decreases consequently.

Answer:

The 300 Ohm resistor

Explanation:

They are in series, so the current has to be the same in both.  P = I^2R, so with the same current, the larger resistor dissipates more power.

Answer:

in the water cycle evaporation occurs when the sunlight warms the surface of the water the heat from the sun makes the water molecules move faster and faster until they move so fast they can escape as a gas once evaporated a molecule from water that vapor spends about 10 days in the air

Answer:

36.87°

Explanation:

Given

Maximum height = 60m

Horizontal distance (range) = 80,m

Required

Initial speed U

Angle of launch

To get the speed, we will use the range formula;

R = U √2H/g

80 = U√2(60)/9.8

80 = U√12.25

80 = 3.5U

U = 80/3.5

U = 22.86m/s

Get the angle of launch

Using the formula

Theta = tan^-1(y/x)

y is the vertical distance

x is the horizontal distance

Theta = tan^-1(60/80)

Theta = tan^-1(0.75)

Theta = 36.87°

Hence the angle of launch is 36.87°

Answer:

The number of photons per second that strike the given area is 2.668 x 10⁸ photons/second

Explanation:

Given;

intensity of the sunlight, I = 2.00 kJ·s−1·m^−2

area of incident, A = 5.2 cm² = 5.2 x 10⁻⁴ m²

Energy of incident photons per second on the given area;

E = IA

E = (2000)( 5.2 x 10⁻⁴)

E = 1.04 J/s

Energy of a photon is given is by;

[tex]E = \frac{hc}{\lambda} \\\\E = \frac{(6.626*10^{-34})(3*10^8)}{(510*10^{-9})}\\\\E = 3.898*10^{-19} \ J/photon[/tex]

The number of photons per second that strike the given area is;

[tex]n = \frac{1.04 \ J/s}{3.898*10^{-19} \ J/photon} \\\\n = 2.668*10^{18} \ photons/second[/tex]

Therefore, the number of photons per second that strike the given area is 2.668 x 10⁸ photons/second

Answer:

Explanation:

The weight of the object can be found by using the formula

[tex]f = \frac{w}{d} \\ [/tex]

w is the workdone

d is the distance

From the question we have

[tex]f = \frac{62500}{25} \\ [/tex]

We have the final answer as

Hope this helps you

Answer:

distance i think

Explanation:

Answer:Final volume after pressure is applied=4,292cm3

Explanation:

Using the bulk modulus formulae

We have that The bulk modulus of waTer is given as  

K =-V dP/dV

Where  K, the bulk modulus of water = 2.15 x 10^9N/m^2

2.15 x 10^9N/m^2= - 4,300 x  4 × 106N/m2 / dV

dV = - 4,300 x  4 × 10^6N/m^2/ 2.15 x 10^9N/m^2

dV (change in volume)= -8.000cm^3

Final volume after pressure is applied,

V= V+ dV

V= 4300cm3 + (-8.000cm3)

=4300cm3 - 8.000cm3

Final Volume, V =4,292cm3

Answer:

Make sure you look at the wording!

Explanation:

if the last word is increased, the answer is increased

if the last word is decreased, the answer is it decreases!

Answer:

10 m/s²

Explanation:

The above question simply indicates motion under gravity.

The acceleration due to gravity (i.e acceleration of free fall) has a constant value of 10 m/s².

Whether the rock is dropped from a height of 5 m or 2.5 m, it will accelerate at 10 m/s² before striking the ground. The only thing that will be different is the time taken for the rock to strike the ground when released from both 5 m and 2.5 m.

Thus, the rock will have a constant acceleration of 10 m/s² irrespective of the height to which it was released.

Since acceleration due to gravity is a constant,  the acceleration of the rock dropped from the 5 m height is the same as that dropped from the 2.5 m height and is equal to 10 m/s²

Acceleration due to gravity is the acceleration a body falling freely from a height above the earth surface which a body experiences due to the gravitational force of attraction of the earth on the body.

Acceleration due to gravity has a constant value which is equal to 10 m/s².

