The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10-11 Nm2/kg2.)
2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Calculate the speed of the satellite.

Answers

Answer 1

Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Explanation: To find the answer, we need to know about the different equations of planetary motion.

How to find the initial speed of the rock as it left the astronaut's hand?We have given with the following values,

                    [tex]m_g=7.88*10^{18}kg\\r_g=6.32*10^4 m.\\h_{max}=1.44*10^3m.\\[/tex]

We have the expression for the initial velocity as,

                    [tex]v=\sqrt{2gh}[/tex]

Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,

                  [tex]g_g=\frac{GM_g}{r_g^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2} =0.132m/s^2.[/tex]

Now, the velocity will become,

                  [tex]v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s.[/tex]

How to find the speed of the satellite?As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,

                [tex]v=\sqrt{ \frac{GM}{r}}=\sqrt{\frac{7.88*10^{18}*6.67*10^{-11}}{1.45*10^5 }} =3.624km/s.[/tex]

Thus, we can conclude that,

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Learn more about the equations of planetary motion here:

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Answer 2

1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.

2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.

Understanding the planetary motion equations is necessary in order to determine the solution.

How to determine the rock's original speed when it left the astronaut's hand?The starting velocity's expression is as follows:

                             [tex]v=\sqrt{2gh}[/tex]  

So, in order to determine v, we must determine the acceleration of glob caused by gravity. We already have,

                    [tex]a=\frac{GM}{R^2} =0.132m/s^2[/tex]

The velocity will now change to,

                  [tex]v=\sqrt{2*9.8*0.132} =19.46m/s[/tex]

How can I determine the satellite's speed?As we are aware, the centripetal force and gravitational force are equivalent, and thus leads to the following satellite speed equation:

                        [tex]v=\sqrt{\frac{GM}{R} } =3.624km/s\\where, M=7.88*10^{18}kg[/tex]

Consequently, we can say that

1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.

2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.

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Related Questions

A boat is moving in a river with a current that has speed vW with respect to the shore. The boat first moves downstream (i.e. in the direction of the current) at a constant speed, vB , with respect to the water. The boat travels a distance D in a time tOut . The boat then changes direction to move upstream (i.e. against the direction of the current) at a constant speed, vB , with respect to the water, and returns to its original starting point (located a distance D from the turn-around point) in a time tIn .
1) What is tOut in terms of vW, vB, and D, as needed?
2) What is tIn in terms of vW, vB, and D, as needed?
3) Assuming D = 120 m, tIn = 170 s, and vW = 0.3 m/s, what is vB, the speed of the boat with respect to the water?
4) Once again, assuming D = 120 m, tIn = 170 s, and vW = 0.3 m/s, what is tOut, the time it takes the boat to move a distance D downstream?

Answers

Answer:

Explanation:

Current  has speed vW with respect to the shore and boat has speed vB with respect to water or current so speed of boat  with respect to shore

vW + vB .

Distance travelled with respect to shore by boat = D

time ( tout ) = distance / speed with respect to shore

tOut = D / ( vW + vB )

When the boat travels upstream , its velocity with respect to shore

= ( vB - vW ) , vB must be higher .

tin = D /  ( vB - vW )

3 ) tin = D /  ( vB - vW )

170 = 120 / (vB - 0.3 )

(vB - 0.3 ) = 12 / 17 = .706

vB = 1.006 m / s

4 )

tOut = D / ( vW + vB )

= 120 / ( .3 + 1.006 )

= 92.26 s

Time taken by a body is ratio of the distance traveled by it to the speed.

1)The expression for [tex]t{out}[/tex] is,

          [tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]

2)The expression for [tex]t{in}[/tex] is,

           [tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]

3) The speed of the boat with respect to the water is 1.006 m/s. 4) The time it takes the boat to move a distance D downstream is 91.9 seconds.

What is upstream and downstream speed?

The net speed of the boat is upstream speed. The difference of the speed of the boat is downstream speed.

