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Tamiku [17]
3 years ago
14

If a 90 kg hockey player slows from 18 m/s to 12 m/s over 0.6 s, what is the force exerted by his skates on the ice? A. -540 N B

. -900 kg-m/s2 C. -1,080 kg-m/s D. -1,800 N
Physics
1 answer:
Zielflug [23.3K]3 years ago
5 0

Answer:

B. -900 kg-m/s2

Explanation:

First of all, we need to calculate the acceleration of the hockey player, which is given by:

a=\frac{v-u}{t}

where

v = 12 m/s is the final speed

u = 18 m/s is the initial speed

t = 0.6 s is the time

Substituting into the formula, we find

a=\frac{12 m/s-18 m/s}{0.6 s}=-10 m/s^2

And now we can calculate the force exerted by using Newton's second law:

F=ma

where m=90 kg is the mass of the hockey player. Substituting into the formula, we find

F=(90 kg)(-10 m/s^2)=-900 N

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Answer:

F = 4.3671 * 10^{-8}\ Newtons

Explanation:

The gravitational force between two corpses is given by the following equation:

F = GMm/d^2

Where F is the force, G is the gravitational constant

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So we have that:

F = 6.67408*10^{-11} * 7.1*10^4 * 73/89^2

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5 0
3 years ago
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aleksklad [387]

Answer:

a)  y = 0.98 t², t=1s y= 0.98 m,  

b) he two blocks must move the same distance

c) v = 1.96 m / s,  d)  a = -1.96 m / s², e)  x = 0.98 m

Explanation:

For this exercise we can use Newton's second law

Big Block

Y axis

             N-W = 0

             N = M g

X axis

             T- fr = Ma

the friction force has the expression

             fr = μ N

             fr = μ Mg

small block

             w- T = m a

             

we write the system of equations

             T - fr = M a

             mg - T = m a

we add and resolved

             mg-  μ Mg = (M + m) a

             a = g \ \frac{m - \mu M}{m+M}

             a = 9.8 \ \frac{10- 0.2 \ 20}{ 10 \ +\ 20}

             a = 9.8 (6/30)

             a = 1.96 m / s²

a) now we can use the kinematic relations

             y = v₀ t + ½ a t²

the blocks come out of rest so their initial velocity is zero

             y = ½ a t²

             y = ½ 1.96 t²

             y = 0.98 t²

for t = 1s y = 0.98 m

       t = 2s y = 1.96 m

b) Time is a scale that is the same for the entire system, the question should be oriented to how far the big block will move.

As the curda is in tension the two blocks must move the same distance

c) the velocity of the block M

           v = vo + a t

           v = 0 + 1.96 t

for t = 1 s v = 1.96 m / s

       t = 2 s v = 3.92 m / s

d) the deceleration if the chain is cut

when removing the chain the tension becomes zero

           -fr = M a

          - μ M g = M a

          a = - μ g

          a = - 0.2 9.8

          a = -1.96 m / s²

e) the distance to stop the block is

         v² = vo² - 2 a x

        0 = vo² - 2a x

        x = vo² / 2a

        x = 1.96² / 2 1.96

        x = 0.98 m

the time to travel this distance is

        v = vo - a t

        t = vo / a

        t = 1.96 /1.96

        t = 1 s

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From these examples, you can see a few things:

-- There's no such thing as "a balanced force" or "an unbalanced force".
It's a <em><u>group</u> of forces</em> that is either balanced or unbalanced.

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