In which direction does the electric field point at a position directly east of a

A 20-cm-long, 190 g rod is pivoted at one end. A 19 g ball of clay is stuck on the other end.

MaRussiya [10]

Answer:

0.76 s

Explanation:

We are given that

Length of rod,L=20 cm= $\frac{20}{100}=0.20m$

1 m=100 cm

Mass of rod,M=190 g= $\frac{190}{1000}=0.19kg$

Mass of ball,m=19 g= $\frac{19}{1000}=0.019 kg$

Using 1 kg=1000g

We have to find the period if the rod and clay swing as a pendulum.

Moment of inertia of rod-clay=Moment of inertia of rod+moment of inertia of clay

$I_{rod-clay}=I_{rod}+I_{clay}$

$I_{rod-clay}=\frac{1}{3}ML^2+mL^2$

Substitute the values then we get

$I_[rod-clay}=\frac{1}{3}(0.19)(0.20)^2+(0.019)(0.20)^2$

$I_{rod-clay}=3.29\times 10^{-3} kgm^2$

Now, the center of mass of the combination of the rod and clay is given by

$y=\frac{Md_1+md_2}{M+m}$

Substitute $d_1=\frac{L}{2}$ =Distance between pivot and the center of the rod

$d_2=L=$ The distance between rod and clay

Using the formula

$y=\frac{0.19\times \frac{0.20}{2}+0.019\times 0.20}{0.19+0.019}$

$y=0.1091$ m

Time period of the oscillation of the system of the rod and the clay is given by

$T=2\pi\sqrt{\frac{I_{rod-clay}}{(M+m)yg}}$

g= $9.8m/s^2$

Using the formula

Time period= $2\times 3.14\sqrt{\frac{3.29\times 10^{-3}}{(0.19+0.019)\times 9.8\times 0.1091}}$

Time-period=0.76 s

Hence, the period =0.76 s

8 0

3 years ago

If a football quarterback can throw the ball at 20.0 m/s, what is the furthest distance the quarterback could throw the ball neg

kvasek [131]

The furthest distance the quarterback could throw the ball neglecting air resistance is 81.63 meters.

The situation given above is an example of projectile motion.

The maximum distance up to which a projectile can be thrown is known as its range, which is given by the formula,

Range(R) = 2u^2 sin(2θ)/g

Here u is the maximum velocity of the projectile= 20 m/s

θ is the angle at which the projectile is launched = 45°

g is the acceleration due to gravity = 9.8 m/s^2

Putting the above values in the given equation,

The range comes out to be 81.63 meters.

Hence, the furthest distance the quarterback could throw the ball neglecting air resistance is 81.63 meters.

To know more about "projectile motion", refer to the following link:

brainly.com/question/11422992?referrer=searchResults

#SPJ4

4 0

1 year ago

A football kicked off at an angle 20° from the ground reaches a maximum height of 4.7 m. What is the initial velocity of the kic

Svetradugi [14.3K]

The motion of the football that is being kicked follows a projectile path and is parabolic in shape. It has a maximum height of meters and the angle with respect to the ground is degrees. The maximum height is the highest position vertically that the ball could reach. From the equations of an object in a projectile motion, we use the equation <span>(V*sin20)^2 = 2gh in order to solve for velocity, V. We do as follows:

</span><span>((V) (sin(20)))^2 = 2gh
(V^2) (0.342^2) = 92.12
V = √ (92.12 / 0.342^2)
V = 28.06 m/s

</span>The initial velocity of the kick would be 28.06 m/s.

6 0

3 years ago

Describe a situation where Newton's third law of motion can be observed

yarga [219]

An example would be a swimmer pushes on the wall with her feet, which causes the wall to push back on her feet due to Newton's third law

3 0

3 years ago

If a person is severely dehydrated you would expect to see all of the following except ________.

Brums [2.3K]

Answer: lower immunity

Explanation:

5 0

3 years ago