could anyone determinate the period knowing that it performs 4000 vibrations in 0.5 minutes, Sorry my english is bad

Please help for 21! will mark brainliest

Katyanochek1 [597]

The answer for this question is D

5 0

3 years ago

When a 12 N horizontal force is applied to a box on a horizontal tabletop, the box remains at rest. The force of friction acting

Mrrafil [7]

Answer:

12N

Explanation:

when a force is applied to a body but still stays at rest or moves at a constant speed , the frictional force is equal to the force applied

3 0

3 years ago

You are coasting on your 12-kg bicycle at 13 m/s and a 5.0-g bug splatters on your helmet. The bug was initially moving at 1.5 m

Brut [27]

Answer:

a) Pi,c = 1066 kgm/s

b) Pi,b = 0.0075 kgm/s

c) ΔV = - 0.0007 m/s

d) ΔV = - 0.0008 m/s

Explanation:

Given:-

- The mass of the bicycle, mc = 12 kg

- The mass of passenger, mp = 70 kg

- The mass of the bug, mb = 5.0 g

- The initial speed of the bicycle, vpi = 13 m/s

- The initial speed of the bug, vbi = 1.5 m/s

Find:-

a.What is the initial momentum of you plus your bicycle?

b.What is the initial momentum of the bug?

c.What is your change in velocity due to the collision the bug?

d.What would the change in velocity have been if the bug were traveling in the opposite direction?

Solution:-

- First we will set our one dimensional coordinate system, taking right to be positive in the direction of bicycle.

- The initial linear momentum (Pi,c) of the passenger and the bicycle would be:

Pi,c = vpi* ( mc + mp)

Pi,c = 13* ( 12+ 70 )

Pi,c = 1066 kgm/s

- The initial linear momentum (Pi,b) of the bug would be:

Pi,b = vbi*mb

Pi,b = 0.005*1.5

Pi,b = 0.0075 kgm/s

- We will consider the bicycle, the passenger and the bug as a system in isolation on which no external unbalanced forces are acting. This validates the use of linear conservation of momentum.

- The bicycle, passenger and bug all travel in the (+x) direction after the bug splatters on the helmet.

Pi = Pf

Pi,c + Pi,b = V*(mb + mc + mp)

Where, V : The velocity of the (bicycle, passenger and bug) after collision.

1066 + 0.0075 = V*( 0.005 + 12 + 70 )

V = 1066.0075 / 82.005

V = 12.9993 m/s

- The change in velocity is Δv = 13 - 12.9993 = - 0.00070 m/s

- If the bug travels in the opposite direction then the sign of the initial momentum of the bug changes from (+) to (-).

- We will apply the linear conservation of momentum similarly.

Pi = Pf

Pi,c + Pi,b = V*(mb + mc + mp)

1066 - 0.0075 = V*( 0.005 + 12 + 70 )

V = 1065.9925 / 82.005

V = 12.99911 m/s

- The change in velocity is Δv = 13 - 12.99911 = -0.00088 m/s

7 0

4 years ago

Read 2 more answers

I was driving along at 20 m/s, trying to change a CD and not watching where I was going. When I looked up, I found myself 45 m f

cricket20 [7]

Answer:

$a=4.44\frac{m}{s^2}$

Explanation:

First we have to find the time required for train to travel 60 meters and impact the car, this is an uniform linear motion:

$t=\frac{d}{v}\\\\t=\frac{60m}{30\frac{m}{s}}=2s$

The reaction time of the driver before starting to accelerate was 0.50 seconds. So, remaining time for driver is 1.5 seconds.

Now, we have to calculate the distance traveled for the driver in this 0.5 seconds before he start to accelerate. Again, is an uniform linear motion:

$d=vt\\d=20\frac{m}{s}(0.5s)=10m$

The driver cover 10 meters in this 0.5 seconds. So, the remaining distance to be cover in 1.5 seconds by the driver are 35 meters. We calculate the minimum acceleration required by the car in order to cross the tracks before the train arrive, Since this is an uniformly accelerated motion, we use the following equation:

$d=v_0t+\frac{1}{2}at^2\\a=\frac{2(d-v_0t)}{t^2}\\a=\frac{2(35m-20\frac{m}{s}*1.5s}{(1.5s)^2}\\a=4.44\frac{m}{s^2}$

7 0

3 years ago

A force of 8,480 N is applied to a cart to decelerate it at a rate of 32.0 m/s 2 . What is the mass of the cart?

Vinvika [58]

The force is given by the product of mass and the acceleration thus the force is given
8480/32 is equal to 265 kg

5 0

3 years ago