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Katena32 [7]
3 years ago
11

A block of 200 g is attached to a light spring with a force constant of 5 N / m and freely in a horizontal plane vibrates. The m

ass is released, separating it 5 cm from its equilibrium position. Find the period of the motion​
Physics
1 answer:
Finger [1]3 years ago
8 0

Answer:

m  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 mm  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 m

Explanation:

That is a reason

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A food packet is dropped from a helicopter during a flood-relief operation from a height of 750 meters. Assuming no drag (air fr
Leviafan [203]

1. Velocity at which the packet reaches the ground: 121.2 m/s

The motion of the packet is a uniformly accelerated motion, with constant acceleration a=g=9.8 m/s^2 directed downward, initial vertical position d=750 m, and initial vertical velocity v_0 = 0. We can use the following SUVAT equation to find the final velocity of the packet after travelling for d=750 m:

v_f^2 -v_i^2 =2ad

substituting, we find

v_f^2 = 2ad\\v_f = \sqrt{2ad}=\sqrt{2(9.8 m/s^2)(750 m)}=121.2 m/s

2. height at which the packet has half this velocity: 562.6 m

We need to find the heigth at which the packet has a velocity of

v_f=\frac{121.2 m/s}{2}=60.6 m/s

In order to do that, we use again the same SUVAT equation substituting v_f with this value, so that we find the new distance d that the packet travelled from the helicopter to reach this velocity:

v_f^2-v_i^2=2ad\\d=\frac{v_f^2}{2a}=\frac{(60.6 m/s)^2}{2(9.8 m/s^2)}=187.4 m

Which means that the heigth of the packet was

h=750 m-187.4 m=562.6 m

3 0
3 years ago
An object has more momentum if it has which of the following?
maks197457 [2]

Answer:

C.) A high velocity and Large mass.

Explanation:

Momentum of any object is defined by following formula

Here :  m = mass of object

v = velocity of object

now we know that since momentum is product of mass and velocity

So in order to have more momentum we need the value of this product to be more.  So this product will me large is both the physical quantity will be more in magnitude. So if mass is large and velocity will be more then the product of them will be large and hence the momentum of object will be more. Btw I had that question too.

6 0
3 years ago
One end of spring is attached to a wall. The other end is attached to a movable block. If you pull the block so as to stretch th
IRINA_888 [86]

Answer: Positive

Explanation:

6 0
3 years ago
A 0.500-kg stone is moving in a vertical circular path attached to a string that is 75.0 cm long. The stone is moving around the
Semenov [28]

Answer:

B. 7.07 m/s

Explanation:

The velocity of the stone when it leaves the circular path is its tangential velocity, v, which is given by

v=\omega r

where \omega is the angular speed and r is the radius of the circular path.

\omega is given by

\omega = 2\pi f

where f is the frequency of revolution.

Thus

v=2\pi fr

Using values from the question,

v=2\pi\times 1.50\times0.75

<em>Note the conversion of 75 cm to 0.75 m</em>

v=2\times3.14\times 1.50\times0.75 = 9.42\times0.75 = 7.065=7.07

6 0
3 years ago
A moving van travels 10km North, then 4 km east, drops off some furniture and then drives 8 km south. (a) Sketch the path of the
Juli2301 [7.4K]

Answer:

4.47 km

Explanation:

If we draw the path of the van then we get a shape with two exposed points A and D. If we draw a line from point D perpendicular to BA we get point E. This gives us a right angled triangle ADE.

From Pythagoras theorem

AD² = AE² + ED²

AD=\sqrt{AE^2+ED^2}\\\Rightarrow AD=\sqrt{2^2+4^2}\\\Rightarrow AD=\sqrt{20}\\\Rightarrow AD=4.47\ km

Hence, the van is 4.47 km from its initial point

3 0
3 years ago
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