6. As distance increases, gravitational force *
(10 Points)
increases
decreases

Answers

Answer 1

Answer: It decreasessssssssss

Related Questions

Nitrogen makes up about what percent of a human's body weight?
What is the maximum distance allowed between the center of hole #2 and datum B as seen in the front view?
At what distance from a long straight wire carrying acurrentof 5.0A is the magnitude of the magnetic field due to thewireequal to the strength of the Earth's magnetic field of about5.0 x10^-5 T?
You're carrying a 3.6-m-long, 21 kg pole to a construction site when you decide to stop for a rest. You place one end of the pole on a fence post and hold the other end of the pole 35 cm from its tip. For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. Part A Part complete How much force must you exert to keep the pole motionless in a horizontal position? Express your answer in newtons. F = 114 N Previous Answers
At the normal boiling temperature of iron, TB = 3330 K, the rate of change of the vapor pressure of liquid iron with temperature is 3.72 x 10-3 atm/K. Calculate the molar latent enthalpy of boiling of iron at 3330 K:

Since fusion and fission are opposite processes that both produce energy,why can we not simply run the process forward and then backwardrepeatedly and have a limitless supply of energy?A. The products of a fission reaction cannot be used for a fusionreaction, and the products of a fusion reaction cannot be used fora fission reaction.B. Fusion reactions can occur cheaply enough, but fission requiresvery high temperatures.C. Fusion produces energy from nuclei larger than iron, and fissionproduces energy from nuclei smaller than iron.D. Fission reactions can occur cheaply enough, but fusion requires very high temperatures

Answers

ANSWER:

D. Fission reactions can occur cheaply enough, but fusion requires very high temperatures

STEP-BY-STEP EXPLANATION:

One of the main reasons fusion power cannot be harnessed is that its power requirements are incredibly high. For fusion to occur, a temperature of at least 100,000,000°C is needed.

Therefore, the correct answer is D. Fission reactions can occur cheaply enough, but fusion requires very high temperatures

Two atoms collide while moving in a dilute gas. The larger atom has a mass M1 = 6 Daltons and a speed v1 = 200 m/s, while the smaller has a mass M2 = 1 Daltons. During the collision both atoms simply bounce off each other. They do not change their speeds, but after the collision they each change their directions, bouncing in the indicated directions. (You may express your results using the mass unit "Daltons". 1 Dalton is approximately equal to the mass of a proton or neutron and is defined as one-twelfth the mass of a single neutral carbon-12 atom in its ground state.)A. What is the magnitude of the change in the momentum, Δp1, of mass M1?
B. What is the change in the total momentum of the pair?
C. What is the magnitude of the change in the momentum Δp2, of mass M2?

Answers

Answer:

a). ΔP1=-2.4 $x10^(3) (D*m)/(s)$

b). Pp=0 F=0

c). ΔP2=2.4 $x10^(3) (D*m)/(s)$

Explanation:

Initial momentum

$P_(1)=m_(1)*v_(i1)$

Final momentum

$P_(1f)=m_(1)*v_(f1)=-m_(1)*v_(i1)$

The change of momentum m1 is:

a).

ΔP1= $P_(1f)-P_(1)$

ΔP1= $-m_(1)*v_(i1)-m_(1)*v_(i1)$

ΔP1= $-2*m_(1)*v_(i1)$

ΔP1= $-2*6 D*200(m)/(s)$

ΔP1= $-2.4x10^(3)(D*m)/(s)$

b).

The law of conservation of energy in this case there is not external forces so the momentum of the pair change is equal to zero

P=0

Fx=0

c).

ΔP1+ΔP2=0

ΔP2=-ΔP1

ΔP2=- $-2.4x10^(3)(D*m)/(s)$

ΔP2= $2.4x10^(3)(D*m)/(s)$

Final answer:

The magnitude of the change in momentum of mass M1 is 2400 Daltons*m/s. The change in the total momentum of the pair is 2000 Daltons*m/s. The magnitude of the change in momentum of mass M2 is -400 Daltons*m/s.

Explanation:

A. To find the magnitude of the change in momentum of mass M1, we use the formula Δp1 = m1 * Δv1, where m1 is the mass of M1 and Δv1 is the change in velocity of M1. Since M1 simply changes direction, its change in velocity is equal to 2 times its original velocity. Therefore, Δp1 = m1 * (2v1) = 6 * (2 * 200) = 2400 Daltons*m/s.

