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Oliga [24]
2 years ago
15

If the average velocity during the athlete's walk back

Physics
1 answer:
goblinko [34]2 years ago
4 0

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

From the question we are told

If the average velocity during the athlete's walk back  to the starting line in Guided Example 2.5 is – 1.50 m/s,

Generally the equation Time spent  is mathematically given as

T=\frac{d}{v}

Therefore

T=\frac{d}{1.50}

Hence ,From the Guide there are other parameters which with this equation will give the exact time the athlete's walk back

T=\frac{d}{1.50}

For more information on this visit

brainly.com/question/22271063

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Student A walks up a 10 meter staircase. Student B runs up the same staircase in less
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Answer: student b ran up the stairs faster

Explanation:if they were going up the same stair case and student b got up there faster that means he was running therefore student b was using more power

6 0
3 years ago
HELP ME ASAP PLEASE
padilas [110]

Answer:

( 1000 × 4 = 4,000) (800×3= 2400) (800×2=1600) the answer is 1600 hope it helps

6 0
3 years ago
the fireman wishes to direct the flow of water from his hose to the fire at b. determine two possible angles u1 and u2 at which
Lelechka [254]

The two possible angles obtained by using the qudratic equation are;

θ_{1} = 15.10° and θ2 = 73.51°

Given, speed of water = v_{A} = 50ft/s

For the motion along x direction, time period can be calculated as follows:

s_{x} = (v_{A}) _x_{} } t

35 = (50 × cosθ) t

t = 0.64 / cosθ

For the motion in y direction, an equation can be obtained as follows:

s_{y} = (v_{A})_{y}  t +\frac{1}{2} (a_{y} )t^{2}

s_{y} = (-v_{A}sinθ) }  t +\frac{1}{2} (a_{y} )t^{2}

Plugging in the values we get:

-20 = (-50_sinθ) }  t +\frac{1}{2} (-32.2} )t^{2}

-20 = -32tanθ - 10.304sec^{2}θ

Upon solving the above quadratic equation, we get,

tanθ = 0.27 , -3.38

Therefore,

tanθ_{1} = 0.27

θ_{1} = 15.10°

and, tanθ_{2} = -3.38

θ_{2} = 73.51

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8 0
11 months ago
. For transverse waves, compare the direction that the particles move to
Elza [17]

Answer:

The particles in transverse waves move perpendicular to the direction of how the wave moves.

Explanation:

5 0
3 years ago
A plate of uniform areal density is bounded by the four curves: where and are in meters. Point has coordinates and . What is the
Natali5045456 [20]

The question is incomplete. The complete question is :

A plate of uniform areal density $\rho = 2 \ kg/m^2$ is bounded by the four curves:

$y = -x^2+4x-5m$

$y = x^2+4x+6m$

$x=1 \ m$

$x=2 \ m$

where x and y are in meters. Point $P$ has coordinates $P_x=1 \ m$ and $P_y=-2 \ m$. What is the moment of inertia $I_P$ of the plate about the point $P$ ?

Solution :

Given :

$y = -x^2+4x-5$

$y = x^2+4x+6$

$x=1 $

$x=2 $

and $\rho = 2 \ kg/m^2$ , $P_x=1 \ $ , $P_y=-2 \ $.

So,

$dI = dmr^2$

$dI = \rho \ dA  \ r^2$  ,           $r=\sqrt{(x-1)^2+(y+2)^2}$

$dI = (\rho)((x-1)^2+(y+2)^2)dx \ dy$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}((x-1)^2+(y+2)^2) dy \ dx$

$I= 2 \int_1^2 \int_{-x^2+4x-5}^{x^2+4x+6}(x-1)^2+(y+2)^2 \  dy \ dx$

$I=2 \int_1^2 \left( \left[ (x-1)^2y+\frac{(y+2)^3}{3}\right]_{-x^2+4x-5}^{x^2+4x+6}\right) \ dx$

$I=2 \int_1^2 (x-1)^2 (2x^2+11)+\frac{1}{3}\left((x^2+4x+6+2)^3-(-x^2+4x-5+2)^3 \ dx$

$I=\frac{32027}{21} \times 2$

  $= 3050.19 \ kg \ m^2$

So the moment of inertia is  $3050.19 \ kg \ m^2$.

4 0
3 years ago
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