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3 years ago

5

during a space shuttle launch about 830,000 kg of fuel is burned in 8 min. the fuel provides the shuttle with a constant thrust,

or forward force. how does newton's second law of motion explain why the shuttle's acceleration in creases as the fuel is burned?

1 answer:

kirill115 [55]3 years ago

8 0

The shuttles acceleration in the creases as the fuel is burned because the acceleration of the obect as produced by net force is directly proportional to the magnitude of the net force.

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A. Draw the electric field lines around a negative charge.

Alborosie

<h2>a. Answer:</h2>

We use Electric field lines for visualizing electric fields, so this helps us to see the problem more real. So an electric field line is an imaginary line or curve drawn through a region of space such that the tangent at any point comes from the direction of the electric-field vector at that point. The electric field lines around a negative charge is shown in the First figure below.

<h2>b. Answer:</h2>

Electric forces can be found by using the Coulomb Law's that states <em>that The magnitude of the electric force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. </em>This can be expressed as follows:

$F=k\frac{\left | q_{1}q_{2} \right |}{r^2} \\ \\ Where: \\ \\ k=9\times 10^9Nm^2/c^2 \\ \\ q_{1}=0.00150 C \ and \ q_{2}=0.00240 C \\ \\ r=0.900 m$

Then:

$F=9\times 10^9\frac{\left | 0.00150 \times 0.00240 \right |}{(0.900)^2} \\ \\ \therefore \boxed{F=40000N}$

This force is repulsive because the two charges are positive and recall that two positive charges or two negative charges repel each other while a positive charge and a negative charge attract each other.

<h2>C. Answer:</h2>

From the statement, we have two charged objects. Let's say that this charges are:

$q_{1} \ and \ q_{2}$

If the amount of charge on one of the objects is tripled, let's say this is the charge $q_{2}$ , then the new charge is:

$q_{N}=3q_{2}$

In the formula of Coulomb:

$F=k\frac{\left | q_{1}q_{N} \right |}{r^2} \\ \\ \therefore F=k\frac{\left | q_{1}(3q_{2}) \right |}{r^2} \\ \\ \therefore \boxed{F=3k\frac{\left | q_{1}q_{2} \right |}{r^2}}$

<em>The conclusion is that if the amount of charge on one of the objects is tripled, the electric force between two charged objects is also tripled</em>

<h2>d. Answer:</h2>

Let's use the Coulomb's Law again to solve this problem. We want to know how the electric force between two charged objects changes if the charges are moved closer together:

$F=k\frac{\left | q_{1}q_{N} \right |}{r^2}$

<em>By saying that the charges are moved closer together, we want to express that r becomes smaller. Since r is in the denominator, this implies that the electric force between these two charged objects becomes greater.</em>

<h2>e. Answer:</h2>

From the figure, we can see a metal sphere on a stand. There we have both positive and negative charges. We can say that the positive charge of this sphere is +10q and the negative and the negative charge is -10q. Since the electric charge is conserved, then the algebraic sum of all the electric charges in any closed system is constant. In conclusion, <em>the sphere has no net charge.</em>

<h2>f. Answer:</h2>

Here we want to know how the negative charges in the same sphere are redistributed when a positively charged rod is brought near it. Therefore, positive charge on rod repels positive charges on the sphere, creating zones of negative and positive charge as indicated in the second Figure.

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3 years ago

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A dog runs at 35 m/s at 45 degrees N of E. What are its x and y components (all answers

pickupchik [31]

Answer:

its x and y component is 24.749m/s

Explanation:

Given

Speed of the dog = 35m/s

x component of the speed = xcos theta

y component of the speed = ycos theta

Given theta =45 degrees

x-component = 35cos45

x-component = 35(0.7071)

x-component = 24.749m/s

y-component = 35sin45

y-component = 35(0.7071)

y-component = 24.749m/s

Hence its x and y component is 24.749m/s

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3 years ago

Light travels at a speed of about 3.0 108 m/s. (a) how many miles does a pulse of light travel in a time interval of 0.1 s, whic

babunello [35]

Speed of light is given as

$c = 3 * 10^8 m/s$

time interval is given as

$\Delta t = 0.1 s$

so the distance covered by the light is given as

$d = v * \Delta t$

$d = 3 * 10^8 * 0.1 = 3 * 10^7 meter$

now as we know that

$1 mile = 1609 meter$

so the distance moved by light is

$d = \frac{3 * 10^7}{1609} = 18645.12 miles$

now for the comparision of this distance with diameter of earth

as we know that radius of earth is

$R = 6.38 * 10^6 m$

so the diameter of earth will be

$d = 2R = 12.76 * 10^6 m$

now the ratio of diameter with the distance that light move will be

$\frac{distance}{diameter} = \frac{3 * 10^7}{12.76 * 10^6}$

$\frac{distance}{diameter} = 2.35$

<em>so it is 2.35 times more than the diameter of earth</em>

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3 years ago

Trong thí nghiệm Y-âng về giao thoa ánh sáng : khoảng cách giữa hai khe là 0,5 mm ; khoảng cách từ mặt phẳng chứa hai khe đến mà

Natalija [7]

không có câu hỏi, trả lời như nào nhỉ?

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3 years ago

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A lunch pail is accidentally kicked off a steel beam on a building under construction. Suppose the initial horizontal speed is 1

vichka [17]

1) 26.6 m

Along the horizontal direction, the lunch pail is moving with a uniform motion (constant speed), since there are no forces acting in this direction.

Therefore, the distance travelled horizontally after a time t is given by:

$d=v_x t$

where we know

$v_x = 1.50 m/s$ is the horizontal velocity

d = 3.50 m is the distance covered horizontally

Solving for t, we find the total time of the motion:

$t=\frac{d}{v_x}=\frac{3.50}{1.50}=2.33 s$

Now we know that the pail takes 2.33 s to fall to the ground. We can now consider the vertical motion of the pail, which is a free fall motion, so the vertical displacement is given by the equation

$s=ut+\frac{1}{2}at^2$

where, taking downward as positive direction:

u = 0 is the initial vertical velocity

$a=g=9.8 m/s^2$ is the acceleration of gravity

Substutting t = 2.33 s, we find how fat the pail has fallen:

$s=\frac{1}{2}(9.8)(2.33)^2=26.6 m$

2) 10.7 m

In this case, we know instead the vertical displacement:

$s=2.50\cdot 10^2 m = 250 m$

Therefore, we can use the same equation again

$s=ut+\frac{1}{2}at^2$

To find the total time of motion:

$t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(250)}{9.8}}=7.14 s$

We know that along the horizontal direction, the velocity is constant:

$v_x = 1.50 m/s$

So, the horizontal distance covered in this time is

$d=v_x t = (1.50)(7.14)=10.7 m$

6 0

4 years ago