Given information-

The speed of the boat with respect to shore is [tex]v_w[/tex].

The speed of the boat in downstream with respect to water is [tex]v_B[/tex].

The distance traveled by the boat is [tex]D[/tex] in time [tex]t_{out}[/tex].

Time taken by a body is ratio of the distance traveled by it to the speed.

1) The net speed of the boat is upstream speed.As the distance traveled by the boat is [tex]D[/tex] in time [tex]t_{out}[/tex]. Thus,

        [tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]

2) The difference of the speed of the boat is downstream speed.As the distance traveled by the boat is [tex]D[/tex] in time [tex]t_{in}[/tex]. Thus,

        [tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]

Now the distance is 120 m, the value of [tex]t_{in}[/tex] is 170 s and [tex]v_W[/tex] 0.3 m/s. Thus,

3) The speed of the boat with respect to the water-Put the values in the formula obtains from the 2nd part of the problem,

         [tex]170=\dfrac{120}{v_B-0.3}\\v_B-0.3=\dfrac{120}{160} \\v_B=0.706+0.3\\v_B=1.006[/tex]

Hence the speed of the boat with respect to the water is 1.006 m/s.

4) The time it takes the boat to move a distance D downstream-Put the values in the formula obtains from the 1st part of the problem,

          [tex]t_{out}=\dfrac{120}{1.006+0.3}\\t{out}=\dfrac{120}{1.306} \\t{out}=91.9[/tex]

Hence the time it takes the boat to move a distance D downstream is 91.9 seconds.

Thus,

1)The expression for [tex]t{out}[/tex] is,

          [tex]t_{out}=\dfrac{D}{v_B+v_W}[/tex]

2)The expression for [tex]t{in}[/tex] is,

           [tex]t_{in}=\dfrac{D}{v_B-v_W}[/tex]

3) The speed of the boat with respect to the water is 1.006 m/s. 4) The time it takes the boat to move a distance D downstream is 91.9 seconds.

Learn more about the upstream and downstream speed here;

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A radio transmitting station operating at a frequency of 125 MHz has two identical antennas that radiate in phase. Antenna B is 9.05 m to the right of antenna A. Consider point P between the antennas and along the line connecting them, a horizontal distance x to the right of antenna A. For what values of x will constructive interference occur at point P?

Answers

Answer:

Explanation:

The formula for frequency is:

[tex]f = \dfrac{c}{\lambda}[/tex]

If we make [tex]\lambda[/tex] the subject; we have:

[tex]\lambda = \dfrac{c}{f}[/tex]

[tex]\lambda = \dfrac{3\times 10^6 \ m/s}{125 \ MHz (\dfrac{10^6 \ Hz}{1 \ MHz})}[/tex]

[tex]\lambda = 2.4 \ m[/tex]

Let assume that there is a point P between antenna A  to B.

where;

A to B = 9.05

A to P = x  and

P to B = 9.05 - x

Then, the condition for the constructive inteference is:

Δx = nλ

x - (9.05 - x) = nλ

2x - 9.05 = n(2.4)

So, we need to start assigning values to n so that the value of x becomes less than or equal to 9.05 m

If n = -1

Then;

2x - 9.05 = (-1)(2.4)

x = 3.325 m

If n = -2

Then;

2x - 9.05 = (-2)(2.4)

x = 2.125 m

If n = -3

Then;

2x - 9.05 = (-3)(2.4)

x = 0.925 m

If n = 0

Then;

2x - 9.05 = (0)(2.4)

x = 4.525 m

If n = 1

2x - 9.05 = (1)(2.4)

x = 5.725 m

If n = 2

Then;

2x - 9.05 = (2)(2.4)

x = 6.925 m

Hence, there exist 7 points in which constructive interference occurs.

How long will it take an object traveling at 90 kilometers per hour to travel 910 kilometers?

