How does rocks and fossils determine Earth's age?

Answer:

Chemical etching is a process of printed circuit board (PCB) manufacturing that provides  many advantages over mechanical methods.

Explanation:

There's one! UvU hope this helped in whatever you're in for! <3

Good Luck!

Answer:

2.605m

Explanation:

Using the formula for calculating Range (distance travelled in horizontal direction)

Range R = U√2H/g

U is the speed = 4.8m/s

H is the maximum height = ?

g is the acc due to gravity = 9.8m/s²

R = 3.5m

Substitute into the formula and get H

3.5 = 4.8√2H/9.8

3.5/4.8 = √2H/9.8

0.7292 = √2H/9.8

square both sides

0.7292² = 2H/9.8

2H = 0.7292² * 9.8

2H = 5.21

H = 5.21/2

H = 2.605m

Hence the height of the ball from the ground is 2.605m

Answer:

divide 10 by 50.

Explanation:

Because its time over speed 10/50

5 meter/ second

I think it's helpful

follow me and don't forget to Mark me as brainlist please

Answer:

People who are in good health at the beginning of a hunger strike "are usually at little risk of dying from malnutrition for at least six to eight weeks," the guidelines state. But people who are ill can die from malnutrition in as little as three weeks. Malnutrition during pregnancy can be quite harmful. Maternal malnutrition during pregnancy can affect the wellbeing and development of the baby. Malnutrition is when the baby and in this case, the mother do not receive adequate nourishment during pregnancy. It means the baby, in the mother’s womb does not get enough nutrients to grow.

Hope this helps!

You need to first measure the angle of descent, i.e. the angle the hill makes with the ground. Then identify the forces acting on the sled, split them up into horizontal and vertical components, or into components that are parallel and perpendicular to the hill, and use Newton's second law to determine the components of the sled's acceleration vector.

There are at least 2 forces acting on the sled:

• its weight, pointing downward with magnitude W = m g

• the normal force, pointing perpendicular to the hill and away from the ground with mag. N

The question doesn't specify, but there might also be friction to consider, indicated in the attachment by the vector F pointing parallel to the slope of the hill and opposing the direction of the sled's motion with mag. F.

Splitting up the forces into parallel/perpendicular components is less work. By Newton's second law, the net force (denoted with ∑ or "sigma" here) in a particular direction is equal to the mass of the sled times its acceleration in that direction:

∑ (//) = W (//) = m a (//)

∑ (⟂) = W (⟂) + N = m a (⟂)

where, for instance, W (//) denotes the component of the sled's weight in the direction parallel to the hill, while a (⟂) denotes the component of the sled's acceleration perpendicular to the hill. If there is friction, you need to add -F to the first equation.

If the hill makes an angle of θ with flat ground, then W makes the same angle with the hill so that

W (//) = -m g sin(θ)

W (⟂) = -m g cos(θ)

So we have

-m g sin(θ) = m a (//)   →   a (//) = -g sin(θ)

-m g cos(θ) + N = m a (⟂)   →   a (⟂) = 0

where the last equality follows from the fact that the normal force exactly opposes the perpendicular component of the weight. This is because the sled is moving along the slope of the hill, and not into the air or into the ground.

Then the acceleration vector is

a = a (//)

with magnitude

||a|| = a = g sin(θ).

Answer:

To find the acceleration, you do 20N/50kg = 0.4 m/s^2

Explanation:

example: a = F/m = 10/2 = 5 m/s2

Answer:b

Explanation:b

Answer:

4 meters a second

Explanation:

100/25

plzz mark brainiest

Answer:

v / √2

Explanation:

A blocked from rest at the top of a hill of height h it there is negligible friction between the block and the hill, the block arrives at the bottom of the hill with speed. The block is released from rest at the top of another hill with a rough surface and height h. if one-half of the initial mechanical energy of the block Earth System is lost due to friction as the block descends the hill, the block will reach the bottom of the hill with a speed of?

Solution:

For the first block block, its potential energy at the top of the hill when it is at rest is converted to kinetic energy after release with a velocity v.

The kinetic energy = (1/2)mv², where m is the mass of the block.

For the other block placed on the top of the rough hill, the mechanical energy (kinetic energy) is halved due to the roughness of the hill. If u  is the speed of this block at the bottom of the hill then:

kinetic energy of block on the rough hill = (1//2)mu²

Hence:

(1//2)mu² = half of main system kinetic energy

(1//2)mu² = 1/2 * (1/2)mv²

mu² = (1/2)mv²

u² = (1/2)v²

u = √(v²/2)

u = v / √2

Hence the speed at the bottom of the rock of the block placed on the rough surface is v / √2

Answer:

B. Longitudinal wave

Explanation:

It's the only option that makes sense. Hope this helped :)

Answer:

2.8 m/s²

Explanation:

v=v0+at <=> 36 = 22 + a×5 <=> 5a = 36-22 <=> 5a = 14 <=> a = 14/5 <=>

<=> a = 2.8 m/s²

Answer:

first one 31

second one 23

Explanation:

on edge ;))

