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3 years ago

Pleaseee someone help me TEST TOMORROW

Physics

2 answers:

Liono4ka [1.6K]3 years ago

4 0

3.A

4.c

5.A
hope this helps

evablogger [386]3 years ago

4 0

3. D
4. A
5. I think its B?
I hope you do good on your test tomorrow!

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a 65 kg skater at rest on a frictionless rink throws a 2 kg ball, giving the ball a velocity of 7 m/s. What is the velocity of t

gayaneshka [121]

The answer to your question is 33

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3 years ago

Oil having a density of 926 kg/m3 floats on water. A rectangular block of wood 3.69 cm high and with a density of 974 kg/m3 floa

zloy xaker [14]

Answer:

the position of the wood below the interface of the two liquids is 2.39 cm.

Explanation:

Given;

density of oil, $\rho _o$ = 926 kg/m³

density of the wood, $\rho _{wood}$ = 974 kg/m³

density of water, $\rho _w$ = 1000 kg/m³

height of the wood, h = 3.69 cm

Based on the density of the wood, it will position across the two liquids.

let the position of the wood below the interface of the two liquids = x

Let the wood be in equilibrium position;

$F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood} -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood} -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm$

Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.

6 0

3 years ago

Block B has mass 6.00 kg and sits at rest on a horizontal, frictionless surface. Block A has mass 2.50 kg and sits at rest on to

zzz [600]

Answer:

Explanation:

Block A sits on block B and force is applied on block A . Block A will experience two forces 1) force P and 2 ) friction force in opposite direction of motion . Block B will experience one force that is force of friction in the direction of motion .

Let force on block A be P . friction force on it will be equal to kinetic friction, that is μ mg , where μ is coefficient of friction and m is mass of block A

friction force = .4 x 2.5 x 9.8

= 9.8 N

net force on block A = P - 9.8

acceleration = ( P - 9.8 ) / 2.5

force on block B = 9.8

acceleration = force / mass

= 9.8 / 6

for common acceleration

( P - 9.8 ) / 2.5 = 9.8 / 6

( P - 9.8 ) / 2.5 = 1.63333

P = 13.88 N .

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4 years ago

How might what is happening in this image affect the nervous system?

Yuri [45]

There is no image!?...was there meant to be something attached?

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3 years ago

The distance, x, covered by a particle in time, t, is given as x=a +bc+ct^2 +dt^3

Sav [38]

Answer:

$a$ has units of distance

$b$ has units of distance over time

$c$ has units of distance over $time^2$

$d$ has units of distance over $time^3$

Explanation:

Since the expression for the distance is:

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then:

$a$ has units of distance

$b$ has units of distance over time

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$d$ has units of distance over $time^3$

because we are supposed to be able to add all of the terms and get a distance. So the products on each term that contains factors of time (t) should be cancelling those time units with units in the denominator of the multiplicative constant s that accompany them.

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3 years ago