Explain the law of inertia in your own words.

Answer: The correct answer is A). Animal burrow because burrow fossils represent the preserved byproducts of behavior rather than physical remains, they are considered a kind of trace fossil. One common kind of burrow fossil is known as Skolithos, and the similar Trypanites, Ophiomorpha and Diplocraterion.

Animal burrow is  an example of a trace fossil.

The preserved remnants of plants and animals that were submerged in sediments like sand and mud beneath ancient seas, lakes, and rivers are known as fossils. Any preserved sign of life that is typically older than 10,000 years is considered a fossil.

After death, soft body parts degrade quickly, while hard body parts, including bones, shells, and teeth, can be replaced by minerals that solidify into rock. Soft components like feathers, ferns, or other signs of life, such footprints or faeces, may also be preserved in extremely rare circumstances. Both less familiar fossils like ammonites and trilobites as well as microscopic-sized ones like single-celled foraminifera or pollen grains can be found in the remains.

A trace fossil, also known as an ichnofossil, is a record of biological activity preserved in the fossil record but without the actual preserved plant or animal remains. Contrasting with body fossils, which are the preserved remains of portions of an organism's body that were typically modified by later chemical activity or mineralization, are trace fossils. Ichnology, the study of such trace fossils, is the field of study of iconologists.

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Answer:

True Because if we observe carefully we will see that some moves of volleyball are similar to handball and tennis

Answer:

6 v

Explanation:

Answer:

Maximum pressure = 4450 N/m²

Explanation:

We'll begin by calculating the volume of the block. This can be obtained as follow:

Volume (V) = 5 cm × 3 cm × 2 cm

V = 30 cm³

Next, we shall determine the mass of the block. This can be obtained as follow:

Density = 8.9 g/cm³

Volume = 30 cm³

Mass =?

Density = mass /volume

8.9 = mass / 30

Cross multiply

Mass = 8.9 × 30

Mass = 267 g

Next, we shall covert 267 g to Kg.

1000 g = 1 Kg

Therefore,

267 g = 267 g × 1 Kg / 1000

267 g = 0.267 Kg

Thus, the mass of the block is 0.267 Kg

Next, we shall determine the force.

Force of gravity (g) = 10 N/Kg

Mass (m) = 0.267 Kg

Force (F) =?

F = mg

F = 0.267 × 20

F = 2.67 N

Next, we shall determine the minimum area since we are trying to obtain the maximum pressure. This can be obtained as follow:

Minimum area = 3 cm × 2 cm

Covert each measurement to metre (m) by dividing each measurement by 100

Minimum area = 0.03 m × 0.02 m

Minimum area = 0.0006 m²

Finally, we shall determine the maximum pressure. This can be obtained as follow:

Force (F) = 2.67 N

Minimum area = 0.0006 m²

Maximum pressure =?

Maximum pressure = Force /minimum area

Maximum pressure = 2.67 / 0.0006

Maximum pressure = 4450 N/m²

Answer:

a) Eg = 3,060.72 j

b) Ek = 2,080 j

c) Etotal = 5,140.72 j

Explanation:

The given parameters are;

The mass of the person, m = 65 kg

The height of the person, h = 4.8 m

The speed of the person, v = 8.0 m/s

a) The gravitational potential energy, [tex]E_g[/tex] = m·g·h

Where;

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

∴ Eg = 65 kg × 9.81 m/s² × 4.8 m = 3,060.72 j

Eg = 3,060.72 j

b) The kinetic energy, Ek = 1/2·m·v²

∴ Ek = 1/2 × 65 kg × (8.0 m/s)² = 2,080 j

Ek = 2,080 j

c) The constant total Mechanical Energy, Etotal = Eg + Ek

∴ Etotal = 3,060.72 j + 2,080 j = 5,140.72 j

Etotal = 5,140.72 j.

Answer:

Following are the solution to the given points:

Explanation:

The shovel shifts the course of the man's force.

Man can use less power.

The strength at the end of the shovel is growing.

