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2 years ago

descirbe the motion of the block oncw the rocket turned off. What would it take to get the block to acclereate again?

Physics

1 answer:

avanturin [10]2 years ago

3 0

Answer:

HW3 Motion (Ldimension _ obiecL acceleralions) Question rocket on the ground is launched with an acceleration of 40 m/s $ upward. seconds inlo the flight the engine shuts off: Assume there no air resistance. Preliminary questions The acceleraton of the rocket i5 40 mls/5. Do you have t0 then adjust this acceleration wth the acceleration due t0 gravity? When Ihe engine shuts ofl, which molion variable(s) changes and how does change? Calculation What is the maximum height that the rocket reaches? At what time atter launch will tne rocket retum to the ground? Queslion You drive into 120 meter long block Irom the Ielt end at = constant speed of 21 mls_ Alter some Ume. you brake with constant acceleralion 0l 3 msls: You want come (0 slop at Ine right end of the block: How much time (Irom when you enter Ihe block) should you wail belore braking? How far down the block (from the left end) d2 you travel before braking? Question You are rest at stop sign; There another stop sign that is 100 meters away: Your maximum acceleration is mists and You maximum braking acceleration misis How much time do you spend acceleraling? How much time do you spend braking? What is the tolal travel lime? Over how much distance do you accelerate? Over how much distance do you brake? What is the tolal travel distanco?

Explanation:

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A charged particle enters a uniform magnetic field B with a velocity v at right angles to the field. It moves in a circle with p

alukav5142 [94]

A) d. 10T

When a charged particle moves at right angle to a uniform magnetic field, it experiences a force whose magnitude os given by

$F=qvB$

where q is the charge of the particle, v is the velocity, B is the strength of the magnetic field.

This force acts as a centripetal force, keeping the particle in a circular motion - so we can write

$qvB = \frac{mv^2}{r}$

which can be rewritten as

$v=\frac{qB}{mr}$

The velocity can be rewritten as the ratio between the lenght of the circumference and the period of revolution (T):

$\frac{2\pi r}{T}=\frac{qB}{mr}$

So, we get:

$T=\frac{2\pi m r^2}{qB}$

We see that this the period of revolution is directly proportional to the mass of the particle: therefore, if the second particle is 10 times as massive, then its period will be 10 times longer.

B) $a. f/10$

The frequency of revolution of a particle in uniform circular motion is

$f=\frac{1}{T}$

where

f is the frequency

T is the period

We see that the frequency is inversely proportional to the period. Therefore, if the period of the more massive particle is 10 times that of the smaller particle:

T' = 10 T

Then its frequency of revolution will be:

$f'=\frac{1}{T'}=\frac{1}{(10T)}=\frac{f}{10}$

6 0

2 years ago

A river flows with a speed of 0.600 m/s. A student first swims upriver 0.500 km, then turns around and returns to his starting p

DerKrebs [107]

Answer:

a) 1111.0 seconds

b) 833.3 s

c) Because of proportions

Explanation:

a) Total time of round trip is the sum of time upriver and time downriver

$t_{total}=t_{up}+t_{down}$

Time upriver is calculated with the net speed of student and 0.500 km:

$t_{up}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=v_{relative to river}+v_{river}=-1.2+0.6=-0.6 m/s\\t_{up}=\frac{500 m}{0.6 m/s}=833.3 s$

(Becareful with units 0.5 km= 500m) Similarly of downriver:

$t_{down}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=1.2+0.6=1.8 m/s\\t_{down}=\frac{500 m}{1.8 m/s}=277.7 s$

So the sum is:

$t_{total}=1111.0s$

b) Still water does not affect student speed, so total time would be simply:

$t_{total}=\frac{1000 m}{1.2 m/s}=833.3 s$

c) For the upriver trip, student moved half the distance in half speed of the calculation in b), so it kept the same ratio and therefore, same time. So the aditional time is actually the downriver.

6 0

2 years ago

A duck with a mass of 0.90 kilograms flies at a rate of 12.0 m/s. What is the kinetic energy of the duck?

bija089 [108]

It should be at about 65J. Not sure, hope I helped.

6 0

3 years ago

Read 2 more answers

The chemical equation provided shows iron rusting to form iron oxide. Use the drop-down menu to choose the coefficients that wil

Brrunno [24]

Answer:

$4 Fe + 3 O_2 \rightarrow 2 Fe_2 O_3$

Explanation:

The original equation is:

$Fe + O_2 \rightarrow Fe_2 O_3$

We notice that:

- we have 1 atom of Fe on the left, and 2 atoms of Fe on the right

- we have 2 atoms of O on the left, and 3 atoms of O on the right

Therefore, the equation is not balanced.

In order to balance it, we can add:

- a coefficient 3 in front of $O_2$

- a coefficient 2 in front of $Fe_2 O_3$

So we have:

$Fe + 3 O_2 \rightarrow 2Fe_2O_3$

Now the oxygen is balanced, but the iron it not balanced yet, since we have 1 Fe on the left and 4 on the right. Therefore, we should add a coefficient 4 on the Fe on the left:

$4 Fe + 3 O_2 \rightarrow 2 Fe_2 O_3$

3 0

2 years ago

Read 2 more answers

An element’s atomic number is 35. how many portions would an atom this element have

kotykmax [81]

35 protons are present in an element whose atomic number is 35.

<u>Explanation: </u>

In an atom of an element, the number of protons = atomic number

Number of protons = number of electrons

Mass Number = number of protons + number of neutrons

Hence, an atom with atomic number 35 will have 35 protons.

4 0

3 years ago