Answers

Explanation:

time = distance / velocity

We know that distance = 910 km and velocity = 90 km/h.

t = d / v

t = 910 km / 90 km/h

t = 10.11 hrs

The object traveled for 10.11 hours long. Hope this helps, thank you !!

Calculate the ratio of the drag force on a passenger jet flying with a speed of 1200 km/h at an altitude of 10 km to the drag force on a prop-driven transport flying at one-fourth the speed and half the altitude of the jet. At 10 km the density of air is 0.38 kg/m3 and at 5.0 km it is 0.67 kg/m3. Assume that the airplanes have the same effective cross-sectional area and the same drag coefficient C. (drag on jet / drag on transport)

Answers

Answer:

[tex]2.267[/tex]

Explanation:

Drag force is given by

[tex]F=\dfrac{1}{2}\rho Av^2C[/tex]

C = Drag coefficient is constant

A = Area is constant

[tex]v_1[/tex] = Velocity of the passenger jet = 1200 km/h = [tex]\dfrac{1200}{3.6}\ \text{m/s}[/tex]

[tex]v_2[/tex] = Velocity of the prop plane = [tex]\dfrac{1}{4}v_1[/tex]

[tex]\rho_1[/tex] = Density of the air where the jet was flying = [tex]0.38\ \text{kg/m}^3[/tex]

[tex]\rho_2[/tex] = Density of the air where the prop plane was flying = [tex]0.67\ \text{kg/m}^3[/tex]

[tex]F\propto \rho v^2[/tex]

[tex]\dfrac{F_1}{F_2}=\dfrac{\rho_1 v_1^2}{\rho_2 v_2^2}\\\Rightarrow \dfrac{F_1}{F_2}=\dfrac{0.38 v_1^2}{0.67 (\dfrac{1}{4}v_1^2)}\\\Rightarrow \dfrac{F_1}{F_2}=2.267[/tex]

The ratio of the drag forces is [tex]2.267[/tex].

A vertical wire carries a current straight up in a region of the magnetic field directed north. What is the direction of the magnetic force on the current due to the magnetic field

Answers

Answer:

The direction of the force on the vertical wire is towards the East or right.

Explanation:

Using Fleming's right hand rule, the current is the middle finger pointing straight up, the magnetic field is the fore-finger pointing Northwards and then the thumb is the direction of the force on the vertical wire.

Following these conventions, the thumb points towards the East. So, the direction of the force on the vertical wire is towards the East or right.

Two students are on a balcony a distance h above the street. One student throws a ball vertically downward at a speed vi; at the same time, the other student throws a ball vertically upward at the same speed. Answer the following symbolically in terms of vi, g, h, and t. (Take upward to be the positive direction.)
(a) What is the time interval between when the first ball strikes the ground and the second ball strikes the ground?
?t = ______
(b) Find the velocity of each ball as it strikes the ground.
For the ball thrown upward vf = ______
For the ball thrown downward vf = ______
(c) How far apart are the balls at a time t after they are thrown and before they strike the ground?
d = _______

Answers

Answer:

Explanation:

 a )

Time for first ball to reach top position

v = u - gt

0 = vi - gt

t = vi / g

Time to reach balcony  while going downwards

= vi /g

Total time = 2 vi / g

Time to go down further to the ground = t₁

Total time = 2 vi / g + t₁

Time for the other ball to go to the ground = t₁

Time difference = ( 2 vi / g + t₁ ) - t₁

= 2vi / g .

( b )

v² = u² + 2gh

For both the throw ,

final displacement = h , initial velocity downwards = vi

( For the first ball also  , when it go down while passing the balcony , it acquires the same velocity vi but its direction is downwards.)

vf² = vi² + 2gh

vf = √ ( vi² + 2gh )

(c )

displacement of first ball after time t

s₁ = - vi t + 1/2 g t²  [ As initial velocity is upwards , vi is negative ]

displacement of second ball after time t

s₂ = vi t + 1/2 g t²

Difference = d =  s₂ - s₁

= vi t + 1/2 g t² - ( - vi t + 1/2 g t² )

d = 2 vi t .