Therefore, the acceleration of the rock dropped from the 5 m height is the same as that dropped from the 2.5 m height and is equal to 10 m/s².

Learn more about acceleration due to gravity at: https://brainly.com/question/11873969

Explanation:

the formula of speed is distance traveled by time it work

31.25 cm/sec

__________________________________________________________

We are given:

Distance travelled = 150 cm

Time taken = 4.8 seconds

Final Speed of the Marble:

Speed of the marble = Distance travelled / Time taken

Speed of the marble = 150 / 4.8

Speed of the marble = 31.25 cm/sec

Answer:

The surface must be a good absorber of light. Spectra of the planets proved that methane in the atmospheres of these planets is what absorbs red light.

Explanation:

Answer:

x' = 0.085 m = 8.5 cm

Explanation:

The law of conservation of energy says that:

Potential Energy Stored in Spring = Loss in Gravitational Potential Energy of Block

(1/2)kΔx² = mgh

where,

k = stiffness of spring = 1070 N/m

Δx = change in length of spring = ?

m = mass of block = 0.5 kg

g = 9.8 m/s²

h = height of block above spring = 0.6 m - 0.15 m = 0.45 m

Therefore,

(1/2)(1070 N/m)Δx² = (0.5 kg)(9.8 m/s²)(0.45 m)

Δx = √[2(2.205 Nm)/(1070 N/m)]

Δx = 0.064 m

but,

Δx = x - x' = 0.15 m - x' = 0.064 m

x' = 0.15 m - 0.064 m

x' = 0.085 m = 8.5 cm

Answer:

Explanation:

Step one:

given data

mass of cable= 4lb/ft

mass of coal= 1000lb

dept of mine= 700ft

Step two:

Required

the work-done to lift the coal and the rope combined

Work-done to lift coal

Wc=1000*700= 700,000 lb-ft

Work-done to lift rope

[tex]Wr=\int\limits^{700} _0 {4(700-y)} \, dx \\\\Wr=4(700y-\frac{1}{2}y^2 )\limits^{700}_0[/tex]

substitute y=700 we have, since y=0 will result to 0

[tex]Wr=4(700*700-\frac{1}{2}*700^2 )\\\\Wr=4(490000-245000)\\\\Wr=4(245000)\\\\Wr=980000ft-lbs[/tex]

Answer: c. incident ray

Explanation:

Answer:

v = 0.823 m/s

Explanation:

A man walks south at a speed of 2.00 m/s for 60.0 minutes.

The distance covered in South = 60 × 60 × 2 = 7200 m

He then turns around and walks north a distance 3000 m in 25.0 minutes.

As they moved in opposite direction, net displacement will be : 7200 - 3000 = 4200 m

Average velocity of the man = net displacement/time

[tex]v=\dfrac{4200\ m}{(60+25)\times 60}\\\\=0.823\ m/s[/tex]

So, the average velocity of the man is 0.823 m/s.

Hi, you've asked an incomplete question. However, I provided some explanation about the replication process that scientists do.

Explanation:

Replication in research involves carefully repeating an original experiment to see whether the same result would be arrived at as in previous research experiments.

For most scientists today, in other to avoid anything that would erroneously affect the results of the replicated experiment they usually follow the same procedures as carried by the previous researchers.

Answer:

B is the best answer for this

Answer:

answer is C on edge 2021

Explanation:

Answer:

Gravity acts on all objects in the universe.

Explanation:

Answer:

220v

Explanation:

Sorry, the question is incomplete

Answer:

on the potential difference applied and on the resistance of the wire.

Explanation:

Ohms law state that the current through a conductor between two points is directly proportional to the potential difference across the two points. Imtroducing the comstant of proportionality, the resistance, one arrives at the usual athematical equation that describes this relationship: I = V/R.

