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2 years ago

5)A 0.50 kg hockey puck is at rest on ice when you hit it with a hockey stick, applying a force of 100 N for

Physics

1 answer:

mojhsa [17]2 years ago

7 0

Answer:

F t = m Δv impulse delivered = change in momentum

Δv = 100 * .1 / .5 = 20 m/s original speed of puck

KE = 1/2 m v^2 = .5 * 20^2 / 2 = 100 J initial KE of puck

E = μ m g d energy lost by puck

Ff = μ m g = m a deceleration of puck due to friction

a = μ g = 9.8 * .2 = 1.96 m/s^2

v2 = a t + v1 = -1.96 * 4 + 20 = 12.2 m/s speed of puck on striking box

m v2 = M V conservation of momentum when puck strikes box

V = m v2 / M = 12.2 * .5 / .8 = 7.63 m/s speed of box after collision

KE = 1/2 M V^2 = .8 * 7.63^2 / 2 = 23.3 J KE of box after collision

KE = μ M g d energy lost by box in sliding distance d

d = 23.3 / (.3 * .8 * 9.8) = 9.91 m distance box slides

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3 years ago

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3 years ago

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If you drive through water, your brakes may become slippery and ineffective. To dry the brakes off, __________.

GenaCL600 [577]

Answer:

Explanation:

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4 years ago

A camera phone called the iPhone X has an image sensor size of 5 mm and a focal length of 4 mm. How long of a selfie-stick must

azamat

Answer:

length of selfie-stick is 1.62 m

Explanation:

Given data

image size h1 = 5 mm = 5 × $10^{-3}$ m

focal length = 4 mm = 4 × $10^{-3}$ m

distance h2 = 2.032 m

to find out

How long of a selfie-stick

solution

here we find first magnification

that is M = h1 /h2

M = 5 × $10^{-3}$ / 2.032

M = 2.46 × $10^{-3}$

and we know M = p/q

so p = Mq = 2.46 × $10^{-3}$ q

so we apply lens formula

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3 years ago

Pam has a mass of 42.4 kg and she is at rest on smooth, level, frictionless ice. Pam straps on a rocket pack. The rocket supplie

fredd [130]

Answer:

F = 2349.6 N

Explanation:

We can solve this exercise using the relationship of momentum and momentum

I = Δp

I = F t

As the woman accelerates at a distance of 29.1 m to go from rest to 56.8 m / s, we can use the kinematics to find the acceleration

v² = v₀² + 2 a x

v₀ = 0

a = v / 2x

a = 56.8 2/2 29.1

a = 55.43 m / s²

Let's look for the time you need to get this speed

v = v₀ + a t

t = v / a

t = 56.8 / 55.43

t = 1,025 s

Let's clear the average force momentum from the momentum

F t = m v- m v₀

F = mv / t

F = 42.4 56.8 / 1.025

F = 2349.6 N

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3 years ago