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2 years ago

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Upon its return voyage from a space mission, the spacecraft has a velocity of 25000 km/h at point A, which is 7130 km from the c

enter of the earth. Determine the velocity of the spacecraft when it reaches point B, which is 6625 km from the center of the earth. The trajectory between these two points is outside the effect of the earth’s atmosphere. The radius of the earth is 6371 km, and g = 9.825 m/s2 on the surface of the earth.

1 answer:

Nadya [2.5K]2 years ago

5 0

Answer:

the velocity at point B is 27120.08 km/hr

Explanation:

check the picture attached for explanation

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How can philosophy help you become a productive citizen<br>

AlexFokin [52]

Answer:

Philosophy is a study that involves the nature of knowledge and truth. It serves as a guide that helps an individual seek which things are valuable and essential in life. ... It gives you a sense of direction, knowing the weight of things, therefore making you more productive.

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2 years ago

Using the superposition method, calculate the current through R5 in Figure 8-71

Vladimir79 [104]

by superposition method we can find current in R5

here first let say only 2V battery is present in the circuit

now the equivalent resistance to be found for which we can say

2.2 k ohm and 1 k ohm is connected in parallel

$r_1 = \frac{2.2 * 1}{2.2 + 1}$

$r_1 = 0.6875 k ohm$

now it is in series with 1 k ohm and then that part is in parallel with 2.2 k ohm

$r_2 = \frac{2.2* (1+0.6875)}{2.2 + (1+0.6875)}$

$r_2 = 0.95 k ohm$

now the current flowing through the battery is

$i = \frac{2}{1 + 0.95} = 1.02 mA$

now this will divide into R3 and R2 so current flowing in R3 will be

$i_1 = \frac{2.2}{2.2+1.6875}*1.02 = 0.58 mA$

now this will again divide in R4 and R5

so current in R5 will be

$i_5 = \frac{R_4}{R_4 + R_5}* i_1$

$i_5 = 0.18 mA$

now when only 3 V battery is present in the circuit

R1 and R2 is in parallel and then it is in series with R3

so parallel combination will be

$r_1 = \frac{1*2.2}{2.2 +1} = 0.6875k ohm$

also after its series with R3

$r_2 = 1 + 0.6875 = 1.6875 k ohm$

now it is in parallel with R5 on other side

$r_3 = \frac{1.6875 * 2.2}{1.6875 + 2.2} = 0.95 k ohm$

now current through the battery will be given as

$i = \frac{3}{1 + 0.95} = 1.53 mA$

now it is divide in r2 and R5

so current in R5 is given as

$i_5 = \frac{r_2}{r_2 + R_5}*i$

$i_5 = \frac{1.6875}{2.2 + 1.6875} * 1.53$

$i_5 = 0.67 mA$

now the total current in R5 will be given by super position which is

$i = 0.67 + 0.18 = 0.85 mA$

so there is 0.85 mA current through R5 resistance

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2 years ago

If you pull a resistant puppy with its leash in a horizontal direction, it takes 80 N to get it going. You can then keep it movi

netineya [11]

Answer:

The coefficient of static friction between the puppy and the floor is 0.7273.

Explanation:

The horizontal force applied to move the puppy from a steady state has to be greater than the force of static friction, after it is moving the force needs to be equal to be greater than the force of dynamic friction in order to maintain its movement. The force of static friction is given by:

$F_s = \mu_s*N$

Where $F_s$ is the static friction force, $\mu_s$ is the coefficient of static friction and $N$ is the normal force. Since there's no angle on the flor the normal force is equal to the weight of the puppy, therefore, $N = 110\text{ N}$ , to make the puppy moving we need to use a force of 80 N, therefore, $F_s = 80 \text{ N}$ , so we can solve for the coefficient as shown below:

$80 = \mu_s*110\\\mu_s = \frac{80}{110} = 0.7273\\$

The coefficient of static friction between the puppy and the floor is 0.7273.

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2 years ago

50 J of work was performed in 20 seconds. How much power was used to do this task?

yuradex [85]

Power=work/time
power=50/20
50/20=2.5
Therefore A. 2.5 W

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2 years ago

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Which of the following planets helped astronomers locate another planet?

worty [1.4K]

Answer: Pluto,Mercury

Explanation:

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2 years ago