Answer:

10J

Explanation:

In this question we have the following information

The charge of the particle is q = 5 C

The equipotenetial level is V1 = 5.5 v

and also the

equipotenetial level is V2 = 3.5 v

So we calculate the

work done W=q x (v1-v2)

workdone = 5 x (5.5-3.5)

= 5x2

=10 J

Workdone = 10 J

So we conclude that the workdone on a particle with these information is 10j

Answer:

m₂ = 1.17 kg

Explanation:

Given that,

Force between two objects, [tex]F=1.01\times 10^{-8}\ N[/tex]

Mass of object 1, [tex]m_1=5.11\ kg[/tex]

Distance between masses, r = 19.9 cm = 0.199 m

The gravitational force between two masses is given by :

[tex]F=\dfrac{Gm_1m_2}{r^2}[/tex]

m₂ is the mass of object 2

[tex]m_2=\dfrac{Fr^2}{Gm_1}\\\\m_2=\dfrac{1.01\times 10^{-8}\times (0.199)^2}{6.67\times 10^{-11}\times 5.11}\\\\=1.17\ kg[/tex]

So, the mass of second object is 1.17 kg.

Answer:

b. identical to the downward force of gravity on the person.

Explanation:

For an object in an elevator,

F = mg - ma       (g > a)

But since the velocity is uniform, a = 0.

Then,

F = mg - 0

F = mg

This is the actual weight of the object.

The object does not feel weightless, so that its actual weight can be measured during the downward motion of the elevator with uniform velocity.

Thus, the upward normal force, N, exerted by the elevator floor on the person is identical to the downward force of gravity on the person.

Answer:

Image result for total velocity definition

The average speed of an object is defined as the distance traveled divided by the time elapsed.

Explanation:

Answer:

B: lodestone

Explanation:

Each magnet has its magnetic poles, north (N) and south (S). Diversified ones are attracted and reptiles of the same name are repelled, similarly to charges, so it was considered possible to separate the magnet at the north and south poles.

Magnetic properties can be lost if the magnet is exposed to high temperatures if it falls or due to some mechanical shocks.

Answer:

60.185 percent of the original kinetic energy is convertible to internal energy.

Explanation:

Let suppose that collision between both particles is entirely inellastic. If there is no external forces exerted on any of the particles, then we can apply the Principle of Linear Momentum Conservation. That is:

[tex]m_{A}\cdot v_{A,o} + m_{B}\cdot v_{B,o} = (m_{A}+m_{B})\cdot v[/tex]

[tex]v = \frac{m_{A}\cdot v_{A,o}+v_{B}\cdot v_{B,o}}{m_{A}+m_{B}}[/tex] (1)

Where:

[tex]m_{A}[/tex] - Mass of the 4.8-g particle, measured in kilograms.

[tex]m_{B}[/tex] - Mass of the 7.4-g particle, measured in kilograms.

[tex]v_{A,o}[/tex] - Initial speed of the 4.8-g particle, measured in meters per second.

[tex]v_{B,o}[/tex] - Initial speed of the 7.4-g particle, measured in meters per second.

[tex]v[/tex] - Final speed of the collided particles, measured in meters per second.

If we know that [tex]m_{A} = 4.8\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.4\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 3\,\frac{m}{s}[/tex] and [tex]v_{B,o} = 0\,\frac{m}{s}[/tex], then the final speed of the system is:

[tex]v = \frac{(4.8\times 10^{-3}\,kg)\cdot \left(3\,\frac{m}{s} \right)+(7.4\times 10^{-3}\,kg)\cdot \left(0\,\frac{m}{s} \right)}{4.8\times 10^{-3}\,kg+7.4\times 10^{-3}\,kg}[/tex]

[tex]v = 1.180\,\frac{m}{s}[/tex]

During the collision part of the initial energy is dissipated in the form of heat, which is related to the internal energy ([tex]\Delta U[/tex]), measured in joules. According to the Principle of Energy Conservation, we have the following model:

[tex]\Delta U = K_{A}+K_{B}-K[/tex] (2)

Where:

[tex]K_{A}[/tex], [tex]K_{B}[/tex] - Initial translational kinetic energies of each particle, measured in joules.

[tex]K[/tex] - Final translational kinetic energy of the collided particles, measured in joules.

