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blagie [28]
2 years ago
15

The acceleration due to gravity on Earth is 9.81 m/s² [down]. A hockey puck was launched from the ice with a velocity of 26.9 m/

s [up]. Assuming no friction, determine the velocity of the hockey puck when it hits the ice.
Physics
1 answer:
charle [14.2K]2 years ago
4 0

Answer:

A hockey puck having a mass of 0.30 kg slides on the frictionless, horizontal surface of an ... magnitude of 5.0 N, and the force F2 has a magnitude of 8.0 N. Determine both ... A large man and a small boy stand facing each other on frictionless ice. They ... A hockey puck on a frozen pond is given an initial speed of 20.0 m/s.

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You are part of a design team assigned the task of making an electronic oscillator that will be the timing mechanism of a micro-
Snowcat [4.5K]

Solution :

We assume that there is a ring having a charge +Q and radius r. Electric field due to the ring at a point P on the axis is given by :

E_P=\int dE \cos

E_P=\int \frac{KdQ}{(\sqrt{r^2+x^2})^2}\times \frac{x}{\sqrt{r^2+x^2}}

\vec{E_P}=\frac{Kx}{r^2+x^2} \int dQ

\vec{E_P}=\frac{KxQ}{(r^2+x^2)^{3/2}} \hat{i}

If we put an electron on point P, then force on point e is :

\vec{F}=-|e|\vec{E_P}

F= \frac{-eKQx}{(r^2+x^2)^{3/2}}= \frac{-eKQx}{r^3[1+\frac{x^2}{r^2}]^{3/2}}

If r >> x , then    $\frac{x^2}{r^2} \approx 0$

Then,  $\frac{-eKQ}{r^3}x$

$ma =\frac{-eKQ}{r^3}x$

$a =\frac{-eKQ}{mr^3}x$

Compare, a = -ω²x

We get,

$\omega^2 = \frac{eKQ}{R^3m}$

$\omega = \sqrt{\frac{eKQ}{r^3m}}$

$2 \pi f = \sqrt{\frac{eKQ}{r^3m}}$

$f = \frac{1}{2 \pi}\sqrt{\frac{eKQ}{mr^3}}$

6 0
3 years ago
The final stages of a star's life depend on its mas. true or false?​
Diano4ka-milaya [45]

Answer:

TRUE

Explanation:

Low mass stars last lots longer.

5 0
3 years ago
You are sitting on a merry-go-round at a distance of 2m from its center. It spins 15 times in 3 min. What distance do you move a
soldier1979 [14.2K]

Answer:

A) 12.57 m

B) 5 RPM

C) 3.142 m/s

Explanation:

A) Distance covered in 1 Revolution:

The formula that gives the relationship between the arc length or distance covered during circular motion to the angle subtended or the revolutions, is given as follows:

s = rθ

where,

s = distance covered = ?

r = radius of circle = 2 m

θ = Angle = 2π radians  (For 1 complete Revolution)

Therefore,

s = (2 m)(2π radians)

<u>s = 12.57 m</u>

B) Angular Speed:

The formula for angular speed is given as:

ω = θ/t

where,

ω = angular speed = ?

θ = angular distance covered = 15 revolutions

t = time taken = 3 min

Therefore,

ω = 15 rev/3 min

<u>ω = 5 RPM</u>

C) Linear Speed:

The formula that gives the the linear speed of an object moving in a circular path is given as:

v = rω

where,

v = linear speed = ?

r = radius = 2 m

ω = Angular Speed in rad/s = (15 rev/min)(2π rad/1 rev)(1 min/60 s) = 1.571 rad/s

Therefore,

v = (2 m)(1.571 rad/s)

<u>v = 3.142 m/s</u>

8 0
3 years ago
A force of 385 N is applied in pushing a stalled automobile at a constant speed for a distance of 150 m. How much work (in J) wa
Helga [31]

Answer:57,750J

Explanation:

3 0
2 years ago
China's GDP passed one trillion USD between Between 2005 and 2010, China's GDP increased by approximately
maksim [4K]

Answer:

1995 and 2000 , 4 trillions

Explanation:

3 0
2 years ago
Read 2 more answers
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