An object of mass 17.0 kg subjected to a non-zero net force moves with an acceleration of 2.2 m/s^2(a) Determine the net force (in N) acting on it. (Enter the magnitude only.)________N(b) What acceleration (in m/s^2) would a 34,0-kg object have if the same net force is applied to it?_______m/s^2

Planetary scientists estimate the ages of the outflow channels and valley networks on mars by craters

Due to impact of meteorite, volcanic activity, or an explosion , there occurs a depression at a certain area , that is called crater .

Ages of the outflow channel and valley network can be estimated by counting the number of impact craters. Because of the almost constant  rate at which impacts have occurred in the solar system .

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The size and aperture of the telescopes determine its ability to gather light

and reveal details.

A) Observatory X can detect dimmer stars and Observatory Y reveals more

details.

Reasons:

The benefits of using several small telescopes included the production

high resolution images of bright objects such as binary stars that reveal

more details similar to a telescope with a very large aperture, with a

process known as interferometry, through which a star's diameter can be

determined.

The benefit of a large telescope is its ability to gather light from a source,

thereby making possible to detect a dim light source, such as dimmer

stars.

Therefore;

The single large telescope in observatory X can detect dimmer stars while

the interferometer consisting of five 10-meter telescopes, spread out over a

region 100 meters across can see more detail in its images.

The correct option is; A) Observatory X can detect dimmer stars and

Observatory Y reveals more details.

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Question options:

A) Observatory X can detect dimmer stars and Observatory Y reveals more

details.

B) Observatory Y can detect dimmer stars and Observatory X reveals more

details.

C) Observatory X both detect dimmer stars and reveals more details

D) Observatory Y both detect dimmer stars and reveals more details

Answer: 34

Explanation:

34

Answer:

Explanation:

Using the formula

v = fλ

v is the speed

f is the frequency

λ is the wavelength

Given the following

v = 332m/s

Frequency is the number of oscillation completed in one second. Since 332 waves passes through a medium in one second, hence the frequency us 222 Hertz

From the formula

λ = v/f

λ = 332/332

λ = 1m

Hence the wavelength is 1m

The difference between the gravitational force experienced by the top and bottom objects is the force between the top objects is 3 times that of the bottom objects.

Newton's law of universal gravitation states the force of attraction between two objects is directly proportional to the product of the masses and inversely proportional to the square of the distance between the objects.

F = Gm1m2/r²

The bigger the distance of separation, the smaller the gravitational force between the object and vice versa.

The bigger the mass of the objects, the bigger the gravitational force between the objects.

Thus, the difference between the gravitational force experienced by the top and bottom objects is the force between the top objects is 3 times that of the bottom objects.

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Answer:

217 N

Explanation:

Answer:217N

Explanation:

First you get the data:

Time conversion: hours to seconds:

\(\bf{2\not{h}*\left(\dfrac{3600 \ seg}{1\not{h}}\right)=7200 \ seg }\)

We calculate the distance, using the formula:

                           \(\bf{d=t*v }\)

                           \(\bf{d=3600\not{seg}*3\frac{m}{\not{m}} }\)

                           \(\bf{d=21600 \ m}\)

Answer: I travel to 21600 meters.

The velocity of the jet is calculated to be 900 miles/hr when the relation between velocities of both propeller and the jet is given.

The distance travelled by a propeller plane and a jet = 3000 miles

Velocity of the plane Vp = 1/3 Velocity of the jet Vj

Vp = 1/3 Vj

Vj = 3 Vp ----(1)

Time taken by the propeller plane to complete the trip = 10 hours

Velocity of the propeller plane = Distance of the propeller plane/Time

⇒ 3000/10 = 300 miles/hr

The velocity of the jet = 3 times the velocity of the propeller plane

⇒ 3 × 300 miles/hr = 900 miles/hr

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Answer:

The answer is a 11.25m/s

Answer:That only applies to highly polished surfaces, eg mirrors.

