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3 years ago

What 3 things can an object with unbalanced forces do?

1 answer:

3 0

Forces are considered balanced when all of the combined forces lead to no change in the motion of the object. For example, when a book is sitting on a table, the force of gravity is pushing downward and the normal force is pushing upward with exactly the same amount of force. Since they are equal and opposite forces, the book does not move.

Unbalanced forces exist when there are unequal forces acting upon the object, which leads to a change in the state of motion. Unbalanced forces can lead to a change in direction, a change in speed, or both a change in direction and in speed.

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Is the following true or false? Continental glaciers fill depressions in the surface with water, where they create cirques.

Lorico [155]

False is the answer to your question

6 0

3 years ago

Pls help me with this fast. I will mark brainiest

denis23 [38]

Answer:

a) 70, 95

b) 95-70= 25cc

c) density= mass/volume

102/25

=4.08g/cc

5 0

3 years ago

A diffraction grating with 600 lines/mmlines/mm is illuminated with light of wavelength 510 nmnm. A very wide viewing screen is

Ksenya-84 [330]

Answer:

A.2.95 m

B.7

Explanation:

We are given that

Diffraction grating=600 lines/mm

d= $\frac{1 mm}{600}=\frac{1\times 10^{-3} m}{600}=1.67\times 10^{-6} m$

Wavelength of light, $\lambda=510 nm=510\times 10^{-9} m$

l=4.6 m

A.We have to find the distance between the two m=1 bright fringes

$sin\theta=\frac{m\lambda}{d}$

For first bright fringe, =1

$sin\theta=\frac{1\times 510\times 10^{-9}}{1.67\times 10^{-6}}=0.305$

$\theta=sin^{-1}(0.305)=17.76^{\circ}$

The distance between two m=1 fringes

$x=2ltan\theta=2\times 4.6 tan(17.76^{\circ})=2.95 m$

Hence, the distance between two m=1 fringes=2.95 m

B.For maximum number of fringes,

$sin\theta=1$

$sin\theta=\frac{m\lambda}{d}$

Substitute the values

$1=\frac{m\times 510\times 10^{-9}}{1.67\times 10^{-6}}$

$m=\frac{1.67\times 10^{-6}}{510\times 10^{-9}}=3.3\approx 3$

Maximum number of bright fringes on the scree= $2m+1=2(3)+1=7$

8 0

3 years ago

At t=0 a grinding wheel has an angular velocity of 25.0 rad/s. It has a constant angular acceleration of 26.0 rad/s2 until a cir

Agata [3.3K]

Answer:

a) The total angle of the grinding wheel is 569.88 radians, b) The grinding wheel stop at t = 12.354 seconds, c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

Explanation:

Since the grinding wheel accelerates and decelerates at constant rate, motion can be represented by the following kinematic equations:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

$\omega = \omega_{o} + \alpha \cdot t$

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

Where:

$\theta_{o}$ , $\theta$ - Initial and final angular position, measured in radians.

$\omega_{o}$ , $\omega$ - Initial and final angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

Likewise, the grinding wheel experiments two different regimes:

1) The grinding wheel accelerates during 2.40 seconds.

2) The grinding wheel decelerates until rest is reached.

a) The change in angular position during the Acceleration Stage can be obtained of the following expression:

$\theta - \theta_{o} = \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\theta-\theta_{o} = \left(25\,\frac{rad}{s} \right)\cdot (2.40\,s) + \frac{1}{2}\cdot \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)^{2}$

$\theta-\theta_{o} = 134.88\,rad$

The final angular angular speed can be found by the equation:

$\omega = \omega_{o} + \alpha \cdot t$

If $\omega_{o} = 25\,\frac{rad}{s}$ , $t = 2.40\,s$ and $\alpha = 26\,\frac{rad}{s^{2}}$ , then:

$\omega = 25\,\frac{rad}{s} + \left(26\,\frac{rad}{s^{2}} \right)\cdot (2.40\,s)$

$\omega = 87.4\,\frac{rad}{s}$

The total angle that grinding wheel did from t = 0 s and the time it stopped is:

$\Delta \theta = 134.88\,rad + 435\,rad$

$\Delta \theta = 569.88\,rad$

The total angle of the grinding wheel is 569.88 radians.

b) Before finding the instant when the grinding wheel stops, it is needed to find the value of angular deceleration, which can be determined from the following kinematic expression:

$\omega^{2} = \omega_{o}^{2} + 2 \cdot \alpha \cdot (\theta-\theta_{o})$

The angular acceleration is now cleared:

$\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot (\theta-\theta_{o})}$

Given that $\omega_{o} = 87.4\,\frac{rad}{s}$ , $\omega = 0\,\frac{rad}{s}$ and $\theta-\theta_{o} = 435\,rad$ , the angular deceleration is:

$\alpha = \frac{ \left(0\,\frac{rad}{s}\right)^{2}-\left(87.4\,\frac{rad}{s} \right)^{2}}{2\cdot \left(435\,rad\right)}$

$\alpha = -8.780\,\frac{rad}{s^{2}}$

Now, the time interval of the Deceleration Phase is obtained from this formula:

$\omega = \omega_{o} + \alpha \cdot t$

$t = \frac{\omega - \omega_{o}}{\alpha}$

If $\omega_{o} = 87.4\,\frac{rad}{s}$ , $\omega = 0\,\frac{rad}{s}$ and $\alpha = -8.780\,\frac{rad}{s^{2}}$ , the time interval is:

$t = \frac{0\,\frac{rad}{s} - 87.4\,\frac{rad}{s} }{-8.780\,\frac{rad}{s^{2}} }$

$t = 9.954\,s$

The total time needed for the grinding wheel before stopping is:

$t_{T} = 2.40\,s + 9.954\,s$

$t_{T} = 12.354\,s$

The grinding wheel stop at t = 12.354 seconds.

c) The deceleration experimented by the grinding wheel was 8.780 radians per square second.

4 0

3 years ago

How is momentum conserved in a newtons cradle <br><br> when one steel ball hits the other

arsen [322]

In Newton's Cradle experiment we know that all cradles of same mass and identical to each other

Now we know that when two identical objects collide elastically then they interchange their velocity

So here we have same illustration

When Newton pulls up a cradle and release it will move hit another cradle which is at rest

Due to elastic collision between them first cradle comes to rest and another cradle will move ahead with same speed this process remains the same and one by one all cradle hit another.

So at the last the cradle at the end will move off with the same speed as the first cradle will hit with the speed.

So in this experiment the cradle at the last end will move off at same distance away from the right end as that of left end we pull the cradle.

So here we can say that in horizontal direction when all cradles are colliding each other there is no external force on the system so momentum is conserved and they all will move off with same speed and hence we observe the above condition.

8 0

3 years ago