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sweet-ann [11.9K]
3 years ago
5

The mass of a rocket decreases as it burns through its fuel. If the rocket engine produces constant force (thrust), how does the

acceleration of the rocket change over time? Answers:- it does not chage- it increases- it decreases
Physics
1 answer:
Serjik [45]3 years ago
6 0

Answer:

it increases-

Explanation:

When the mass of a rocket decreases as it burns through its fuel and the force ( thrust) is constant then by newtons second law of motion

F= ma  here F is constant this means that   ma= constant

⇒ m= F /a    this implies that mass is inversely proportional to  acceleration.

its means when the mass decreases the acceleration must increase. hence the acceleration increases

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A steady current I flows through a wire of radius a. The current density in the wire varies with r as J = kr, where k is a const
grin007 [14]

Answer:

Explanation:

we can consider an element of radius r < a and thickness dr.  and Area of this element is

dA=2\pi r dr

since current density is given

J=kr

then , current through this element will be,

di_{thru}=JdA=(kr)(2\pi\,r\,dr)=2\pi\,kr^2\,dr

integrating on both sides between the appropriate limits,

\int_0^Idi_{thru}=\int_0^a2\pi\,kr^2\,dr&#10;\\\\&#10;I=\frac{2\pi\,ka^3}{3} -------------------------------(1)

Magnetic field can be found by using Ampere's law

\oint{\vec{B}\cdot\,d\vec{l}}=\mu_0\,i_{enc}

for points inside the wire ( r<a)

now, consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius r.

by applying the Ampere's law, we can write

\oint{\vec{B}_{in}\cdot\,d\vec{l}}=\mu_0\,i_{enc}&#10;

by symmetry \vec{B} will be of uniform magnitude on this loop and it's direction will be tangential to the loop.

Hence,

B_{in}\times2\pi\,l=\mu_0\int_0^r(kr)(2\pi\,r\,dr)=&#10;\\\\2\pi\,B_{in} l=2\pi\mu_0k \frac{r^3}{3}&#10;\\\\B_{in}=\frac{\mu_0kl^2}{3}&#10;

now using equation 1, putting the value of k,

B_{in} = \frac{\mu_{0} l^2 }{3 } \,\,\, \frac{3I}{2 \pi a^3}&#10;\\\\B_{in} = \frac{ \mu_{0} I l^2}{2 \pi a^3}&#10;

B)

now, for points outside the wire ( r>a)

consider a point at a distance 'r' from the center of wire. The appropriate Amperian loop is a circle of radius l.

applying the Ampere's law

\oint{\vec{B}_{out}\cdot\,d\vec{l}}=\mu_0\,i_{enc}&#10;

by symmetry \vec{B} will be of uniform magnitude on this loop and it's direction will be tangential to the loop. Hence

B_{out}\times2\pi\,r=\mu_0\int_0^a(kr)(2\pi\,r\,dr)&#10;\\\\2\pi\,B_{out}r=2\pi\mu_0k\frac{a^3}{3}&#10;\\\\B_{out}=\frac{\mu_0ka^3}{3r}&#10;

again using,equaiton 1,

B_{out}= \mu_0 \frac{a^3}{3r} \times \frac{3 I}{2 \pi a^3}&#10;\\\\B_{out} = \frac{ \mu_{0} I}{2 \pi r}

8 0
3 years ago
15. What is a thermograph?
Genrish500 [490]

Answer: A device that uses infrared sensors.

Explanation:

8 0
3 years ago
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Light travels at a speed of 2.998*108 m/s. Light takes approximately 3.25 minutes to travel from the Sun to reach a planet. Calc
ANTONII [103]

Answer:

585×10⁸ m

Explanation:

Distance = rate × time

d = (2.998×10⁸ m/s) (3.25 min) (60 s/min)

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5 0
3 years ago
When heat is removed from a gas the gas will eventually
Lostsunrise [7]
The answer should be B, Condense.

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Evaporation is when a liquid transforms into a gas.
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Calculate the force needed to accelerate a car of a mass 1000 kg by 3 m/s2
beks73 [17]

Answer:

<h2>3000 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 1000 × 3

We have the final answer as

<h3>3000 N</h3>

Hope this helps you

5 0
3 years ago
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