If you take a high quality laser (ie with low divergence) and aim it at a wall, you can see the spot where the laser beam reaches the wall from anywhere with a direct line-of-sight to the spot where the laser beam reaches the wall. This due to micro imperfections on the surface of the wall. At a microscopic level, the wall surface is very rough and pointing in all directions.

As to why, a beam of light bounces of a highly polished surface, I can only surmise that it is essentially due to kinematics, ie the only force opposing the light beam is normal to the surface, hence there no forces along the reflective surface. Since there are no forces along the reflective surface, the speed component of light along the reflective surface remains unchanged. However, on the plane perpendicular to the reflective surface the, the light photons bounce off at the same speed at which the hit the reflective surface because the mass of the reflective surface is much much much larger than the mass of the photons, which means that the reflective surface won’t move at all. Since conservation of momentum requires that momentum after the collision be the same as the momentum before the collision then the only way for that to happen is if the velocity of the photon perpendicular to the reflective surface is of exactly the same magnitude but in the opposite direction. Vector resolution of the speed component of the reflected beam means that the angle of reflection must be the same as the angle of incidence.

Explanation:

B

They store the radioactive waste in pools of water until it is no longer dangerous. It is then transferred to a dry cask. :))

The pressure exerted by both on the trampoline is 25,480 N/m².

The pressure exerted by both on the trampoline is the force per unit area which is calculated as follows;

P = F/A

where;

Substitute the given parameters and solve for the pressure exerted.

P = (mg)/A

A = 0.25 m x 0.1 m

A = 0.025 m²

P = (65 x 9.8) / (0.025)

P = 25,480 N/m²

Thus, the pressure exerted at any given surface depends on the applied force and the area of the given surface.

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Answer:

yes that is correct

Explanation:

Answer: D.

Explanation:

a p e x confirmed its correct

Water Skiing is the only sport in the option above which is not an areobic outdoor sport. (Option D)

Aerobic exercise also known as cardio or cardio-respiratory exercise is physical exercise of low to high intensity that depends primarily on the aerobic energy-generating process. Examples of aerobic sports are swimming, climbing, running, kayaking etc.

Therefore, the outdoor sports which is not an areobic sport is water skiing.

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Answer:

to stretch their muscles so they don't strain or hurt themselves

Explanation:

hi!

Answer:

it's kinda like a warm up. they're getting their body ready for extreme physical activity.

Explanation:

hope it helps, and hi

Answer:

242.4 V

Explanation:

Applying,

Q = cm(t₂-t₁)................ Equation 1

Where Q = amount of  heat, c = specific heat capacity, m = mass, t₁ = initial temperature, t₂ = final temperature.

But this heat is produced by electrical power

Therefore,

P = Q/t

Where P = Electrical power, t = time

Also,

P = V²/R

Where V = voltage, R = resistance

Therefore

V²/R = Q/t......................... Equation 2

Substitute equation 1 into equation 2

V²/R = cm(t₂-t₁)/t................. Equation 3

Given: R = 250 ohms, m = 4.6 kg, t₁ = 22°C, t₂ = 32°C, t = 15 min. = (15×60) = 900 seconds

Constant: c = 4600 J/kg.°C

Substitute these values into equation 2

V²/250 = 4600(4.6)(32-22)/900

V²/250 = 235.11

V² = 250×235.11

V² = 58777.78

V = √58777.78

V = 242.4 V

Answer:

A) Conduction

Explanation:

The mitt keeps the heat from reaching your hand through conduction

Answer: A.conduction

Explanation: i got it right on my test!

Answer:

Mass they are formed with

Explanation:

All stars are born with the same composition such as one quarter of helium, three quarter of hydrogen, and heavier element does not found more than two except their mass. Mass that differ all stars from other stars they born with. How much the mass of a star is large , the star would be as messy.

For example our sun is a star and have large mass four times greater than other living stars. It is found that smallest mass is around 1/12th of sun mass. The masses of a star is determined by its orbit binary. Sun is called a brown dwarf.