Answer:

winter solstice i think

Explanation:

Answer:

a/ solar eclipse

Explanation:

because the moon is on the left side of earth

Answer:

encyclopedia most reliable I think

Answer:

The value is  [tex]w = 7.54 \ m[/tex]        

Explanation:

From the question we are told that

     The length of the crack is  [tex]a = 0.3 \ m[/tex]

     The  frequency is  [tex]f = 30.0 \ kHz = 30 *10^{3} \ Hz[/tex]

      The distance outside the cave that is being consider is  [tex]D = 100 \ m[/tex]

      The speed of sound is [tex]v_s = 340 \ m/s[/tex]

Generally the wavelength of the wave is mathematically represented as

        [tex]\lambda = \frac{v}f}[/tex]

=>     [tex]\lambda = \frac{340 }{30*10^{3}}[/tex]

=>     [tex]\lambda = 0.0113 \ m/s[/tex]

Generally for a  single slit the path difference between the interference patterns of the sound wave and the center  is mathematically represented as  

          [tex]y = \frac{ n * \lambda * D}{a}[/tex]

=>     [tex]y = \frac{ 1 * 0.0113 * 100}{0.3}[/tex]

=>     [tex]y = 3.77 \ m[/tex]

Generally the width of the sound beam is mathematically represented as

         [tex]w = 2 * y[/tex]

=>      [tex]w = 2 * 3.77[/tex]

=>      [tex]w = 7.54 \ m[/tex]        

Answer:

b. bird

i am pretty sure this is right, but i cant be sure bcause i  know what the sentence is.  

Answer:

a) 2.5secs

b) 31.25m

Explanation:

a) From the question, we are given;

Initial speed u = 15m/s

Final speed v = 10.0m/s

Acceleration =-2.0m/s²

Required

How long it takes take the car to accelerate (Time t)

Using the equation of motion

v = u+ at

10 = 15+(-2)t

10 = 15-2t

10-15 = -2t

-5 = -2t

t = 5/2

t = 2.5secs

Hence it takes the car 2.5secs to accelerates.

b) We are to find distance S

Using the equation

S = ut + 1/2at²

S = 15(2.5)+1/2(-2)(2.5)²

S = 37.5-6.25

S = 31.25m

Hence the car as moved by 31.25m during the breaking period.

Hiii I think the answer might be 3.3 seconds,not exaclty sure tho

Answer:

35

Explanation:

c7z ya boiiiiiiiiiiiiiiiiii

Answer:

Because they want attention

Answer:

Explanation:

a)

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100[tex]cm^{2}[/tex] x 10cm = 1000[tex]cm^{3}[/tex]

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

[tex]\frac{1g}{cm^{3} }[/tex] x 1000[tex]cm^{3}[/tex] = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

b)

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ [tex]\frac{8g}{cm^{3} }[/tex] = 2.5 [tex]cm^{3}[/tex]

c)

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5[tex]cm^{3}[/tex] overflowed. So now we the same process as in number a) just with a few adjustments.

[tex]\frac{1g}{cm^{3} }[/tex] x (1000[tex]cm^{3}[/tex] - 2.5[tex]cm^{3}[/tex]) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

Answer:

3. When the number of turns, N is doubled, the strength of the electromagnet is also doubled

4. Doubling the voltage, doubles the strength of the electromagnet

5. The number of paper clips a 7.5 V battery would pick is approximately 28 paper clips

The number of paper clips a 7.5 V battery would pick is 59 paperclips

6. For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is approximately 7 paperclips

For the 50-coil electromagnet, the average number of paper clips a 1 V battery would pick is 16 paperclips

Explanation:

3. The Magnetomotive Force, MMF = The Number of Turns on the Coil, N × The Current I Flowing in the Coil, I

∴ MMF = N × I

When the number of turns, N is doubled, the magnetomotive force, MMF is also doubled, and the strength of the electromagnet is doubled

4. Given that the voltage, V applied to the coil = The current, I flowing × The resistance, R of the coil, we have

V = I × R

Therefore, for a given constant resistance in the coil, doubling the voltage, doubles the current and therefore doubles the strength of the electromagnet

5. The average slope for the 25-coil electromagnet = (23 - 12)/(6 - 3) = 3.[tex]\bar 6[/tex]

The number of paper clips a 7.5 V battery would pick = 12 + (7.5 - 3) × 11/3 = 28.5 paperclips  ≈ 28 paper clips

The average slope for the 50-coil electromagnet = (48 - 26)/(6 - 3) = 7.[tex]\bar 3[/tex]

The number of paper clips a 7.5 V battery would pick = 26 + (7.5 - 3) × 22/3 = 59 paperclips

