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kenny6666 [7]
3 years ago
12

a car is going around a roundabout.explain why we say that it is accelerating even if it is moving at a constant speed.

Physics
1 answer:
kiruha [24]3 years ago
4 0

Because 'acceleration' does NOT mean 'speeding up'. 
It means ANY change in motion ... speeding up, slowing down,
or changing DIRECTION.

When traveling a roundabout, or any curved path, the direction
is constantly changing even if the speed is constant, so there is
constant acceleration going on.

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A solid metal ball of radius 1.5 cm bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing a
blsea [12.9K]

We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is

E=7*10^{9}N/C

From the question we are told that

A solid metal ball of radius 1.5 cm

bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing

uniformly distributed charge of -7 nC

The distance between the centers of the balls is 9 cm

Generally the equation for the electric field  is mathematically given as

E=\frac{kq_2}{d^2}\\\\E=\frac{(9*10^9)7*10^{-2}}{9*10^{-2}}\\\\

E=7*10^{9}N/C

For more information on this visit

brainly.com/question/21811998

4 0
3 years ago
A position vector with magnitude 10 m points to the right and up. its x-component is 6.0 m. part a what is the value of its y-co
lidiya [134]

The position vector can be transcribed as:

A<span> = 6 i + y j                           </span>

i <span>points in the x-direction and j points in the y-direction.</span>

The magnitude of the vector is its dot product with itself:

<span>|A|2 = A·A</span>

<span>102  = (6 i + y j)•(6 i+ y j)            Note that i•j = 0, and  i•i  = j•j = 1 </span>

<span>100  = 36 + y2       </span>

<span>64    = y2</span>

<span>get the square root of 64 = 8</span>

<span>The vertical component of the vector is 8 cm.</span>

3 0
3 years ago
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3 years ago
In preparation for building a space station, an astronaut removes a self-telescoping uniform rod from the cargo bay and releases
emmainna [20.7K]

To solve this problem it is necessary to apply the concepts related to the conservation of angular momentum. This can be expressed mathematically as a function of inertia and angular velocity, that is:

L = I\omega

Where,

I = Moment of Inertia

\omega= Angular Velocity

For the given object the moment of inertia is equivalent to

I = \frac{mr^2}{12}

Considering that the moment of inertia varies according to distance, and that there are two of these without altering the mass we will finally have to

L_i = L_f

I_i \omega_1 = I_f \omega_2

(\frac{mr_{initial}^2}{12})(\omega_1)=(\frac{mr_{final}^2}{12})(\omega_2)

(r_{initial}^2})(\omega_1)=(r_{final}^2)(\omega_2)

Our values are given as,

r_{initial} = 3m\\\omega_1 = 0.05rad/s \\r_{final}=1.5m

Replacing we have,

(3^2})(0.05)=(1.5^2)(\omega_2)

\omega_2 = 0.2rad/s

Therefore the angular speed after the catch slips is 0.2rad/s

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Answer:

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