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ddd [48]
2 years ago
9

A car experiences a centripetal acceleration of 4.4 m/s ^2 as ur rounds a corner with a speed of 15 m/s. What is the radius of t

he corner?
Physics
1 answer:
damaskus [11]2 years ago
3 0
The calculation of the centripetal acceleration of an object following a circular path is based on the equation,

                  a = v² / r

where a is the acceleration, v is the velocity, and r is the radius.

Substituting the known values from the given above,

             4.4 m/s² = (15 m/s)² / r

The value of r from the equation is 51.14 m.

Answer: 51.14 m
You might be interested in
In Paul Hewitt's book, he poses this question: "If the forces that act on a bullet and the recoiling gun from which it is fired
Sauron [17]
They have different accelerations because of their masses. According to Newton's Second Law, an objects acceleration is inversely proportional to its mass. Therefore the object with the larger mass, in this case the gun, will have a smaller acceleration. In the same way, the less massive object, being the bullet, will have a higher acceleration.

Hope this helps :)
4 0
2 years ago
A charge of uniform volume density (40 nC/m3) fills a cube with 8.0-cm edges. What is the total electric flux through the surfac
GREYUIT [131]

Answer:

The flux through the surface of the cube is 2.314\ Nm^{2}/C

Solution:

As per the question:

Edge of the cube, a = 8.0 cm = 8.0\times 10^{- 2}\ m

Volume Charge density, \rho_{v} = 40 nC/m^{3} = 40\times {- 9}\ C/m^{3}

Now,

To calculate the electric flux:

\phi = \frac{q}{\epsilon_{o}}                                                      (1)

where

\phi = electric flux

\epsilon_{o} = 8.85\times 10^{- 12}\ F/m = permittivity of free space  

Volume Charge density for the given case is given by the formula:

\rho_{v} = \frac{Total\ charge, q}{Volume of cube, V}                  (2)

Volume of cube, V = a^{3}

Thus

V = (8.0\times 10^{- 2})^{3} = 5.12\times 10^{- 4}\ m^{3}

Thus from eqn (2), the total charge is given by:

q = \rho_{v}V = 40\times {- 9}\times 5.12\times 10^{- 4}

q = 2.048\times 10^{-11}\ F = 20.48\ pF

Now, substitute the value of 'q' in eqn (1):

\phi = \frac{2.048\times 10^{-11}}{8.85\times 10^{- 12}} = 2.314\ Nm^{2}/C

5 0
2 years ago
You slip a wrench over a bolt. Taking the origin at the bolt, the other end of the wrench is at x=18cm, y=5.5cm. You apply a for
mart [117]

Answer:

The torque on the wrench is 4.188 Nm

Explanation:

Let r = xi + yj where is the distance of the applied force to the origin.

Since x = 18 cm = 0.18 cm and y = 5.5 cm = 0.055 cm,

r = 0.18i + 0.055j

The applied force f = 88i - 23j

The torque τ = r × F

So, τ = r × F = (0.18i + 0.055j) × (88i - 23j) = 0.18i × 88i + 0.18i × -23j + 0.055j × 88i + 0.055j × -23j

= (0.18 × 88)i × i + (0.18 × -23)i × j + (0.055 × 88)j × i + (0.055 × -22)j × j  

= (0.18 × 88) × 0 + (0.18 × -23) × k + (0.055 × 88) × (-k) + (0.055 × -22) × 0   since i × i = 0, j × j = 0, i × j = k and j × i = -k

= 0 - 4.14k + 0.0484(-k) + 0

= -4.14k - 0.0484k

= -4.1884k Nm

≅ -4.188k Nm

So, the torque on the wrench is 4.188 Nm

8 0
2 years ago
John is carrying a shovelful of snow. The center of mass of the 3.00 kg of snow he is holding is 15.0 cm from the end of the sho
Andru [333]

Answer:

James is correct here as the force of hand pushing upwards is always more than the force of hand pushing down

Explanation:

Here we know that one hand is pushing up at some distance midway while other hand is balancing the weight by applying a force downwards

so here we can say

Upwards force = downwards Force + weight of snow

while if we find the other force which is acting downwards

then for that force we can say that net torque must be balanced

so here we have

F_{down} L_1 = W_{snow} L_2

so here we have

F_{down} = \frac{L_2}{L_1} (W_{snow})

so here we can say that upward force by which we push up is always more than the downwards force

8 0
2 years ago
Kiting during a storm. The legend that Benjamin Franklin flew a kite as a storm approached is only a legend — he was neither stu
Dvinal [7]

Answer:

The current is   I  =  1.1434*10^{-5}}\  A

Explanation:

From the question we are told that

   The radius of the kite string is  R =  2.02 mm =  0.00202 \ m

   The  distance it extended upward is   D =  0.823 km = 823 \  m

   The thickness of the water layer is d = 0.506 mm  =  0.000506 \  m

   The resistivity is  \rho =  159\ \Omega  \cdot m

   The potential  difference is  V  =   186 MV =  186 *10^{6} \  V

Generally the cross sectional area of the water layer is mathematically represented as

      A =  \pi r^2

Here  r is mathematically represented as

      r =  [(R + d ) - R]

=>   r =  [(0.00202 +  0.000506 ) - 0.00202]

=>  r =  0.000506

=>     A = 3.142 *  [0.000506]^2  

=>     A = 8.0447*10^{-7}\ m^2  

Generally the resistance of the water is mathematically represented as

    R =  \frac{\rho  * D }{A}

=>   R =  \frac{159  *823 }{8.0447*10^{-7}}

=>   R = 1.62662 * 10^{11} \  \Omega

Generally the current is mathematically represented as

      I  =  \frac{V}{R}

=>    I  =  \frac{186 *10^{6} }{1.62662 * 10^{11}}

=>    I  =  1.1434*10^{-5}}\  A

8 0
2 years ago
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