Water freezes more quickly when it is exposed to colder temperatures and when it is in a container with a large surface area, which allows for faster heat transfer.

Water freezes at 32 degrees Fahrenheit (0 degrees Celsius). The time it takes for water to freeze at this temperature depends on several factors such as the volume of water, the container it is in, and the environment in which it is located.

Assuming that the water is in a typical home freezer and the volume of water is not too large, it would typically take several hours for the water to freeze completely. The exact amount of time it takes for the water to freeze would depend on the volume of water and the temperature of the freezer.

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Given that,

Mass of satellite = 500 kg

Gravitational force = 3000 N

We need to calculate the radius of the circular orbit

Using formula of gravitational force

\(F_{g}=\dfrac{GMm}{(R+h)^2}\)

Where, G = gravitational constant

R = radius of earth

h = radius of the circular orbit

M = mass of earth

m = mass of satellite

Put the value into the formula

\(3000=\dfrac{6.67\times10^{-11}\times500\times6\times10^{24}}{(6.4\times10^{6})^2+h^2}\)

\(h^2=\dfrac{6.67\times10^{-11}\times500\times6\times10^{24}-3000\times(6.4\times10^{6})^2}{3000}\)

\(h=\sqrt{\dfrac{6.67\times10^{-11}\times500\times6\times10^{24}-3000\times(6.4\times10^{6})^2}{3000}}\)

\(h=5073460.35\ m\)

\(h=5.1\times10^{6}\ m\)

(II). We need to calculate the speed of the satellite

Using formula of velocity

\(v=\sqrt{\dfrac{GM}{r}}\)

Put the value into the formula

\(v=\sqrt{\dfrac{6.67\times10^{-11}\times6\times10^{24}}{5.1\times10^{6}}}\)

\(v=8858.36\ m/s\)

\(v=8.8\times10^{3}\ m/s\)

\(v=8.8\ km/s\)

(III). We need to calculate the period of the orbit

Using formula of  time period

\(T=2\pi\sqrt{\dfrac{r^3}{GM}}\)

Put the value into the formula

\(T=2\pi\sqrt{\dfrac{(5.1\times10^{6})^3}{6.67\times10^{-11}\times6\times10^{24}}}\)

\(T=3617.40\ sec\)

\(T=1.00\ hr\)

Hence, (I). The radius of the circular orbit is \(5.1\times10^{6}\ m\)

(II). The speed of the satellite is 8.8 km/s.

(III). The period of the orbit is 1.00 hr.

Answer:

Grandma will arive \(\frac{1}{2}\)hr later

Explanation:

\(s=20\\d=210 mi|\\ s=70 mi|h \\ \\ \frac{60}{1} =\frac{210}{+} = 3.5 hrs\)

Answer: the landers velocity DECRECES AWAY the refrence, D

Explanation:

The  velocity between approximately 11 seconds and 15 seconds, the lander's velocity increases toward the reference.

Given:

The lander's velocity-time graph

To find:

The velocity of the lander between the time frame of 11 seconds to 15 seconds.

Solution:

So, from this, we can conclude that the velocity between approximately 11 seconds and 15 seconds, the lander's velocity increases toward the reference.

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Answer:

The mass of the air in a room can be calculated by multiplying the volume of the room by the density of the air. The volume of the room can be calculated by multiplying its length, width, and height. So, if the room has dimensions of 1m x 2m x 3m, its volume would be 1m x 2m x 3m = 6 cubic meters. If the density of the air in the room is 1.3kg/m³, the mass of the air in the room would be 6m³ x 1.3kg/m³ = 7.8kg.

Answer:

34kg

Explanation:

Answer:

b

Explanation:

Answer: less than 9.8 m/s

Explanation:

Answer:

To find the average speed after 2 minutes, we need to calculate the total distance covered in 2 minutes and divide it by 2.

