What would happen to a 100N gravitational force if both masses were doubled and the radius were
doubled

Answer: The processes are conduction, convection and radiation.

Explanation: Unless people interfere, thermal energy — or heat — naturally flows in one direction only: from hot toward cold. Heat moves naturally by any of three means. The processes are known as conduction, convection and radiation. Sometimes more than one may occur at the same time.

Answer:

This is because some kinetic energy had been transferred to something else.

Explanation:

An inelastic collision is a collision in which there is a loss of kinetic energy. While momentum of the system is conserved in an inelastic collision, kinetic energy is not.

The force that cannot be the resultant of these forces is 10N since it is less than both given forces

Given the forces with a magnitude of 12N and 24N, the resultant of these forces must not be less than any of the two forces.

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Answer:

1,967 kilojoules.

Explanation:

1 kilo = a thousand.

There are 1.967 kilojoules in 1,967,256 joules.

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Answer:

C.Compacting the waste and burying it in a shallow landfill.

Explanation:

It is buried in a shallow landfill because it is less reactive.

Answer:

100 J work was done.

Hope you could get an idea from here.

Doubt clarification - use comment section.

An example of the use of expansion of materials when they are heated is in

Thermometers. Thermometer is an instrument which is used to measure the

temperature of a place.

It contains mercury which expands as temperature increases. This is used

to accurately determine the temperature of an object.

The disadvantage of expansion of materials when they are heated can be

seen in roads as there is expansion of the surface as a result of heat

exposure leads to roads being rough thereby incurring more costs for

repairs.

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Answer:

Although the Earth has a lesser mass than the Sun, it is far closer to you, allowing for a stronger pull.

Explanation:

Answer: D

Explanation: moogle helps lol.

Hi there!

Recall the conservation of momentum for an inelastic collision:

[tex]\large\boxed{m_1v_1 + m_2v_2 = (m_1 + m_2)v_f}}[/tex]

Remember, velocity is a VECTOR and direction must be accounted for. Let the 1000 kg car have a positive velocity and the 1500 kg car have a negative velocity (opposite direction).

Plug in the given values:

[tex](1000)(20) + (1500)(-10) = (1000 + 1500)v_f}}[/tex]

Solve:

[tex]20000 -15000 = 2500v_f}}\\\\5000 = 2500v_f\\\\v_f =\boxed{ 2 m/s}[/tex]

So, your science teacher has given your class the classic "mousetrap car" assignment: to make, design, and build a small vehicle powered by the snapping action of a mousetrap to make your car travel as far as possible. If you want to come out ahead of all the other students in your class, you'll need to make your car as efficient as possible so you can squeeze every last inch out of your "car". With the right approach, it's possible to streamline your car's design for maximum distance using only common home materials. You could also buy a mousetrap car kit from any craft store and skip wondering if it will work.

Use large rear wheels. Large wheels have greater rotational inertia than small wheels. In practice, this means that once they start rolling, they're harder to stop rolling. This makes large wheels perfect for distance-based contests — theoretically, they'll accelerate less quickly than smaller wheels, but they'll roll much longer and they'll travel a greater distance overall. So, for maximum distance, make the wheels on the drive axle (the one the mousetrap is tied to, which is usually the rear one) very large. The front wheel is a little less important — it can be large or small. For a classic drag racer look, you'll want big wheels in the back and smaller ones in front.

Use thin, light wheels. Thinner wheels have less friction and may go farther if the distance is what you want or need with your mousetrap racer. It's also important to take the weight of the wheels themselves into account — any unneeded weight will ultimately slow your car down or lead to added friction. In addition, it's worth noting that wide wheels can even have a small negative effect on the car's drag due to air resistance. For these reasons, you'll want to use the thinnest, lightest wheels available for your car.

Old CDs or DVDs work fairly well for this purpose — they're large, thin, and extremely light. In this case, a plumbing washer may be used to reduce the hole size in the middle of the CD (to fit the axle better).

If you have access to old vinyl, these also work extremely well, though they may be too heavy for the smallest mousetraps.

Use a narrow rear axle. Assuming your car is a rear-wheel-drive car, each time your rear axle turns, the rear wheels turn. If your rear axle is extremely skinny, your mousetrap car will be able to turn it more times for the same length of string than it would if it were wider. This translates to turning your rear wheels more times, meaning greater distance! For this reason, it's a wise idea to make your axle out of the skinniest material available that can still support the weight of the frame and wheels.

Narrow wooden dowel rods are a great, easily-accessible choice here. If you have access to thin metal rods, these are even better — when lubricated, they usually have less friction.

Create traction by giving the edges of the friction of the wheels. If the wheels slip against the ground when the trap is sprung, energy is wasted — the mousetrap works to make the wheels turn, but you don't get any extra distance. If this happens with your car, adding a friction-inducing material to the rear wheels may reduce their slippage. To keep your weight requirements down, use only as much as is necessary to give the tips of the wheels some grip and no extra. Some suitable materials are:[1]

Electrical tape

Rubber bands

Additionally, placing a piece of sandpaper under the rear wheels at the start line can reduce slippage as the car begins to move (when it is most likely)

Answer:

he was a random act like you to be the first time I see is a great day and night to get a very happy to see you soon I hope that the world is not the same thing to say about this

a) When a box slides across a rough surface, eventually coming to rest, its kinetic energy is used during work done against frictional resistance force.

b) This work done is stored in the box as a potential energy. Thus the motion of the box is consistent with the Law of Conservation of Energy.