B. The change in the total momentum of the pair is equal to the sum of the changes in momentum of M1 and M2. Since M2 also changes direction, its change in velocity is equal to 2 times its original velocity. Therefore, the change in the total momentum is Δp1 + Δp2 = 2400 Daltons*m/s + (-400 Daltons*m/s) = 2000 Daltons*m/s.

C. To find the magnitude of the change in momentum of mass M2, we use the same formula as in part A, but with the values for M2. Δp2 = m2 * Δv2 = 1 * (2 * (-200)) = -400 Daltons*m/s.

Learn more about Momentum here:

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A 1.5m wire carries a 7 A current when a potential difference of 87 V is applied. What is the resistance of the wire?

Answers

Answer:

$R\approx12.43 \,\, \Omega$

Explanation:

We can use Ohm's Law to find the resistance R of a wire that carries a current I under a given potential difference:

$V=I\,\,R\nR = (V)/(I) \nR=(87)/(7) \nR\approx12.43 \,\, \Omega$

Answer:

Ohm's law states that I=V/R (Current=volts divided by resistance). Since we're looking for resistance, we'll rewrite it as R=V/I. Then just plug in the numbers; R=84/9, R= 9 1/3 or 28/3. The resistance of the wire is 9.33... or 9 1/3 ohm's, depending on how you wanna write it.

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Beverage can is thrown upward and then falls back down to the floor. As usual, a y axis extends upward (positive direction). Which of the following best describes the acceleration of the can during its free flight?a) -9.8 m/s^2, then momentarily zero, then +9.8 m/s^2
b) +9.8 m/s^2 throughout
c) -9.8 m/s^2 throughout
d) zero throughout
e) +9.8 m/s^2, then momentarily zero, then -9.8 m/s^2

Answers

a) -9.8 m/s^2, then momentarily zero, then +9.8 m/s^2
Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s

A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.(a) Find the electric force on charge q2.
F12 = ? mN
(b) Find the electric force on q1.
F21 = ? mN
(c) What would your answers for Parts (a) and (b) differ if q2 were -6.0 µC?

Answers

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = (kq_1q_2)/(d^2)$

Here

k = Coulomb's Constant

$q_(1,2) =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_(12) = ( (9*10^9 Nm^2/C^2)(1*10^(-6) C)(6*10^(-6) C))/((1 m)^2)$

$F_(12) = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_(21) = F_(12)$

$F_(21) = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_(12) = ((9*10^9 Nm^2/C^2)(1*10^(-6) C)(-6*10^(-6) C))/((1 m)^(2))$

$F_(12) = F_(21) = -54 mN$

Force is negative i.e. attractive

A typical sugar cube has an edge length of 1 cm. If you had a cubical box that contained a mole of sugar cubes, what would its edge length be? (One mole = 6.02 ✕ 1023 units.)

Answers

Since volume of each cube is 1 cm^3
Then we can get the volume of 1 mole of cubes, which is $1 * 6.02 * 10^23 cm^3$
The edge $edge = v^1/3$
And the new adge that we are looking for: $new edge = (6.02*10^23)^1/3$ = $= 1.8191 * 46415888.336 = 84435142.472$
So, the final soution for the edge length of cube is 844km.

Do hope it helps!
Regards.

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In the circuit shown in the figure above, suppose that the value of R1 is 100 k ohms and the value of R2 is 470 k ohms. At which of the following locations in the circuit would you measure the highest voltage with your meter? A. Between points A and B B. Between points A and C C. Between points B and E D. Between points B and C

The position of a particle as a function of time is given by x = (2.0 m/s)t + (-3.0 m/s2)t2. (a) plot x-versus-t for time from t = 0 to t = 1.0 s. (do this on paper. your instructor may ask you to turn in this plot.) this answer has not been graded yet. (b) find the average velocity of the particle from t = 0.45 s to t = 0.55 s. m/s (c) find the average velocity from t = 0.49 s to t = 0.51 s.

Point charge 3.5 μC is located at x = 0, y = 0.30 m, point charge -3.5 μC is located at x = 0 y = -0.30 m. What are (a)the magnitude and (b)direction of the total electric force that these charges exert on a third point charge Q = 4.0 μC at x = 0.40 m, y = 0?