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Monica [59]
3 years ago
7

Providing trademark protection​

Physics
2 answers:
beks73 [17]3 years ago
7 0
The 5 Things You Must Do to Protect Your Trademark
1. Do Your Homework. The USPTO won't register your trademark if there is a “likelihood of confusion" with another registered trademark. ...
2. Prepare and File a Trademark Application. ...
3. Respond Promptly to Office Actions or Oppositions. ...
4. Monitor Your Trademark. ...
5. Maintain Your Trademark.
romanna [79]3 years ago
5 0

Answer:

what are you asking exactly?

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Calculate Speed The 2-kg metal ball moving at a speed of 3 m/s strikes a 1-kg wooden ball that is at rest. After the collision,
enot [183]

Answer:

53466kg.

Explanatiokn:

5 0
2 years ago
a baby carriage is sitting on the top of a hill that is 21 m high. the carriage with a baby 75kg. the carriage has what energy?
lara31 [8.8K]

Answer:

It has potential energy

PE = 15450.75 J

Explanation:

we know its potential energy because the carriage is not moving, although it can

PE = mgh

in this case, m is 75, h is 21, and g is always 9.81

plug these in: PE = (75) (21) (9.81)

your answer: PE = 15450.75 J

4 0
3 years ago
Three identical stars of mass M form an equilateral triangle that rotates around the triangle’s center as the stars move in a co
zheka24 [161]

Answer:

v=\sqrt{\dfrac{2GM}{L}}

Explanation:

M = Mass of planets

R = Radius of circle

v = Velocity

\theta = Angle

The circle is inside the triangle

cos\theta=\dfrac{\dfrac{L}{2}}{R}\\\Rightarrow R=\dfrac{L}{2cos\theta}

The centripetal acceleration

\dfrac{Mv^2}{R}=2\dfrac{GM^2}{L^2}cos\theta\\\Rightarrow \dfrac{Mv^2}{\dfrac{L}{2cos\theta}}=2\dfrac{GM^2}{L^2}cos\theta\\\Rightarrow \dfrac{Mv^22cos\theta}{L}=2\dfrac{GM^2}{L^2}cos\theta\\\Rightarrow v^2=\dfrac{2GM}{L}\\\Rightarrow v=\sqrt{\dfrac{2GM}{L}}

The speed of the stars is v=\sqrt{\dfrac{2GM}{L}}

5 0
4 years ago
A military helicopter on a training mission is flying horizontally at a speed of 90.0 m/s when it accidentally drops a bomb (for
Elena-2011 [213]

Answer:

1) 10.1 s  2) 909 m 3) 90.0 m/s 4) -99m/s 5) just over the bomb.

Explanation:

1)

  • In the vertical direction, as the bomb is dropped, its initial velocity is 0.
  • So, we can find the time required for the bomb to reach the earth, applying the following kinematic equation for displacement:

       \Delta y = \frac{1}{2}*a*t^{2} (1)

  • where Δy = -500 m (taking the upward direction as positive).
  • a=-g=-9.8 m/s²
  • Replacing these values in (1), and solving for t, we have:

       t =\sqrt{\frac{2*\Delta y}{-g}} = \sqrt{\frac{2*(-500m)}{-9.8m/s2}} = 10.1 s

  • The time required for the bomb to reach the earth is 10.1 s.

2)

  • In the horizontal direction, once released from the helicopter, no external influence acts on the bomb, so it will continue moving forward at the same speed. that it had, equal to the helicopter.
  • As the time must be the same for both movements, we can find the horizontal displacement just as the product of this speed times the time, as follows:

       x = v_{0x} * t = 90.0 m/s * 10.1 s = 909 m.

3)

  • The horizontal component of the bomb's velocity is the same that it had when left the helicopter. i.e. 90 m/s.

4)

  • In order to find the vertical component of the bomb's velocity just before it strikes the earth, we can apply the definition of acceleration, remembering that v₀ = 0, as follows:

        v_{f} = -g*t = -9.8 m/s2*10.1 s = -99 m/s

5)

  • If the helicopter keeps flying horizontally at the same speed, it will be always over the bomb, as both travel horizontally at the same speed.
  • So, when the bomb hits the ground, the helicopter will be exactly over it.

8 0
4 years ago
What do the spheres in the model represent
Sophie [7]

Answer: no

Explanation: we need a picture

4 0
3 years ago
Read 2 more answers
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