According to the work-energy theorem, the work done by the net force acting on a body equals the change in kinetic energy.

It can simply be written as:

Work done = initial kinetic energy - final kinetic energy

The work energy theorem equation is the one presented above.

Now when a box slides across a rough surface, eventually coming to rest, frictional force comes into play. This force is opposite to the direction of motion of the box. Hence, the kinetic energy of the box is used during work done against frictional resistance force and stored in the box as a potential energy and thus  Law of Conservation of Energy followed.

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For the front glass of the car to get wet, [tex]V_c \geq 10 \ m/s[/tex].

The given parameters:

The relative velocity of the car with respect to the falling rain is calculated as;

[tex]V_{C/R} = V_C- V_R[/tex]

Thus, for the front glass of the car to get wet, [tex]V_c \geq 10 \ m/s[/tex].

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Answer:

potential energy = mgh = 300 × 10 × 6m = 18000 joule or 18 kilo joule.

Explanation:

Answer:

the slope of velocity-time graph represent an object acceleration

Las mascotas son una buena fuente de compañía y socias en la soledad.

La protagonista ama a su gato porque las mascotas son una buena fuente de compañía y es un antídoto para la soledad. Estas mascotas como perros, gatos, loros, etc. dan placer y felicidad a las personas.

La gente tiene mascotas para pasar tiempo con ellas y jugar con ellas para refrescar sus mentes. Tener mascotas es una actividad maravillosa que hace que nuestra mente esté relajada y llena de felicidad, por lo que podemos concluir que las mascotas son una buena fuente de compañía y compañera en la soledad.

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Answer:

Option C is the correct answer

Explanation:

Distance travelled by car during reaction time

[tex]=15\times0.4\\\\=6m[/tex]

The car stopped before hitting the animal by [tex]1 m[/tex]

Distance travelled during deceleration is [tex]21-6-1=14m[/tex]

Hence by [tex]v^2=u^2+2as[/tex]

We have

[tex]0^2=15^2+2 \cdot a \cdot 14\\\\a=\frac{-225}{28} \\\\=-8.03m/s^2[/tex]

Option C is the correct answer

Distance traveled during reaction time

Total distance

[tex]\\ \sf\longmapsto v^2-u^2=2as[/tex]

[tex]\\ \sf\longmapsto -(15)^2=2(14)a[/tex]

[tex]\\ \sf\longmapsto -225=28a[/tex]

[tex]\\ \sf\longmapsto a=-8.0m/s^2[/tex]

Answer:

friction. or gravity both ofbthem .....................

..

Answer:

fossils found on Earth

Explanation:

i just did the test 100%

Answer:

Fossils on Earth

Explanation:

Fossils can show that the continents were all together by putting them together on a map you can see the fit. and when you look where fossils were, they would be right next to each other.

Answer:

1. 60 kg m/s

2. 2.4 kg

3. none they both have same momentum

Both carts experience the same force but Cart A has a greater speed after the recoil.

The given parameters;

Apply the principle of conservation of linear momentum to determine the velocity of the carts after collision;

[tex]m_Av_0_A\ + m_Bv_0_B = m_Av_f_A \ + m_Bv_f_B\\\\the \ initial \ velocity \ of \ both \ carts = 0\\\\0.4(0) + 0.8(0) = 0.4v_f_A + 0.8v_f_B\\\\0 = 0.4v_f_A + 0.8v_f_B\\\\0.4v_f_A = -0.8v_f_B\\\\v_f_A= \frac{-0.8 v_f_B}{0.4} \\\\v_f_A = - 2 \ v_f_B \ \ m/s[/tex]

According to Newton's third law of motion, action and reaction are equal and opposite. The force exerted on cart A is equal to the force exerted on cart B but in opposite direction.

[tex]F_A = -F_B[/tex]

Thus, the correct statement that compares the motion and forces acting on the two carts is "Both carts experience the same force but Cart A has a greater speed after the recoil."

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The planet that Punch should travel to in order to weigh 118 lb is Pentune.

