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3 years ago

13

During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel? Use the formula =1/2 * t2 to so

lve your answer.

2 answers:

murzikaleks [220]3 years ago

6 0

D = 70,560 meters this is the answer need more characters there we go:)

Ulleksa [173]3 years ago

3 0

8/9 score.....................

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If an object absorbs all colors but red, we see

julia-pushkina [17]

[I researched for you, since I am not in that particular level to know that knowledge yet. I assure this is accurate info :)]

The answer is A, red.
"Remember, the color you see is light REFLECTING off the surface of that object. If all colors are absorbed in to the surface EXCEPT red, red must be reflected, and you'll see red." - Yahoo User @Chap

8 0

3 years ago

A parked car's horn (belonging to a musician) emits a concert A of frequency 440 on a day when the speed of sound is 342 m/s. Yo

julsineya [31]

Answer:

19.08 m/s

Explanation:

f = actual frequency emitted by the parked car's horn = 440 Hz

V = speed of sound = 342 m/s

f' = frequency of the horn observed by you = 466 Hz

v = speed of your car moving towards the parked car = ?

frequency of the horn observed by you is given as

$f' = \frac{Vf}{V - v}$

$466 = \frac{(342)(440)}{342 - v}$

v = 19.08 m/s

3 0

3 years ago

Two workers are sliding 300 kg crate across the floor. One worker pushes forward on the crate with a force of 400 N while the ot

almond37 [142]

Answer:

The kinetic coefficient of friction of the crate is 0.235.

Explanation:

As a first step, we need to construct a free body diagram for the crate, which is included below as attachment. Let supposed that forces exerted on the crate by both workers are in the positive direction. According to the Newton's First Law, a body is unable to change its state of motion when it is at rest or moves uniformly (at constant velocity). In consequence, magnitud of friction force must be equal to the sum of the two external forces. The equations of equilibrium of the crate are:

$\Sigma F_{x} = P+T-\mu_{k}\cdot N = 0$ (Ec. 1)

$\Sigma F_{y} = N - W = 0$ (Ec. 2)

Where:

$P$ - Pushing force, measured in newtons.

$T$ - Tension, measured in newtons.

$\mu_{k}$ - Coefficient of kinetic friction, dimensionless.

$N$ - Normal force, measured in newtons.

$W$ - Weight of the crate, measured in newtons.

The system of equations is now reduced by algebraic means:

$P+T -\mu_{k}\cdot W = 0$

And we finally clear the coefficient of kinetic friction and apply the definition of weight:

$\mu_{k} =\frac{P+T}{m\cdot g}$

If we know that $P = 400\,N$ , $T = 290\,N$ , $m = 300\,kg$ and $g = 9.807\,\frac{m}{s^{2}}$ , then:

$\mu_{k} = \frac{400\,N+290\,N}{(300\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}$

$\mu_{k} = 0.235$

The kinetic coefficient of friction of the crate is 0.235.

5 0

3 years ago

A boy pulls a 28.0-kg box with a 230-N force at 35° above a horizontal surface. If the coefficient of kinetic friction between t

ddd [48]

Answer:

1977.696 J

Explanation:

Given;

Weight of the box = 28.0 kg

Force applied by the boy = 230 N

angle between the horizontal and the force = 35°

Therefore,

the horizontal component of the force = 230 × cosθ

= 230 × cos 35°

= 188.405 N

Coefficient of kinetic friction, μ = 0.24

Force by friction, f = μN

here,

N = Normal force = Mass × acceleration due to gravity

or

N = 28 × 9.81 = 274.68 N

therefore,

f = 0.24 × 274.68

or

f = 65.9232 N

Now,

work done by the boy, W₁ = 188.405 N × Displacement

= 188.405 N × 30

= 5652.15 J

and,

the

work done by the friction, W₂ = - 65.9232 N × Displacement

= - 65.9232 N × 30 m

= - 1977.696 J

[ since the friction force acts opposite to the direction of motion, therefore the workdone will be negative]

8 0

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