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Zielflug [23.3K]
2 years ago
5

A hockey puck attached to a horizontal spring oscillates on a frictionless, horizontal surface. The spring has force constant 4.

50 N/m and the oscillation period is 1.20 s. What is the mass of the puck
Physics
1 answer:
bekas [8.4K]2 years ago
3 0

Answer:

m = 0.164 kg

Explanation:

T (period)

k (force/spring constant)

m (mass)

T = 2*Pi*sqrt(m/k)

T/(2*Pi) = sqrt(m)/sqrt(k)

(T/(2*Pi))*sqrt(k) = sqrt(m)

m = ((T/(2*Pi))*sqrt(k))^2

m = 4.5*((1.2/(2*Pi)))^2

m = 0.1641403175

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A plane is traveling at 300 m/s. How far will it travelin 1 hour?
Airida [17]

Answer:

1080000

Explanation:

300 x 60s=18000m/minute

18000 x 60min=1080000m/h

Explanation:

3 0
2 years ago
Asteroids are between 1000 km and less than 10 m in diameter. What is the diameter of most asteroids?
e-lub [12.9K]

This question needs research to be answered. From the given information alone it can't be answered without making wild assumptions.

Ideally, you need to take a look at a distribution (or a histogram) of asteroid diameters, identify the "mode" of such a distribution, and find the corresponding diameter. That value will be the answer.

I am attaching one such histogram on asteroid diameters from the IRAS asteroid catalog I could find online. (In order to get a single histogram, you need to add the individual curves in the figure first). Eyeballing this sample, I'd say the mode is somewhere around 10km, so the answer would be: the diameter of most asteroid from the IRAS asteroid catalog is about 10km.

7 0
3 years ago
A soccer field is viewed from above, while a ball is kicked eastward with an initial speed of 10.0 m/s. The ball experiences a c
satela [25.4K]

Answer:

the ball travelled approximately 60 m towards north before stopping

Explanation:

 Given the data in the question;

First course : a_{x} = 0.75 m/s², d_{x} = 20 m, u_{x} = 10 m/s

now, form the third equation of motion;

v² = u² + 2as

we substitute

v_{x}² = (10)² + (2 × 0.75 × 20)

v_{x}² = 100 + 30

v_{x}² = 130

v_{x} = √130

v_{x} = 11.4 m/s

for the Second Course:

u_{y} =  11.4 m/s,  a_{y} = -1.15 m/s²,  v_{y} = 0

Also, form the third equation of motion;

v² = u² + 2as

we substitute

0² = (11.4)² + (2 × (-1.15) ×  d_{y} )

0 = 129.96 - 2.3d_{y}

2.3d_{y}  = 129.96

d_{y} = 129.96 / 2.3

d_{y} = 56.5 m

so;

|d| = √( d_{x}² + d_{y}² )

we substitute

|d| = √( (20)² + (56.5)² )

|d| = √( 400 + 3192.25 )

|d| = √( 3592.25 )

|d| = 59.9 m ≈ 60 m

Therefore, the ball travelled approximately 60 m towards north before stopping

7 0
2 years ago
Define energy and provide examples of potential and kinetic energy
In-s [12.5K]

Energy is the capacity for doing work..

Kinetic energy - Moving car

Potential energy - flowing water up the hill

6 0
2 years ago
A 72-kg boy is surfing and catches a wave which gives him an initial speed of 1.6 m/s. He then drops through a height of 1.61 m,
chubhunter [2.5K]

Answer:

2849.98 J

Explanation:

From the question,

Work done by the boy = change Potential energy of the boy + change in kinetic energy of the boy

W = ΔP + ΔK..................... Equation 1

Where W = work done by the boy, ΔP = change in potential energy of the boy, ΔK = Change in kinetic energy of the boy.

But,

ΔP = mgΔh.................... Equation 2

ΔK = 1/2mΔv²................. Equation 3

Where m = mass of the boy, Δh = change in height of the boy, Δv = change in velocity of the boy.

Substitute equation 2 and 3 into equation 1

W = mgΔh+1/2mΔv²................. Equation 4

Given: m = 72 kg, Δh = 1.61 m, Δv = 8.5-1.6 = 6.9 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 72(9.8)(1.61)+1/2(72)(6.9²)

W = 1136.016+1713.96

W = 2849.98 J

8 0
3 years ago
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