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2 years ago

How would reversing the wheel’s initial direction of rotation affect the result??

Physics

1 answer:

d1i1m1o1n [39]2 years ago

4 0

It would change the sign on the vector quantities and have no change to the scalar quantities

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Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim

crimeas [40]

Answer:

Centripetal acceleration of the car is 17.4 m/s²

Explanation:

It is given that,

Radius of the track, r = 57 m

Speed of car, v = 31.5 m/s

We need to find the centripetal acceleration of the race car. The formula for the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(31.5\ m/s)^2}{57\ m}$

$a=17.4\ m/s^2$

So, the centripetal acceleration of the race car is 17.4 m/s². Hence, this is the required solution.

5 0

1 year ago

A metal sphere of radius 10 cm carries a charge of +2.0 μC uniformly distributed over its surface. What is the magnitude of the

frozen [14]

Answer:

$8.0\cdot 10^5 N/C$

Explanation:

Outside the sphere's surface, the electric field has the same expression of that produced by a single point charge located at the centre of the sphere.

Therefore, the magnitude of the electric field ar r = 5.0 cm from the sphere is:

$E=k\frac{q}{(R+r)^2}$

where

$k=8.99\cdot 10^9 N m^2C^{-2}$ is the Coulomb's constant

$q=2.0 \mu C=2.0 \cdot 10^{-6}C$ is the charge on the sphere

$R=10 cm = 0.10 m$ is the radius of the sphere

$r=5.0 cm = 0.05 m$ is the distance from the surface of the sphere

Substituting, we find

$E=(8.99\cdot 10^9 Nm^2 C^{-2})\frac{2.0\cdot 10^{-6} C}{(0.10 m+0.05 m)^2}=8.0\cdot 10^5 N/C$

3 0

2 years ago

Julius competes in the hammer throw event. The hammer has a mass of 7.26 kg and is 1.215 m long. What is the centripetal force o

nevsk [136]

In the circular motion of the hammer, the centripetal force is given by
$F=m \frac{v^2}{r}$
where m is the mass of the hammer, v its tangential speed and r is the distance from the center of the motion, i.e. the length of the hammer.
Using the data of the problem, we find:
$F=m \frac{v^2}{r}=(7.26 kg) \frac{(31.95 m/s)^2}{1.215 m}=6100 N$

4 0

2 years ago

Read 2 more answers

Person lifting a chair convert from what energy to another

vagabundo [1.1K]

A person lifting a chair is converting chemical energy to mechanical energy.

4 0

2 years ago

A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit

makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

2.87 = -1.442 + 1.8 u

u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

7 0

2 years ago