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Kazeer [188]
3 years ago
14

List at least 5 examples of civil rights

Physics
1 answer:
mariarad [96]3 years ago
6 0

Answer:

right to vote, the right to a fair trial, the right to government services, the right to a public education, and the right to use public facilities.

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(I) A novice skier, starting from rest, slides down an icy frictionless 8.0° incline whose vertical height is 105 m. How fast is
Vlad1618 [11]

Answer:

v = 45.37 m/s

Explanation:

Given,

angle of inclination = 8.0°

Vertical height, H  = 105 m

Initial K.E. = 0 J

Initial P.E. = m g H

Final PE = 0 J

Final KE = \dfrac{1}{2}mv^2

Using Conservation of energy

KE_i + PE_i + KE_f + PE_f

0 + m g H = \dfrac{1}{2}mv^2 + 0

v = \sqrt{2gH}

v = \sqrt{2\times 9.8 \times 105}

v = 45.37 m/s

Hence, speed of the skier at the bottom is equal to v = 45.37 m/s

3 0
3 years ago
Why are circuit breakers and fuses important?
Lesechka [4]

Answer:

A fuse and circuit breaker both serve to protect an overloaded electrical circuit by interrupting the continuity, or the flow of electricity. ... Fuses tend to be quicker to interrupt the flow of power, but must be replaced after they melt, while circuit breakers can usually simply be reset.

6 0
3 years ago
All objects near the earths surface - regardless of size and weight - hhave the same force of gravityvacting on them.
masya89 [10]

Answer:

B. False

Explanation:

Not all objects near the earths surface - regardless of size and weight - have the same force of gravity on them.

4 0
3 years ago
Lucia raced her car on a raceway
denpristay [2]

Answer:

Good question to ask in physics, sir maam

8 0
2 years ago
Read 2 more answers
Assume that the electric field E is equal to zero at a given point. Does it mean that the electric potential V must also be equa
lyudmila [28]

Answer:

  • No, this doesn't mean the electric potential equals zero.

Explanation:

In electrostatics, the electric field \vec{E} is related to the gradient of the electric potential V with :

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r})

This means that for constant electric potential the electric field must be zero:

V(\vec{r}) = k

\vec{E} (\vec{r}) = - \vec{\nabla} V (\vec{r}) = - \vec{\nabla} k

\vec{E} (\vec{r}) = -  (\frac{\partial}{\partial x} , \frac{\partial}{\partial y } , \frac{\partial}{\partial z}) k

\vec{E} (\vec{r}) = -  (\frac{\partial k}{\partial x} , \frac{\partial k}{\partial y } , \frac{\partial k}{\partial z})

\vec{E} (\vec{r}) = -  (0,0,0)

This is not the only case in which we would find an zero electric field, as, any scalar field with gradient zero will give an zero electric field. For example:

V(\vec{r})= (x+2)^2 (y+4)^3 (z+5)^4

give an electric field of zero at point (0,0,0)

8 0
3 years ago
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