Explain the difference in the function of plant and animal cell organelles, including cell membrane, cell wall, nucleus, cytoplasm, mitochondria, chloroplast, and vacuole

Answers

Answer:

Plant cell Animal cell

2. Have a cell membrane. 2. Have no chloroplasts.

3. Have cytoplasm. 3. Have only small vacuoles.

4. Have a nucleus. 4. Often irregular in shape.

5. Often have chloroplasts

containing chlorophyll. 5. Do not contain plastids.

pls help me this is a major SOS pls help pls btw this is IXL

Answers

Explanation:

the object with the higher temperature has greater thermal energy

So the answer is

the stick of butter with less thermal energy.

Hope it will help :)

Answer:

The stick of butter with less thermal energy

Explanation:

I am pretty sure

Which one of Newton’s Laws best explains a bottle flip?

Answers

Answer:

the 2nd law-

Hope this helps <3

Explanation:

Which two statements help explain why digital storage of data is so reliable?

A. Memory chips are sturdy.

U B. Digital data usually deteriorate over time.

C. It is usually possible to recover data from a memory chip even

when the device containing it is broken.

D. Digital data are easier to copy than analog data are, making them

more accessible to thieves.

Answers

Answer:

A. Memory chips are sturdy.

C. It is usually possible to recover data from a memory chip even when the device containing it is broken.

Explanation:

Digital storage of data refers to the process which typically involves saving computer files or documents on magnetic storage devices usually having flash memory. Some examples of digital storage devices are hard drives, memory stick or cards, optical discs, cloud storage, etc.

A reliable storage ensures that computer files or documents are easily accessible and could be retrieved in the event of a loss.

The two statements which help explain why digital storage of data is so reliable are;

A. Memory chips are sturdy: they are designed in such a way that they are compact and firm.

C. It is usually possible to recover data from a memory chip even when the device containing it is broken.

Answer:

A and C

Explanation:

got it right on a p e x

PLZZZZ HELPPPPPPPPPppppp​

Answers

What grade are you in?!!??

Which of the following choices is the best example of potential energy?

Answers

Answer:

A basketball sitting still in a players hands

Explanation:

The other 3 answers have the ball in motion (going towards the basket, bouncing, and rolling) so that would be kinetic energy.

When the basketball is sitting in the player's hands, it has the potential to be in motion.

Answer:

it is D not B it D

Explanation:

If a wave has a speed of 1000 m/s and frequency of 500 Hz, what is the wavelength?

• 1500 Hz
• 2 m
• 0.05 m

Answers

Answer:

2 m

Explanation:

speed=frequency×wavelength

wavelength=speed/frequency

wavelength=1000/500

=2 m

If you stand on a trampoline, it depresses under your weight. When you stand on a hard stone floor, __________. If you stand on a trampoline, it depresses under your weight. When you stand on a hard stone floor, __________. the floor deforms very slightly under your weight only if you are heavy enough does the floor deform at all under your weight the floor does not deform at all under your weight

Answers

Answer:

the floor deforms very slightly under your weight

Explanation:

A trampoline is made up of a large piece of strong cloth held by springs on which you jump up and down as a sport. So, If you stand on a trampoline, it depresses under your weight.  However, the floor does not deform under your weight as it is too stiff.

Therefore,

when you stand on a hard stone floor, the floor deforms very slightly under your weight.

Communication satellites are placed in a geosynchronous orbit, i.e., in a circular orbit such that they complete one full revolution about the earth in one sidereal day (23.934 h), and thus appear stationary with respect to the ground. Determine the altitude of these satellites above the surface of the earth in both SI and U.S. customary units.

Answers

Answer:

Explanation:

Let the radius of orbit of geostationary satellite be R .

