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JulsSmile [24]
3 years ago
15

A girl sits on a tire swing and is pushed in a circular motion by her father. Her tangential speed is 2.8 m/s and the girl trave

ls in a circle with a diameter of 4.0 m. What is the girls centripetal acceleration?
Physics
2 answers:
Ber [7]3 years ago
4 0

Answer:

The girl's centripetal acceleration is 3.92 m/s².

Explanation:

Given that,

Tangential speed = 2.8 m/s

Diameter = 4.0 m

We need to calculate the centripetal acceleration

Using formula of centripetal acceleration

\alpha=\dfrac{v^2}{r}

Where, v = tangential speed

r = radius

Put the value into the formula

\alpha=\dfrac{2.8^2}{2.0}

\alpha=3.92\ m/s^2

Hence, The girl's centripetal acceleration is 3.92 m/s².

Jet001 [13]3 years ago
3 0
Formula: Ca=Vt^2/r

centripetal acceleration(Ca)= ?
tangential speed(Vt)= 2.8m/s
Radius(r)=2m

Substitute: Ca=2.8^2/2
                  Ca=3.92m/s^2
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ipn [44]

Answer:

No, not necessarily

Explanation:

If an object is moving with an acceleration that causes its speed to be reduced, there will be a moment in which it reaches v = 0, but this doesn't necessarily mean that the acceleration isn't acting anymore. If the object continues its movement with the same acceleration, it's velocity will become negative.

An example of an object that has zero velocity but non-zero acceleration:

If you throw an object in the air with a certain velocity, it will move vertically, reducing its velocity in a 9,8 m/s^{2} rate (which is the acceleration caused by gravity). At a certain point, the object will reach its maximum height, and will start to fall. In the exact moment that it reaches the maximum height, before it starts falling, its velocity is zero, but gravity is still acting on the object (this is the reason why it starts falling instead of just being stopped at that point). Therefore, at that point, the object has zero velocity but an acceleration of 9,8 m/s^{2}.

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3 years ago
How are size and mass of the moon different from that of the earth?
Elodia [21]
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3 0
3 years ago
What is impulse? How does this relate to momentum?
lawyer [7]
Impulse is a force acting briefly on a body and producing a finite change of momentum.
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4 0
3 years ago
The electric field in a region is uniform (constant in space) and given by E-( 148.0 1 -110.03)N/C. An additional charge 10.4 nC
enyata [817]

Answer:

The y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

Explanation:

<u>Given:</u>

  • Electric field in the region, \vec E = (148.0\ \hat i-110.0\ \hat j)\ N/C.
  • Charge placed into the region, q = 10.4\ nC = 10.4\times 10^{-9}\ C.

where, \hat i,\ \hat j are the unit vectors along the positive x and y axes respectively.

The electric field at a point is defined as the electrostatic force experienced per unit positive test charge, placed at that point, such that,

\vec E = \dfrac{\vec F}{q}\\\therefore \vec F = q\vec E\\=(10.4\times 10^{-9})\times (148.0\ \hat i-110.0\ \hat j)\\=(1.539\times 10^{-6}\ \hat i-1.144\times 10^{-6}\ \hat j)\ N.

Thus, the y-component of the electric force on this charge is F_y = -1.144\times 10^{-6}\ N.

3 0
3 years ago
A point charge A of charge +4micro coloumb and another B of -1 micro coloumb are placed at a distance in air 1m apart then the d
andrew11 [14]

Answer:

Explanation:

Given that,

A point charge is placed between two charges

Q1 = 4 μC

Q2 = -1 μC

Distance between the two charges is 1m

We want to find the point when the electric field will be zero.

Electric field can be calculated using

E = kQ/r²

Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.

Then, the magnitude of the electric at point x is zero.

E = kQ1 / r² + kQ2 / r²

0 = kQ1 / x²  - kQ2 / (1-x)²

kQ1 / x² = kQ2 / (1-x)²

Divide through by k

Q1 / x² = Q2 / (1-x)²

4μ / x² = 1μ / (1 - x)²

Divide through by μ

4 / x² = 1 / (1-x)²

Cross multiply

4(1-x)² = x²

4(1-2x+x²) = x²

4 - 8x + 4x² = x²

4x² - 8x + 4 - x² = 0

3x² - 8x + 4 = 0

Check attachment for solution of quadratic equation

We found that,

x = 2m or x = ⅔m

So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.

5 0
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