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NeX [460]
3 years ago
9

How do the components of a vector transform under a translation of coordinates?

Physics
1 answer:
Rus_ich [418]3 years ago
8 0
Vectors are invariant under translation. This means that they are unaffected or unchanged even if the coordinates are translated. So while the coordinates of two points are changed or moved during translation, the distance between the two points does not change. The best way to picture this out is using real life scenario. For example, you and I are at rest relative to each other but sitting on a plane at different locations, then we will have a fixed distance. Now the plane flies and its overall location is changed, similarly our individual coordinates also change but the distance between us will remain the same.<span>

</span>
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A hockey puck is struck so that it slides at a constant speed and strikes the far side of the rink, 58.2 m away. The shooter hea
DENIUS [597]

Answer:

v = 33.66 m/s

Explanation:

Let hockey puck is moving at constant speed v

so here we have

d = vt

so time taken by the puck to strike the wall is given as

t = \frac{58.2}{v}

now time taken by sound to come back at the position of shooter is given as

t_2 = \frac{58.2}{340}

t_2 = 0.17s

so we know that total time is 1.9 s

1.9 = t + t_2

1.9 = t + 0.17

1.9 - 0.17 = t

t = 1.73 s

now we have

1.73 = \frac{58.2}{v}

v = 33.66 m/s

7 0
3 years ago
Please help me on 3,4 and 6. <br> Thanks
rusak2 [61]
3. <span>The second piston will experience the same force as compared with the first. This is because since the </span>pressure is the same everywhere inside the fluid system,<span> the force is proportional to the surface area. We are told that both the first and the second piston have the same surface area, therefore, they will both experience the same force/pressure.


4. </span>The situation is much the same as number 3 above, with the exception that the second piston is twenty times larger than the first. Again, since the pressure is the same everywhere inside the fluid system, the force is proportional to the surface area. We are told that the second piston is 20 times larger than the first, therefore, the larger piston will experience 20 times larger the force of the small one.


6. The answer is TRUE. The <span>hydraulic </span>braking system<span> of most cars makes use of a vacuum servo (or booster), which is located between the </span>brake pedal<span> and the master cylinder piston. </span><span>This vacuum servo amplifies the force applied </span><span>from the </span>brake pedal<span>.</span>

3 0
4 years ago
Can a body having zero velocity move with uniform speed? Give an example.
GaryK [48]
Sure. Body can move with uniform speed, and having zero velocity, when velocity becomes zero due to change in direction over time t.

For Example. - An Object is moving with uniform speed in a circular path, then after one complete revolution, it's velocity is zero, but speed still remains uniform 

Hope this helps!
3 0
4 years ago
Cause and Effect: What would happen to a space vehicle in orbit around Earth if it sped up?
gulaghasi [49]

Answer:

G M m / R^2 = m v^2 / R     gravitational force = centripetal force

G M = v^2 R   =    constant

As v increases R will must decrease

Take the moon as an example

S = 2 pi R   where R is about 240,000 miles for one orbit

S / 1 day = 54,000 miles/day  for a 28 day circuit

S / 1 hr = 54000 / 24 = 2200 mph  which is much less than a satellite in orbit

4 0
3 years ago
To initiate a nuclear reaction, an experimental nuclear physicist wants to shoot a proton into a 5.50-fm-diameter 12C nucleus. T
Fantom [35]

Answer:

V_1= 3.4*10^7m/s

Explanation:

From the question we are told that

Nucleus diameter d=5.50-fm

a 12C nucleus

Required kinetic energy K=2.30 MeV

Generally initial speed of proton must be determined,applying the law of conservation of energy we have

            K_2 +U_2=K_1+U_1

where

K_1 =initial kinetic energy

K_2 =final kinetic energy

U_1 =initial electric potential

U_2 =final electric potential

mathematically

   U_2 = \frac{Kq_pq_c}{r_2}

where

r_f=distance b/w charges

q_c=nucleus charge =6(1.6*10^-^1^9C)

K=constant

q_p=proton charge

Generally kinetic energy is know as

         K=\frac{1}{2}  mv^2

Therefore

         U_2 = \frac{Kq_pq_c}{r_2} + K_2=\frac{1}{2}  mv_1^2 +U_1

Generally equation for radius is d/2

Mathematically solving for radius of nucleus

         R=(\frac{5.50}{2}) (\frac{1*10^-^1^5m}{1fm})

         R=2.75*10^-^1^5m

Generally we can easily solving mathematically substitute into v_1

   q_p=6(1.6*10^-^1^9C)

   K_1=9.0*10^9 N-m^2/C^2

   U_1= 0

   R=2.75*10^-^1^5m

   K=2.30 MeV

   m= 1.67*10^-^2^7kg

   V_1= (\frac{2}{1.67*10^-^2^7kg})^1^/^2 (\frac{(9.0*10^9 N-m^2/C^2)*(6(1.6*10^-^1^9C)(1.6*10^-^1^9C)}{2.75*10^-^1^5m+2.30 MeV(\frac{1.6*10^-^1^3 J}{1 MeV}) }

    V_1= 3.4*10^7m/s

Therefore the proton must be fired out with a speed of V_1= 3.4*10^7m/s

8 0
3 years ago
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