Time period of satellite = 2πR / V₀ where V₀ is orbital velocity

T = 2πR / √gR

T= 2πR / √(GM / R )

T = 2πR¹°⁵ / √GM  

R¹°⁵ = T x √GM  / 2π

T = 23.934 h = 23.934 x 60 x 60 s = 86126.4 s

R¹°⁵ = 86126.4 x √ ( 6.67 x 10⁻¹¹ x 5.972 x 10²⁴ )  / 2π

= 86126.4 x √ ( 398.33 x 10¹²  )  / 2π

= 86126.4 x 19.95 x 10⁶  / 2π

= 273.428 x 10⁹

R = 42.92 x 10⁶ m

= 42920 km

Radius of orbit = 42920 km

radius of earth = 6370 km

Altitude of satellite = 42920 - 6370 = 36550 km .

In US customary unit = 36550 x 10³ /.9144 yards

= 36550 x 10³ /(.9144 x 1760 ) miles

= 22771 miles .

who is bill cypher and what is his origin?

Answers

Answer:

Bill Cipher is the true main antagonist of Gravity Falls. He is a Dream-Demon with mysterious motives and seems to have a vendetta against the Pines family, especially his old rival Stanford Pines

Explanation:



The electric field from two charges in the plane of the paper is represented by the dashed lines and arrows below.

Select a response for each statement below. (Use 'North' towards top of page, and 'East' to the right)


The magnitude of the E-field at Ris .... than at M.

The force on a (+) test charge at P is zero.

The magnitude of the charge on the left is .... that on the right.

The force on a (+) test charge at L is directed ....

The force on a (-) test charge at J is directed

The force on a (-) test charge at N is directed ....

The sign of the charge on the right is negative.

Answers

Answer:

a) electric field at point P must be zero

b) harged must be positive

c) force ais in the direction of the electric field

d)  force is in the opposite direction to the electric field

e)  force is in the opposite direction to the field

Explanation:

After reading your exercise, it is unfortunate that the diagram did not come out, but we are going to answer the questions in general.

a) force on a charge (+) is zero

this implies that the electric field at point P must be zero

        F = q E

b) the magnitude of the charge on the left is on the right

this indicates that the charged must be positive since the lines must exit the charge

c) force on load directed towards (direction not indicated)

since the charge is positive the force at point L is in the direction of the electric field at this point

d) force on test load (-) does not indicate direction

The force on a negative charge is in the opposite direction to the electric field at point J

e) Force on a test load (-) at point N

the force is in the opposite direction to the field at point N

An insulated, vertical piston-cylinder assembly contains 50 L of steam at 105 oC. The outside pressure is 101 kPa. The piston has a diameter of 20 cm and the combined mass of the piston and the load is 75 kg. The electrical heater and the paddle wheel are turned on and the piston rises slowly by 25 cm with a constant pressure. The total internal energy increases by 3.109 kJ.

Determine:

a. The pressure of air inside the cylinder during the process.
b. The boundary work performed by the gas.
c. The combined work transfer by the shaft and electricity.

Answers

Answer:

Explanation:

From the given information:

The pressure of the air during the process = [tex]P_{atm} + P_{due \ to \ wt \ of \ piston}[/tex]

[tex]= 101 \ kPa + \dfrac{75 \ kg \times 9.8 \ m/s^2 \times \dfrac{1 \ N }{1 \ kg.m/s^2} }{\dfrac{\pi}{4}(0.2 \ m)^2} ( \dfrac{1 \ N }{m^2} \times \dfrac{1 \ kPa}{1000 \ n/m^2})[/tex]

The pressure of the air during the process = 124.42 kPa

The boundary work = P × ΔW

The boundary work = 124.42  kPa × (π/4) × (0.2 m)² × 0.25 m × (1 kJ/1 kPa.m³)

The boundary work = 0.977 kJ

The combined work transfer = [tex]W_{boundary} + \Delta U[/tex]