6. The slope calculated from a start point of approximately 0.4 V, is given as follows;

The slope for the 25-coil electromagnet = (12 - 6)/(3 - 0.4) = 30/13

Therefore, for the 25-coil electromagnet,  the average number of paper clips a 1 V battery would pick = 6 + (1 - 0.4) × 30/13) = 96/13 ≈ 7 paperclips

The slope for the 50-coil electromagnet = (26 - 13)/(3 - 0.4) = 5

Therefore, for the 50-coil electromagnet,  the average number of paper clips a 1 V battery would pick = 13 + (1 - 0.4) × 5 = 16 paperclips

3. When the number of revolutions, N, is twice, the electromagnet's strength is doubled as well.

4. Doubling the voltage doubles the electromagnet's strength.

5. The number of paper clips selected by a 7.5 V battery is 59.

6. For a 50-coil electromagnet, a 1 V battery would pull up around 7 paperclips on average. The average number of paper clips picked by a 1 V battery for a 50-coil electromagnet is 16 paperclips.

The magnetic field of an electromagnet is created by an electric current. Electromagnets are caused from conducting wire wriggled into a coil.

A magnetic field is created in the hole by a current passing through the wire.

The magnetic field production is increased by;

1) increasing the number of turns

2) increase the area of the loop

3) by moving the magnet faster

Answer for the following given options as follows;

3. When the number of revolutions, N, is twice, the electromagnet's strength is doubled as well. Because the no of coils in the circuit is directly proportional to the strength of electromagnet.

4. From the ohm's law principal doubling the voltage twice the current and hence the electromagnet's strength for a given constant resistance in the coil increases.

5. For the 25-coil electromagnet, the average slope is ;

[tex]\rm m_{avg}= \frac{23-12}{6-3} \\\\ m_{avg}= 3[/tex]

The number of paper clips selected by a 7.5 V battery is;

[tex]\rm n = 12+ (7.5-+3)\times \frac{11}{3}\\\\n = 28.5[/tex]

For the 50-coil electromagnet, the average slope;

[tex]\rm (m_{avg})_{50}= \frac{48-26}{6-3} \\\\\ \rm (m_{avg})_{50}= 7[/tex]

The number of paper clips selected by a 7.5 V battery is;

[tex]\rm N = 26+(7.5-3) \times \frac{22}{3} \\\\ N= 59[/tex]

The following is the slope determined from a starting position of around 0.4 V:

For the 25-coil electromagnet, the slope is;

[tex]\rm m_{25}=\frac{12-6}{3-0.4} \\\\\ m_{25}=\frac{6}{2.6} \\\\ m_{25}=\frac{30}{13}[/tex]

As a result, the average number of paper clips a 1 V battery would choose for the 25-coil electromagnet is;

[tex]\rm N_{25}=6+ (1-0.4)\times \frac{30}{13} \\\\ N_{25}=\frac{96}{13} \\\\ N_{25}= 7 \ paperclip[/tex]

For the 50-coil electromagnet, the slope is;

(26 - 13)/(3 - 0.4) = 5.

As a result, the average number of paper clips picked by a 1 V battery is;

[tex]\rm N'_{50}= 13+ (1-0.14)\times 5 \\\\ N'_{50}=16[/tex]

Hence,the average number of paper clips picked by a 1 V battery for a 50-coil electromagnet is 16 paperclips.

To learn more about the electromagnets, refers to the link:

https://brainly.com/question/23727978

#SPJ5

Note: Your question inputs seem a little odd. But, I am assuming that you really mean '10km in 5 ms'.

Answer:

The Average speed = 15000 / 0.005 = 3000000 m/s

Explanation:

Average speed = Total Distance / Total time

Given

Thus,  

Average speed = Total Distance / Total time

Average speed = 15000 / 0.005 = 3000000 m/s

Answer:

okay

Explanation:

okay

Answer:

Cool

Explanation:

Answer:

Buoyant force = 3.0 N

The object will not float.

Explanation:

Apparent weight of a body immersed in water is the actual weight of object minus buoyant force

Given in the question that;

Weight of object in air = 7.0 N

Apparent weight of object = 4.0 N

4.0 N = 7.0 N - Buoyant force

Buoyant force = 7.0 - 4.0 = 3.0 N

In this case, the buoyant force is less than weight of the object thus the object will sink.

Answer:

The momentum of the ball at the maximum height is zero (0 kg.m/s).

Explanation:

Given;

mass of the ball, m = 0.17 kg

initial velocity of the ball, u = 12 m/s

acceleration due to gravity, g = 9.81 m/s²

At maximum height the final velocity of the ball, v = 0

Momentum of the ball at the maximum height is given as;

P = mv

P = (0.17)(0)

P = 0 kg.m/s

Therefore, the momentum of the ball at the maximum height is zero (0).

Answer:

1. Tsunami

2.Divergent

3.Sensor

4.Convergent

5.Magnitude

Explanation:

Answer:

Explanation:

The momentum of an object can be found by using the formula

momentum = mass × velocity

From the question we have

momentum = 0.85 × 19 = 16.15

We have the final answer as

Hope this helps you