Total Distance after 2 minutes = 75m

Total Time after 2 minutes = 2 minutes

Average Speed after 2 minutes = Total Distance / Total Time

Average Speed after 2 minutes = 75m / 2 min = 37.5 m/min

Therefore, the average speed after 2 minutes is 37.5 m/min.

I Hope This Helps!

Considering a 0.95 MeV y-ray photon and Randomized Variables E = 0.95 MeV so the  frequency of light will be 2.294×10²⁰ Hz.

Given,

The energy of photon

E = 0.95 MeV = 0.95× 1.6×10⁻¹³ J or 1.52 × 10⁻¹³ J  

Using E = hυ  

We have:  

υ = E/h = 1.52×10⁻¹³/(6.626×10⁻³⁴)  

= 2.294×10²⁰ Hz  

Thus the frequency of light will be 2.294×10²⁰ Hz.

Associated to each wave vector ~k and polarization state σ there is an electromagnetic normal mode of oscillation with frequency ω(k) = kc and complex amplitude aσ( ~k). In the absence of sources, the behavior of the electromagnetic field is completely described by these normal modes. We will assume the normal modes are defined using periodic boundary conditions.

For each ~k and σ there is a Hilbert space of square integrable functions and a Hamiltonian( ~k). For each ~k and σ there is a basis of energy eigenvectors; the corresponding ladder operators act on this basis by adding and subtracting quanta of energy ω(k).

We can use this basis to induce a basis for the tensor product over all the oscillator Hilbert spaces, which is the state space for the composite system of normal modes, i.e., the electromagnetic field.

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Answer:

3.5 Hours

Explanation:

I believe all that has to be done is dividing 84 by 24, which results in 3.5, or 3 and a 1/2 hours.

The pressure in a tire is observed to rise after a car has been driven 5 miles. This is an example of Charles' Law.

Charles' Law describes the relationship between the temperature and volume of a gas when the pressure is kept constant. Charles' Law is stated as:V α TT α VP= k, where k is a constant.In this case, the temperature of the gas is proportional to the volume.

As the temperature of the tire increases due to friction with the road, the volume of air in the tire expands, resulting in an increase in pressure. This is precisely what happens when a tire is driven for an extended period. Charles' Law may also be used to describe the behavior of an ideal gas in general.

It implies that as the temperature of a gas increases, the volume of the gas increases, and vice versa. The pressure in a tire is an excellent example of Charles' Law because it occurs when the temperature of the gas inside the tire rises, resulting in an increase in pressure.

The pressure in a tire increases when a car is driven because the friction between the tire and the road raises the tire's temperature, causing the air inside the tire to expand, which increases the pressure in the tire.In conclusion, Charles' Law is an important gas law that describes the relationship between the temperature and volume of a gas when the pressure is kept constant.

This law is relevant to understanding how the pressure in a tire changes when a car is driven and the temperature of the gas inside the tire increases. Therefore, the pressure in a tire is an example of Charles' Law.

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The image distance is 5.85 cm, the magnification is -0.45, and the image height is -1.0125 cm (inverted and smaller than the object).

Using the thin lens equation, we can find the image distance (di),

1/f = 1/d₀ + 1/d₁ where f is the focal length of the lens and do is the object distance. Substituting the given values,

1/8 = 1/13 + 1/d₁

Solving for di,

di = 5.85 cm

The magnification (M) is given by,

M = - d₁/d₀ where the negative sign indicates that the image is inverted. Substituting the given values,

M = -5.85/13

M = -0.45 (to two decimal places)

The image height (hi) can be found using the magnification equation,

M = h₁/h₀ where h₀ is the object height. Substituting the given values and solving for hi,

h₁ = M x h₀

h₁ = -0.45 x 2.25 cm

= -1.0125 cm

The negative sign indicates that the image is inverted and smaller than the object. Therefore, the image distance is 5.85 cm, the magnification is -0.45, and the image height is -1.0125 cm (inverted and smaller than the object).

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