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3 years ago

I hit a 2.5kg hockey puck which results in an acceleration of 10 m/s squared what is the force i exerted on the puck

Physics

1 answer:

Studentka2010 [4]3 years ago

8 0

newton's law 2 ... f=ma ... f=2.5x10=25N

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Without the wheels, a bicycle frame has a mass of 8.29 kg. Each of the wheels can be roughly modeled as a uniform solid disk wit

trapecia [35]

Answer:

69.66 Joule

Explanation:

mass of bicycle frame, mf = 8.29 kg

mass of wheel, mw = 0.820 kg

radius, r = 0.343 m

velocity, v = 3.6 m/s

There are two wheels in the bicycle.

There are two types of kinetic energy of the system one is kinetic energy of rotation and another is rotational kinetic energy.

$K = \frac{1}{2}m_{f}v^{2}+ 2\times \frac{1}{2}m_{w}v^{2}+ 2\times \frac{1}{2}I_{w}\omega^{2}$

$K = \frac{1}{2}m_{f}v^{2}+m_{w}v^{2}+ \frac{1}{2}\times m_{w}v^{2}$

$K = \frac{1}{2}m_{f}v^{2}+ \frac{3}{2}\times m_{w}v^{2}$

$K = \frac{1}{2}\times 8.29\times 3.6^{2}+ \frac{3}{2}\times 0.820\times 3.6^{2}$

K = 69.66 J

3 0

4 years ago

A 0.080-kg remote control 22.0 cm long rests on a table, as shown in the figure below, with a length L overhanging its edge. To

Lesechka [4]

Base in your question that ask for the distance that the remote control would extend beyond the edge where as a remote control is 22cm long and has a mass of 0.08kg =, base on my calculation the answer would be 1.27cm i hope you understand my answer

4 0

3 years ago

A 4kg watermelon is dropped from a height of 45m. What is the velocity of the watermelon just before it hits the ground?

Makovka662 [10]

Answer:

v=30 m/s

Explanation:

h - height

g - acceleration due to gravity=10

t - time

v- velocity

$h = \frac{1}{2} \times g \times t {}^{2}$

45 = 5t²

t² = 9

t=3 seconds

v=g×t

v=10×3

v=30 m/s

3 0

3 years ago

The number of wavelengths that pass a given point/second is called

myrzilka [38]

Answer:

I believe the answer is frequency

Hope this helps!! (let me know if it is right)

Explanation:

6 0

3 years ago

A 330 kg piano slides 3.6 m down a 28o incline and is kept from accelerating by a man who is pushing back on it parallel to the

e-lub [12.9K]

Answer:

a. 652.68N

b. -2349.65J

c. -3116.12J

d. 5465.77J

e. Zero

Explanation:

a. According to equilibrium of forces, the force of gravity is equal to the sum of the frictional force and force exerted by the man in the opposite direction (since they're both resistant forces).

Fg = Fm + Fr

Fm = Fg - Fr

Fm = mgsin(28°) - umgcos(28°)

u = coefficient of frictional force.

Fm = 330*9.8*sin28 - 0.4*330*9.8*cos28

Fm = 1518.27 - 865.59

Fm = 652.68N

b. Work done by man is:

Wm = -Fm * d

Wm = -652.68 * 3.6

Wm = -2349.65J

c. Work done by friction force:

W(Fr) = -Fr * d

W(Fr) = -865.59 * 3.6

W(Fr) = -3116.12J

d. Work done by gravity:

Wg = Fg * d

Wg = 1518.27 * 3. 6

Wg = 5465.77J

e. Net work done on the piano is:

Work done by friction + work done by gravity + work done by man

= -3116.12 + 5464.77 + (-2349.65)

= 0J

7 0

3 years ago