The mass of Punch will be constant in every planet;

[tex]W = mg\\\\m = \frac{W}{g}\\\\m = \frac{236}{g}[/tex]

The acceleration due to gravity of each planet with respect to Earth is calculated by using the following relationship;

[tex]F = mg = \frac{GmM}{R^2} \\\\g = \frac{GM}{R^2}[/tex]

where;

For Planet Tehar;

[tex]g_T =\frac{G \times 2.1M}{(0.8R)^2} \\\\g_T = 3.28(\frac{GM}{R^2} )\\\\g_T = 3.28 g[/tex]

For planet Loput:

[tex]g_L =\frac{G \times 5.6M}{(1.7R)^2} \\\\g_L = 1.94(\frac{GM}{R^2} )\\\\g_L = 1.94g[/tex]

For planet Cremury:

[tex]g_C =\frac{G \times 0.36M}{(0.3R)^2} \\\\g_C = 4(\frac{GM}{R^2} )\\\\g_C = 4 g[/tex]

For Planet Suven:

[tex]g_s =\frac{G \times 12M}{(2.8R)^2} \\\\g_s = 1.53(\frac{GM}{R^2} )\\\\g_s = 1.53 g[/tex]

For Planet Pentune;

[tex]g_P =\frac{G \times 8.3 }{(4.1R)^2} \\\\g_P = 0.5(\frac{GM}{R^2} )\\\\g_P = 0.5 g[/tex]

For Planet Rams;

[tex]g_R =\frac{G \times 9.3M}{(4R)^2} \\\\g_R = 0.58(\frac{GM}{R^2} )\\\\g_R = 0.58 g[/tex]

The weight Punch on Each Planet at a constant mass is calculated as follows;

[tex]W = mg\\\\W_T = mg_T\\\\W_T = \frac{236}{g} \times 3.28g = 774.08 \ lb\\\\W_L = \frac{236}{g} \times 1.94g =457.84 \ lb\\\\ W_C = \frac{236}{g}\times 4g = 944 \ lb \\\\ W_S = \frac{236}{g} \times 1.53g = 361.08 \ lb\\\\W_P = \frac{236}{g} \times 0.5 g = 118 \ lb\\\\W_R = \frac{236}{g} \times 0.58 g = 136.88 \ lb[/tex]

Thus, the planet that Punch should travel to in order to weigh 118 lb is Pentune.

The complete question is below:

Which planet should Punch travel to if his goal is to weigh in at 118 lb? Refer to the table of planetary masses and radii given to determine your answer.

Punch Taut is a down-on-his-luck heavyweight boxer. One day, he steps on the bathroom scale and "weighs in" at 236 lb. Unhappy with his recent bouts, Punch decides to go to a different planet where he would weigh in at 118 lb so that he can compete with the bantamweights who are not allowed to exceed 118 lb. His plan is to travel to Xobing, a newly discovered star with a planetary system. Here is a table listing the planets in that system (find the image attached).

In the table, the mass and the radius of each planet are given in terms of the corresponding properties of the earth. For instance, Tehar has a mass equal to 2.1 earth masses and a radius equal to 0.80 earth radii.

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Answer:

9758 how many significant figures

Answer:

The earth travels around the sun in a (parabolic) path known as the ecliptic.

The axis of the earth's rotation is tilted about 23 deg to the ecliptic.

At the winter solstice (around Dec. 21), the sun will appear to be at its farthest south in its orbit. (This marks the beginning of winter for observers in the northern hemisphere and the beginning of summer for an observer in the southern hemisphere)

The tilt of the axis of the earth's rotation causes the change in the daylight hours.

From the graph, net work done on the box and  the speed of the box when it reaches position x = 8.0 m are 28 Nm and 3.2 m/s respectively

From the question, these are the given parameters;

Part A

The net work done will be the area under the graph.

From position X = 2m to X = 8m gives us the shape of a trapezium.

A = 1/2( a + b )h

A = 1/2( 2 + 6 ) x 8

A = 8 x 4

A = 32 Nm

From X = 0 to X = 2 gives us the shape of a triangle.

A = 1/2bh

A = 1/2 x 2 x (-4)

A = -4 x 1

A = -4 Nm

Net Work done = 32 - 4

Net work done = 28 Nm

Part B

Applying the concepts of work and energy to solve for the speed of the box when it reaches position x = 8.0 m

Net Work done = 1/2m[tex]V^{2}[/tex]

Substitute all the necessary parameters

28 = 1/2 x 5.5 x [tex]V^{2}[/tex]

5.5[tex]V^{2}[/tex] = 56

[tex]V^{2}[/tex] = 56/5.5

[tex]V^{2}[/tex] = 10.18

V = [tex]\sqrt{10.1818}[/tex]

V = 3.19 m/s

Therefore, net work done on the box and  the speed of the box when it reaches position x = 8.0 m are 28 Nm and 3.2 m/s respectively

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Answer:

So what you do is you add all of them then you put an n with the answer.

Explanation:

Hi there!

Recall Newton's Second Law:

[tex]\large\boxed{\Sigma F = ma}}[/tex]

∑F = Net force (N)

m = mass (kg)

a = acceleration (m/s²)

We are already given the mass and acceleration, so we can plug these values into the equation:

∑F = 1600 · 4.5 = 7200 N

Answer:

We are required to analyze the conservation of kinetic energy in a perfectly inelastic collision.

kinetic energy:K=1/2mv²

momentum:p=mv

Explanation:

Using the equations for final velocity in terms of masses and initial velocity for a perfectly inelastic collision, work out the final kinetic energy. Calculate the initial kinetic energy and show that kinetic energy is not conserved. Use variables, no numbers

You need to follow law of conservation of energy

We know

You may use momentum formula mass×velocity

In both you need to form equations then solve for final velocity