Its shovel is a tool that is used to excavate and transfer its large particles. This streamlines the job for all the men to find the pit.

Its machines modify their method that uses force. Its man must therefore use less force to lift the bulk material.

Answer:

The shovel changes the direction of the force the man applies.

The man can apply less force.

The force applied by the end of the shovel increases.

(a,b,e)

Explanation:

edge 2021

Answer:

the plant died because of lack of sunlight

Explanation:

A plant gets its food from the photosynthesis process, which combines CO₂ and H₂0 into carbohydrates, the two substances are in the box, but for this they need energy from sunlight.

           CO₂ + H₂O + Sunlight /chlorophyll → (CH₂0)ₙ + O₂

the extra oxygen is released into the atmosphere.

Therefore the plant died because of lack of sunlight

Answer:

77%

Explanation:

efficiency= work output/work input X 100%

e = 1,200j/ 1,550 j x100%

e = 1,200/1,550= 0.77

e = 0.77 x 100%

e = 77%

Answer:

Lemme think rq

*inserts writing noises*

*insert intese thinking*

*insert big brain moment*

Well the weight IS 2.3 kg so I think the weight of a squirrel is 2.3 kg

Answer:

Silicon Dioxide

Answer:

Silicon dioxide

Explanation:

Answer:

12 N

Explanation:

Given that a 12 kg object's velocity changes from 5 m/s to 10 m/s in 5 s, what amount of force was applied to increase the speed of the object?

Solution

M = 12 kg

T = 5 s

The formula to use will be

F = ma

Where

Acceleration a = change in velocity/ time

Acceleration a = ( 10 - 5 ) / 5

Acceleration a = 5 / 5

Acceleration a = 1 m/s^2

Substitutes a into the formula above

F = ma

F = 12 × 1

F = 12 N

Therefore, the amount of force applied to increase the speed of the object is 12 N.

ANSWER:

Answer:

C

Explanation:

Answer:

Hello, I believe it would have a negative charge considering protons have a positive charge while elctrons have a negative charge

Explanation:

Answer:

i think it would be either b or c because....

Explanation:

"The desert biome is best adapted for aloe because aloe plant usuallly grows in open area under sunlight exposure where the intensity of sunlight is very high. Its grows in very rocky, dry and open areas."

Answer:=7x−7

Explanation:

=2x+5x+−7

=(2x+5x)+(−7)

=7x+−7

Answer:

Newton's Third Law

Explanation:

When one object exerts a force on another object, the second object exerts a force on the first object that is equal in magnitude, but opposite in direction; for every action, there is an equal, but opposite reaction; called the law of action-reaction.

Answer:

0.37 kg

Explanation:

I'm not a professor myself, but this is how I worked it out:

using the graph, after 100 seconds, the temperature is 100 degrees Celsius.

If we now substitute everything into the specific heat capacity equation, making the mass "m", we would come up with:

4200 = 155000/(m x 100)

If we rearrange and solve for m, we get 0.37 kg.

I'm not sure if I have done this correctly, feel free to correct me.

Hope this helps!

By using the specific heat capacity formula, the  mass of water in the kettle is 0.47 kg

Give that the energy transferred to the water in 100 seconds was 155 000 J and the specific heat capacity of water = 4200 J/kg °C

From the graph;

The initial temperature of the water at time zero is 22 degree Celsius

The final temperature of the water at time 100s is 100 degree Celsius

Change in temperature (T) = 100 - 22 = 78 degree Celsius

To determine the mass of water in the kettle, use the heat energy formula below

E = mcT

Substitute all the parameters into the equation

155000 = 4200 x 78 M

Make M the subject of the formula

M = 155000/327600

M = 0.47 kg

Therefore, the  mass of water in the kettle is 0.47 kg

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Answer:

(1) 2 ohms

(2) 12 Volts Across each resistor

(3) I₁ = 3 A, I₂ = 2 A, I₃ = 1 A

Explanation:

From the question,

(1) Equilvalent Resistance (Rt) for parallel connection is

1/Rt =(1/R₁)+ (1/R₂) + (1/R₃)

Where R₁ = 4 ohms, R₂ = 6 ohms, R₃ = 12 ohms

1/Rt = 1/4 +1/6 +1/12

1/Rt = (3+2+1)/12

1/Rt = 6/12

1/Rt = 1/2

Rt = 2 ohms.