The combined work transfer = 0.977 + 3.109 kJ

The combined work transfer = 4.086 kJ

You are driving a car behind a truck. Both your car and the truck are moving at a speed of 80km/hr. If the driver of the truck suddenly slams on the brakes, what minimum distance betweenyour car and the truck is needed so that your car does not crash into the truck’s rear end? (This is called the "​minimum trailing distance​".) To simplify this problem, assume that the truck andthe car have the same braking acceleration.

a. In order to simplify the calculations for this problem, you are told to assume that the braking acceleration of the car and the truck are the same. What other reasonable assumptions do you need to make in order to solve this problem?
b. For both the truck and the car, draw an acceleration- and velocity-versus-time graph.
c. Find an expression for the minimum trailing distance. (Your expression should only contain symbols of physical quantities. No numbers are needed here.)
d. Find the numerical value for the minimum trailing distance (Plug the values of physical quantities into your expression from part A (do not forget units!))

Answers

Answer:

Explanation:

Let the velocity of car and truck be u and breaking acceleration be a .

We shall have to assume the reflex time of the driver of the car . By the time he applies brake , his car will cover some distance . There will be some time tag between the time the truck starts decelerating and the driver of the car responding to that . During this period the car will not start decelerating . It will keep on moving with uniform velocity of u .

Let this time lag be t .

b )

For answer see the attached file

c )

The minimum trailing distance will be the distance covered by car before it starts decelerating in response to truck's deceleration .

minimum trailing distance d = u x t

d )  u = 80 km / h = 22.22 m /s

reflex action time t = 0.1 s  ( assumed time )

d = 22.22 x .1

= 2.2 m

Which of the following statements is true?
A. Friction primarily affects objects that contain iron.
B. Friction pulls objects toward the center of the Earth
C.
Friction does not affect objects in motion.
D.
Friction slows down or stops objects in motion.

Answers

Answer:

D. Friction slow down or stop objects in motion.

You are trying to push a 30 kg canoe across a beach to get it to a lake. Initially, the canoe is
at rest, and you exert a force over a distance of 3 m until it has a speed of 1.2 m/s.

a. How much work was done on the canoe?

b. The coefficient of kinetic friction between the canoe and the beach is 0.2. How much work was done by friction on the canoe?

c. How much work did you perform on the canoe?

d. What force did you apply to the canoe?

Answers

Answer:

m = 30, g = 9.8, coefficient = 0.2, so force due to friction = 30 x 9.8 x 0.2 = 58.8 N, so work done by friction = 58.8 x 1.2 = 70.56 J

Explanation:

Dereck is looking at how electrically charged objects can attract other objects without touching. What control would he need to use?

An electrically charged object
An uncharged object
A positively charged object
A negatively charged object

Answers

Answer:

its An uncharged object.

if its not charged the electrically wont go on it

Answer:

uncharged object

Explanation:

Concept Simulation 4.1 reviews the central idea in this problem. A boat has a mass of 4490 kg. Its engines generate a drive force of 4520 N due west, while the wind exerts a force of 890 N due east and the water exerts a resistive force of 1210 N due east. Take west to be the positive direction. What is the boat's acceleration, with correct sign

Answers

Answer:

-0.54m/s²

Explanation:

According to Newton's second law of motion

F = ma

Force = mass * acceleration

Given

Mass m = 4490kg

Take the sum of forces

Sum of force along the east = 890+1210 = 2100N

Sum of forces along the west = -4520N

Net force = -4520+2100

Net force = -2420N

Acceleration = Net force/Mass

Acceleration = -2420/4490

Acceleration = -0.54m/s²

Hence the boat acceleration is -0.54m/s²

|:Give one word answer
1. An object that allows whole light to pass through it _______​

Answers

Answer:

Translucent object

Explanation:

attraction is seen between the poles of two bar magnet in the case of​

Answers

Answer:

he magnetic field of a bar magnet is strongest at either pole of the magnet. It is equally strong at the north pole when compared with the south pole. The force is weaker in the middle of the magnet and halfway between the pole and the center

Explanation:

What is the acceleration of an object going from O m/s to 25 m/s in 5s?

Answers

Answer:

5m/s^2 is the acceleration.

