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nataly862011 [7]
3 years ago
13

When the spacecraft is at the halfway point, how does the strength of the gravitional force on the spaceprobe by Earth compre wi

th the strength
Physics
1 answer:
mixer [17]3 years ago
5 0

Solution :

When the spacecraft is at halfway point, the distance from the Earth as well as Mars are same. We have to account the masses of the planets. The gravitational force that is exerted by the Earth is greater because of its combined mass with the space probe.

The mass of Earth is greater than the mass of Mars. Therefore, the force of Earth is more than Mars.

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Science please helppp
Daniel [21]

The red box must way more. Gravitational potential energy is the product of a an objects mass times the acceleration due to gravity (which is constant on earth) times its height. Since the objects are on the same shelf they are at the same height, and since gravitational acceleration is constant as long as we stay on planet earth, then the mass is the only possible thing that could have changed. This means that the red box must weigh more than the blue box.

3 0
3 years ago
A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the
BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

F_g=\frac{Gm_1m_2}{r^2} where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2} I'm going to do the math on the top and then on the bottom and divide at the end.

F_g=\frac{2.4012*10^{40}}{3.721*10^{23}} and now when I divide I will express my answer to the correct number of sig dig's:

Fg= 6.45 × 10¹⁶ N

8 0
3 years ago
Relay of thermostate is not working why
Colt1911 [192]

Check the power source. Your thermostat may just not be connected right or at all. A blown fuse, tripped circuit breaker or dead batteries will prevent the thermostat from turning on your furnace.

Dirty thermostat? That’ll cause issues. Clean up any dust, dirt, spider webs and other debris. Any of these things can coat the inside of the thermostat and interfere with both electrical and mechanical functions of the thermostat. Put this on your get-ready-for-winter cleaning list. Just use a soft, clean brush to clean the inside components gently. Don’t get anything wet. Also you can use a can of compressed air, such as is used for electronics, to clear debris.

Check for any loose wires or terminal screws inside the thermostat. Make sure wires aren’t corroded or detached. Never remove the thermostat cover without removing the batteries or turning off the power at the fuse or breaker box. Tighten screws and secure loose wires if needed.

It may be time to replace your thermostat is it’s old. They aren’t meant to last forever and an old thermostat may be costing you a lot of money in wasted energy and time spent tinkering with an outdated model. There are great programmable thermostats available now that are easy to use and simple to connect to your existing HVAC system. Click here for more info on programmable thermostats.

7 0
3 years ago
A charged sphere with 1 × 10 8 units of negative charge is brought near a neutral metal rod. The half of the rod closer to the s
jeyben [28]

Answer:

-4*10⁴ units.

Explanation:

As the metal rod was initially neutral (which means that it has the same quantity of positive and negative charges), after being close to the charged sphere, as charge must be conserved, the total charge of the metal rod must  still remain to be zero.

So, if due to the influence of the negative charge in the sphere, the half of the road closer to the sphere has a surplus charge of +4*10⁴ units, the charge on the half of the rod farther from the sphere must be the same in magnitude but of the opposite sign, i.e., -4*10⁴ units.

5 0
3 years ago
A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc
marta [7]
This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so: 

v_{d} =  \frac{J}{n\left | q \right |}

<span>The current I is any motion of charge from one region to another, so this is given by:

</span>I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)

The magnitude of the current density is:

J = \frac{I}{A} = \frac{3189.73}{2x10^{-6}} = 1594.86(A/m^{2})

Being:

A=2mm^{2} = 2x10^{-6}m^{2}
<span>
Finally, for the drift velocity magnitude vd, we find:

</span>v_{d} = \frac{1594.86}{5.8x1028\left |1.60x10^{-19}|\right } = 1.67x10^{18}(m/s)

Notice: The current I is very high for this wire. The given values of the variables are a little bit odd
6 0
3 years ago
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