(2) Since the resistors are connected in parallel, They will have the same potential difference across them,

Hence the P.d across each resistor = 12 Volts.

(3) For R₁,

I₁ = V/R₁ = 12/4

I₁ = 3 A.

For R₂,

I₂ = V/R₂

I₂ = 12/6

I₂ = 2 A

For R₃,

I₃ = V/R₃

I₃ = 12/12

I₃ = 1 A

Answer:

How does the weight of the nose on a paper

airplane affect the distance it will travel

when measured in centimeters?

EXPERIMENT CONCLUSION: (Explain whether or

not your data supports, or fails to support your identified

prediction. Explain why, including facts and details!)

The distance did increase when I added 1 paper clip.

However, I was surprised that the distance decreased

when I added a second paper clip. I believe this

happened because the nose of the plane became too

heavy. When the plane was released, the nose of the

plane descended much quicker. I think the plane was

more balanced in weight when only 1 clip was on the

nose.

Answer:

5 m/s

Explanation:

use speed=distance/time

300/60=5

Answer:

A neutron star is the most intensely dense object in all the universe. Of course, the argument can be made that a black hole is the most dense, but considering that a black hole is technically beyond the event horizon, it is neutron stars that get the top spot for the being the ‘most dense’.

Explanation:

:) please mark me as brainliest, thanks

Answer:

1. troposphere

2. troposphere (still)

3. The most abundant gas in the atmosphere is nitrogen, with oxygen second. Argon, an inert gas, is the third most abundant gas in the atmosphere.

4.The stratosphere is where you'll find the very important ozone layer. The ozone layer helps protect us from ultraviolet radiation (UV) from the sun. In fact, the ozone layer absorbs most of the UV radiation the sun sends to us. Life as we know it wouldn't be possible without this layer of protection.

5. It contains the ozone, which protects Earth from the sun's radiation. Why aren't there many meteors in the troposphere? They burn up before they reach the troposphere.

Explanation:

1.The troposphere is the atmospheric layer closest to the planet and contains the largest percentage (around 80%) of the mass of the total atmosphere. Temperature and water vapor content in the troposphere decrease rapidly with altitude.

2.The layer closest to Earth's surface is the troposphere, reaching from about seven and 15 kilometers (five to 10 miles) from the surface. The troposphere is thickest at the equator, and much thinner at the North and South Poles.

3. Explained in the answer.

4. Explained in the answer.

5. Explained in the answer.

Answer:

a. i His weight and the normal force on him due to the ground

ii The weight and the normal force are balance

b. There are four forces acting on him while he is walking. These are the backward push due to his leg on the ground and the frictional force between his feet in the horizontal direction and, his weight and normal force on his feet in the vertical direction.

Explanation:

a. i Describe the forces that are acting on a person who is standing still on a sidewalk

His weight and the normal force on him due to the ground act on him while the person is standing on the sidewalk

ii. identify whether the forces are balanced or unbalanced.

The weight and the normal force are balance because, the person does not move in the vertical or horizontal direction.

b. Describe how the forces would change if the person started walking.

When the person starts walking, the backward push due to his leg on the ground and the frictional force between his feet and the ground now acts on him in the horizontal direction. The frictional force acts in the forward direction and tends to move him forward. His weight and normal force on his feet still act vertically and are both balanced since he does not move in the vertical direction.

So, there are four forces acting on him while he is walking. These are the backward push due to his leg on the ground and the frictional force between his feet in the horizontal direction and, his weight and normal force on his feet in the vertical direction.