Answer:

[tex]\boxed {\boxed {\sf a= 5 \ m/s^2}}[/tex]

Explanation:

Acceleration is the change in speed over time.

[tex]a=\frac{ v_f-v_i}{t}[/tex]

The object accelerates from 0 meters per second to 25 meters per second in 5 seconds.

[tex]v_f= 25 \ m/s\\v_i= 0 \ m/s \\t= 5 \ s[/tex]

Substitute the values into the formula.

[tex]a=\frac{ 25 \ m/s -0 m/s }{ 5 \ s}[/tex]

Solve the numerator.

[tex]a=\frac{25 \ m/s}{5 \ s}[/tex]

Divide

[tex]a= 5 \ m/s/s= 5 \ m/s^2[/tex]

The object's acceleration is 5 meters per square second.

How can you drop two eggs the fewest amount of times, without them breaking?

Answers

Because the shell didn’t crack.

I hope this helps.

Answer:

get 2 jugs of water put an egg in each one drop the jugs with parachutes on them in long grass on a sunny non windy day

Explanation:

egg+ground=broken

egg-ground= egg+air

egg+air=unbroken

egg+water= egg+wet

egg+water= unbroken

egg+egg= 2 egg

egg+egg+air= egg+egg+unbroken+unbroken

egg+egg+unbroken+unbroken=(egg+unbroken)2

longgrass+egg= 40%unbroken+60broken+egg

longgrass+egg+egg=20%unbroken+80%broken+2egg

ground+water=mud

mud+egg=unbroken+egg+muddy

air+water=raining

egg+raining+air=wet+egg+slip+50%broken+50%unbroken

ask if need more proof

A shuttle bus slows down with an average acceleration of -2.4 m/s2. How long does it
take the bus to slow from 9.0 m/s to rest?

Answers

Answer:

[tex]\boxed {\boxed {\sf 3.75 \ seconds }}[/tex]

Explanation:

Average acceleration is found by dividing the change in acceleration by the time.

[tex]a=\frac{ v_f-v_i}{t}[/tex]

The shuttle bus has an acceleration of -2.4 meters per square second. It slows from 9.0 meters per second to rest, or 0 meters per second. Therefore:

[tex]a= -2.4 \ m/s^2 \\v_f= 0 \ m/s \\v_i= 9 \ m/s[/tex]

Substitute the values into the formula.

[tex]-2.4 \ m/s^2=\frac{0 \ m/s - 9 \ m/s}{t }[/tex]

Solve the numerator.

[tex]-2.4 \ m/s^2 = \frac{-9 \ m/s}{t}[/tex]

We want to solve for t, the time. We have to isolate the variable. Let's cross multiply.

[tex]\frac{-2.4 \ m/s^2}{1} = \frac{-9 \ m/s}{t}[/tex]

[tex]-9 \ m/s *1= -2.4 \ m/s^2 *t[/tex]

[tex]-9 \ m/s=-2.4 m/s^2*t[/tex]

t is being multiplied by -2.4. The inverse of multiplication is division, so divide both sides by -2.4

[tex]\frac{-9 \ m/s }{-2.4 \ m/s^2} =\frac{ -2.4 \ m/s^2*t}{-2.4 \ m/s^2}[/tex]

[tex]\frac{-9 \ m/s }{-2.4 \ m/s^2} =t[/tex]

[tex]3.75 \ s=t[/tex]

It takes 3.75 seconds.

How far can a bus carrying small children, travel at a rate of 60 km per hour travel in 2 1/2 hours?

Answers

150km

Explanation:

speed = 60km/hr.time = 2¹/2 hr = 5/2 hrdistance = speed × time = 60 ×5/2 = 150km

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Q5. Use Superposition to V. in the circuit below? (5 points)
4 mA
12V
2 ΚΩ
2 mA
1 ΚΩ
2 ΚΩ

Answers

Answer:

4va

12va

2jk

1jk

2jk

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