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3 years ago

13

If we want to see our full image then the minimum size of the plane mirror:

1 answer:

raketka [301]3 years ago

8 0

It would be D I believe! Depending on the angle of the mirror and distance positioned!

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A bicycle wheel with radius 0.3 m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total accelera

AlekseyPX

Answer:

Explanation:

Given

Radius of bicycle wheel $r=0.3\ m$

Initial angular velocity $\omega _0=0$

It rotates 3 revolution in 5 s therefore

$\omega =2\pi 3=\6\pi =18.85\ rad/s$

using $\omega =\omega _0+\alpha t$

where $\alpha =angular\ acceleration$

$\omega =Final\ angular\ velocity$

$t=time$

$\alpha =\frac{18.85}{5}=3.77 rad/s^2$

Total acceleration of any point will be a vector sum of tangential acceleration and centripetal acceleration

$\omega at t=1$

$\omega =0+3.77\times 1=3.77 rad/s$

$a_c=\omega ^2\cdot r$

$a_c=(3.77)^2\cdot 0.3=4.26 m/s^2$

Tangential acceleration $a_t=\alpha \times r$

$a_t=3.77\times 0.3=1.13 m/s^2$

$a_{net}=\sqrt{a_t^2+a_c^2}$

$a_{net}=\sqrt{(1.13)^2+(4.26)^2}$

$a_{net}=4.41 m/s^2$

7 0

3 years ago

A key falls from a bridge that is 45 m above the water. the key falls straight down and lands in a model boat traveling at a con

erastova [34]

Let the key is free falling, therefore from equation of motion

$h = ut +\frac{1}{2}gt^2.$ .

Take initial velocity, u=0, so

$h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2$ .

$h = 0\times t + \frac{1}{2}g t^2= \frac{1}{2}gt^2 \\\ t =\sqrt{\frac{2h}{g} }$

As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula

$d= v \times t$

From above substituting t,

$d = v \times \sqrt{\frac{2h}{g} }$ .

Now substituting all the given values and g = 9.8 m/s^2, we get

$d = 3.5 \ m/s \times \sqrt{\frac{2 \times 45 m}{9.8 m/s^2} } = 10.60 m$ .

Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.

7 0

3 years ago

The diagram below shows a food web.

Mekhanik [1.2K]

i think it’s B. sorry if i’m wrong

8 0

2 years ago

Read 2 more answers

A 1000-kilogram car traveling due east at 15 meters per second hit from behind and receives a forward impulse of 6000 newton-sec

strojnjashka [21]

The change in momentum of the car is 6000 kg m/s

Explanation:

According to the impulse theorem, the change in momentum of an object is equal to the impulse exerted on the object, therefore:

$\Delta p = I$

where

$\Delta p$ is the change in momentum

I is the impulse exerted

For the car in this problem, the impulse received is

I = 6000 kg m/s (in the forward direction)

Therefore, the change in momentum of the car is equal to this value:

$\Delta p = I = 6000 kg m/s$ (in the forward direction)

We can also calculate what is the new momentum of the car. In fact, the initial momentum is

$p_i = mu = (1000 kg)(15 m/s)=15,000 kg m/s$

And so, the new momentum is

$p_f = p_i + \Delta p = 15,000 + 6,000 = 21,000 kg m/s$

Learn more about impulse and momentum:

brainly.com/question/9484203

#LearnwithBrainly

8 0

3 years ago

Please answer all of these questions for brainly!

Nana76 [90]

Answer:

1. C
2. B
3. D
4. A

3 0

3 years ago