Answer:

La altura máxima que alcanzan el mono y el acróbata es [tex]z = \frac{1}{2}\cdot \left(\frac{M}{M+m} \right)\cdot v_{o}^{2}+\left(1-\frac{M\cdot g}{M+m} \right)\cdot h[/tex].

Explanation:

Asumamos que tanto el acróbata, el mono y el sistema acróbata-mono son conservativos y que el acróbata comienza su acción a una altura de cero. El estudio se divide en dos etapas: (i) El acróbata se dirige al mono, (ii) El acróbata recoge al mono y alcanzan una altura máxima.

Para resolver esta cuestión, nos valemos del Principio de Conservación de la Energía.

Parte I

La energía cinética traslacional inicial ([tex]K_{1,a}[/tex]) es igual a la suma de la energía cinética traslacional final ([tex]K_{2, a}[/tex]) y la energía potencial gravitacional final ([tex]U_{g,2,a}[/tex]).

[tex]K_{1,a} = K_{2,a} + U_{g,2,a}[/tex] (1)

[tex]\frac{1}{2}\cdot M \cdot v_{o}^{2} = \frac{1}{2}\cdot M\cdot v_{1}^{2} + M\cdot g \cdot h[/tex] (1b)

Donde:

[tex]M[/tex] - Masa del acróbata.

[tex]g[/tex] - Constante gravitacional.

[tex]v_{o}[/tex] - Rapidez inicial del acróbata.

[tex]v_{1}[/tex] - Rapidez del acróbata justo antes de recoger al mono.

[tex]h[/tex] - Altura inicial del mono.

Parte II

La suma de las energías iniciales cinética traslacional ([tex]K_{2, a}[/tex]) y potencial gravitacional de acróbata ([tex]U_{g,2,a}[/tex]) y la energía inicial potencial gravitacional del mono ([tex]U_{g,2,m}[/tex]) es igual a la suma de las energías potenciales gravitacionales iniciales del sistema acróbata-mono ([tex]U_{g,3,a+m}[/tex]), es decir:

[tex]K_{2,a} + U_{g,2,a}+U_{g,2,m} = U_{g,3,a+m}[/tex] (2)

[tex]\frac{1}{2}\cdot M\cdot v_{1}^{2} + (M+m)\cdot g \cdot h = (M+m)\cdot g \cdot z[/tex] (2b)

Donde:

[tex]m[/tex] - Masa del mono.

[tex]z[/tex] - Altura máxima del sistema acróbata-mono.

De (1b) tenemos que la rapidez del acróbata justo antes de recoger al mono es:

[tex]\frac{1}{2}\cdot M \cdot v_{o}^{2} = \frac{1}{2}\cdot M\cdot v_{1}^{2} + M\cdot g \cdot h[/tex]

[tex]v_{o}^{2} = v_{1}^{2}+2\cdot g\cdot h[/tex]

[tex]v_{1} = \sqrt{v_{o}^{2}-2\cdot g\cdot h}[/tex]

Finalmente, la altura máxima alcanzada por el sistema acróbata-mono es:

[tex]\frac{1}{2}\cdot M\cdot v_{1}^{2} + (M+m)\cdot g \cdot h = (M+m)\cdot g \cdot z[/tex]

[tex]z = \frac{M\cdot v_{1}^{2}}{2\cdot (M+m)\cdot g}+h[/tex]

[tex]z = \frac{1}{2}\cdot \left(\frac{M}{M+m} \right)\cdot (v_{o}^{2}-2\cdot g\cdot h)+ h[/tex]

[tex]z = \frac{1}{2}\cdot \left(\frac{M}{M+m} \right)\cdot v_{o}^{2}+\left(1-\frac{M\cdot g}{M+m} \right)\cdot h[/tex]

La altura máxima que alcanzan el mono y el acróbata es [tex]z = \frac{1}{2}\cdot \left(\frac{M}{M+m} \right)\cdot v_{o}^{2}+\left(1-\frac{M\cdot g}{M+m} \right)\cdot h[/tex].