By applying the definition of translational kinetic energy, we expand and simplify the equation above:

[tex]\Delta U = \frac{1}{2}\cdot m_{A}\cdot v_{A,o}^{2}+\frac{1}{2}\cdot m_{B}\cdot v_{B,o}^{2} -\frac{1}{2}\cdot (m_{A}+m_{B})\cdot v^{2}[/tex] (3)

If we get that  [tex]m_{A} = 4.8\times 10^{-3}\,kg[/tex], [tex]m_{B} = 7.4\times 10^{-3}\,kg[/tex], [tex]v_{A,o} = 3\,\frac{m}{s}[/tex], [tex]v_{B,o} = 0\,\frac{m}{s}[/tex] and [tex]v = 1.180\,\frac{m}{s}[/tex], the internal energy associated with the system is:

[tex]\Delta U = \frac{1}{2}\cdot (4.8\times 10^{-3}\,kg)\cdot \left(3\,\frac{m}{s} \right)^{2}+ \frac{1}{2}\cdot (7.4\times 10^{-3}\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}-\frac{1}{2}\cdot (4.8\times 10^{-3}\,kg+7.4\times 10^{-3}\,kg)\cdot \left(1.180\,\frac{m}{s} \right)^{2}[/tex]

[tex]\Delta U = 0.013\,J[/tex]

And the initial energy of both particles is:

[tex]E_{o} = \frac{1}{2}\cdot (4.8\times 10^{-3}\,kg)\cdot \left(3\,\frac{m}{s}\right)^{2}+\frac{1}{2}\cdot (7.4\times 10^{-3}\,kg)\cdot \left(0\,\frac{m}{s} \right)^{2}[/tex]

[tex]E_{o} = 0.0216\,J[/tex]

Lastly, the percentage of the original kinetic energy that is convertible to internal energy is: ([tex]\Delta U = 0.013\,J[/tex], [tex]E_{o} = 0.0216\,J[/tex])

[tex]\%e = \frac{\Delta U}{E_{o}}\times 100\,\%[/tex] (4)

[tex]\%e = \frac{0.013\,J}{0.0216\,J}\times 100\,\%[/tex]

[tex]\%e = 60.185\,\%[/tex]

60.185 percent of the original kinetic energy is convertible to internal energy.

Answer:

e

Explanation:

cause you have any attachments are not sure if you can get it

When a block slides down an inclined plane, normal force does zero work. Hence, option (a) is correct.

The definition of force in physics is: The push or pull on a massed object changes its velocity.

An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude. A spring balance can be used to calculate the Force. The Newton is the SI unit of force.

The component of a contact force in mechanics known as the normal force is perpendicular to the surface that an item encounters.

As normal force always acts perpendicular to the motion, this force does zero work.

But friction force the weight acts in opposite direction of the motion, this force does work against the motion.

In an inclined plane, the parallel component of  weight does work along the motion.

Learn more about normal force here:

https://brainly.com/question/18799790

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Answer:

2.9 cm

Explanation:

Assuming that the rear wheel has a radius of 0.330 m

Given that

r(a) = 12 cm -> 0.12 m

w(a) = 0.6 rev/s -> 3.77 rad/s

v = 5 m/s

r(w) = 0.330 m

The speed on any point on the rim at the sprocket in the front is

v(a) = w(a).r(a) = 3.77 * 0.12 = 0.4524 m/s

Also,

v(a) = speed at any point on the chain

v(b) = speed at any point on the rim of the rear sprocket

v(a) = v(b)

where v(b) = w(b).r(b)

Recall that the speed at any point on the rear wheel is v, where

v = w(b).r(w)

5 = w(b) * 0.330

w(b) = 5/0.330

w(b) = 15.15 rad/s

On substituting this in the equation, we have

v(b) = w(b).r(b).

Remember also, that v(a) = v(b), so

0.4524 = 15.15 * r(b)

r(b) = 0.4524 / 15.15

r(b) = 0.029 m -> 2.9 cm

Therefore, the radius of the rear sprocket needed is 2.9 cm