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4 years ago

Which best describes a concave lens?

Physics

1 answer:

-Dominant- [34]4 years ago

4 0

Answer:

Explanation:

Concave lens also called as diverging lens i.e. it diverges ray of light coming towards it.

Concave lens is thicker at the edges and thinner at the center.

Concave lens formed the virtual image of the object i.e. it cannot be trace on screen. This lens is used to treat the nearsightedness or myopia in which a person is unable to see the far object clearly.

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The maximum tension the wire can withstand without breaking is 300 N . A 0.800 kg projectile traveling horizontally hits and emb

topjm [15]

Answer:

Let's investigate the case where the cable breaks.

Conservation of angular momentum can be used to find the speed.

$\vec{L}_1 = \vec{L}_2\\\vec{L}_1 = m\vec{v_0} \\\vec{L}_2 = I\vec{\omega}\\$

The projectile embeds itself to the ball, so they can be treated as a combined object. <u>The moment of inertia of the combined object is equal to the sum of the moment of inertia of both objects. </u>

$I = I_{projectile} + I_{ball}\\I = mr^2 + mr^2\\I = 2mr^2$

where r is the length of the cable.

<u>After the collision, the ball and the projectile makes a circular motion because of the cable.</u> So, the force (tension) in circular motion is

$F = \frac{mv^2}{r}$

The relation between linear velocity and the angular velocity is

$v = \omega r$

So,

$F = \frac{m(\omega r)^2}{r} = m\omega^2 r = 300\\mv_0r =I\omega\\\\\omega = mv_0r/I\\300 = m(\frac{mv_0r}{I})^2r = m(\frac{mv_0r}{2mr^2})^2r = m(\frac{v_0}{2r})^2r = \frac{mv_0^2r}{4r^2} = \frac{mv_0^2}{4r}\\300 = \frac{0.8v_0^2}{4r}\\1500 = v_0^2/r\\v_0 = \sqrt{1500r}$

As can be seen, the maximum velocity for the projectile without breaking the cable is $\sqrt{1500r}$ , where r is the length of the cable.

6 0

3 years ago

The acceleration of a particle is defined by the relation a 5 28 m/s2. Knowing that x 5 20 m when t 5 4 s and that x 5 4 m when

mamaluj [8]

Explanation:

It is given that,

The acceleration of a particle, $a=-8\ m/s^2$ (negative as the particle is decelerating)

Initial distance, x₁ = 20 m

Initial time, t₁ = 4 s

New distance x₂ = 4 m

Velocity, v = 10 m/s

(A) Calculating initial distance using second equation of motion as :

$x_1=ut_1+\dfrac{1}{2}at^2$

$20=4u+\dfrac{1}{2}(-8)\times 4^2$

u = 21 m/s

When velocity of the particle is zero, time taken is t (say). Using first equation of motion as :

$v=u+at$

$0=21+(-8)t$

t = 2.62 seconds

So, the velocity of the particle is zero at t = 2.62 seconds.

(B) Velocity at t = 11 s

$v=21+(-8)\times 11$

v = 13 m/s

Total distance covered at t = 11 s. The overall path travelled by the particle during its entire journey is called total distance covered.

$d=ut+\dfrac{1}{2}at^2+|ut+\dfrac{1}{2}at^2|$

$d=21\times 2.62+\dfrac{1}{2}\times (-8)(2.62)^2+|21\times 8.38+\dfrac{1}{2}\times (-8)(8.38)^2|$

d = 132.48 m

So, the distance travelled by the particle at t = 11 seconds is 132.48 meters.

5 0

3 years ago

A 1.89 kg object on a frictionless horizontal track is attached to the end of a horizontal spring whose force constant is 4.77 N

Maksim231197 [3]

Answer:

Magnitude F(t)=26.6 N

Direction: -x

Explanation:

Given data

Spring constant K=4.77 N/m

Mass m=1.89 kg

Displace A=5.56m

Time t=3.96s

To find

Magnitude of force F

Solution

The angular frequency is given as

$w=\sqrt{\frac{K}{m} } \\w=\sqrt{\frac{4.77N/m}{1.89kg} }\\w=1.59rad/s$

Force on object is

$F(t)=-mAw^{2}Cos(wt)$

Substitute given values

$F(t)=-(1.89kg)(5.56m)(1.59rad/s)^{2}Cos(1.59*3.96)\\F(t)=-26.6N$

Magnitude F(t)=26.6 N

Direction: -x

4 0

3 years ago

The cannon on a battleship can fire a shell a maximum distance of 26.0 km.

Alla [95]

It is possible to demonstrate that the maximum distance occurs when the angle at which the projectile is fired is $\theta = 45^{\circ}$ .
In fact, the laws of motions on both x- and y- directions are
$S_x(t)= v_0 cos \theta t$
$S_y(t)= v_0 \sin \theta t - \frac{1}{2} gt^2$
From the second equation, we get the time t at which the projectile hits the ground, by requiring $S_y(t)=0$ , and we get:
$t= \frac{2 v_0 \sin \theta}{g}$
And inserting this value into Sx(t), we find
$S_x(t) = 2 \frac{v_0^2}{g} \sin \theta \cos \theta= \frac{v_0^2}{g} \sin (2\theta)$
And this value is maximum when $\theta=45^{\circ}$ , so this is the angle at which the projectile reaches its maximum distance.

So now we can take again the law of motion on the x-axis
$S_x(t)= \frac{v_0^2}{g} \sin (2\theta)$
And by using $S_x = 26 km=26000 m$ , we find the value of the initial velocity v0:
$v_0 = \sqrt{ \frac{S_x g}{\sin (2\theta)} } = \sqrt{ \frac{(26000m)(9.81m/s^2)}{\sin (2\cdot 45^{\circ})} } =505 m/s$

8 0

3 years ago

A bowling ball and a ping-pong ball are each tied to a string and hung from the ceiling. The distance from the ceiling to the CM

zhannawk [14.2K]

Both would have the same period

3 0

3 years ago