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pentagon [3]
2 years ago
9

A grasshopper jumps at a 65.0 degree angle at 5.42m/s. At what time does it reach its maximum height?

Physics
2 answers:
dimulka [17.4K]2 years ago
8 0

Answer:

A grasshopper reached the maximum height 1.23 m.

Explanation:

Given that,

Velocity = 5.42 m/s

Angle = 65.0°

We need to calculate the maximum height

Using formula of maximum height

y_{max}=\dfrac{v^2\sin^2\theta}{2g}

Where, v = 5.42 m/s

g = acceleration due to gravity

\theta  =Angle

Put the value into the formula

y_{max}=\dfrac{5.42^2\times\sin^2(65.0)}{2\times9.8}

y_{max}=1.23\ m

Hence, A grasshopper reached the maximum height 1.23 m.

crimeas [40]2 years ago
6 0
When the grasshoppers vertical velocity is exactly zero.
v = -g•t + v0.
v: vertical part of velocity. Is zero at maximum height.
g: 9.81
t: time you are looking for
v0: initial vertical velocity
Find the vertical part of the initial velocity, by using the angle at which the grasshopper jumps.
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ElenaW [278]

Answer:

Speed of the wave is 7.87 m/s.

Explanation:

It is given that, tapping the surface of a pan of water generates 17.5 waves per second.

We know that the number of waves per second is called the frequency of a wave.

So, f = 17.5 Hz

Wavelength of each wave, \lambda=45\ cm=0.45\ m

Speed of the wave is given by :

v=f\lambda

v=17.5\times 0.45

v = 7.87 m/s

So, the speed of the wave is 7.87 m/s. Hence, this is the required solution.

5 0
2 years ago
a 1.50*10^-5 C charge feels a 2.89*10^-3 N force when it moves 288m/s perpendicular (90) deg to a magnetic field. how strong is
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6.68, -1
Explanation: correct for acellus
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2 years ago
The first law of thermodymaic stae the <br> 5ktiopheuithkjhuiguihaoitg
pentagon [3]

I'm assuming you want the first law of thermodynamics.

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3 0
2 years ago
A circular pipe of 25-mm outside diameter is placed in an airstream at 25C and 1-atm pressure. The air moves in cross flow over
kifflom [539]

Answer:

f_D = =3.24 N/m

Explanation:

data given

properties of air\nu\ of air =19.31*10^{-6} m2/s

\rho = 1.048 kg/m3

k = 0.0288 W/m.K

WE KNOW THAT

Reynold's number is given as

Re =\frac{VD}{\nu}

      = \frac{ 15*0.025}{19.31*10^{-6}}

      = 1.941 *10^4

drag coffecient is given as

C_D = \frac{f_D}{A_f\frac{\rho v^2}{2}}

solving for f_D

f_D = C_D A_f*\frac{\rho v^2}{2}

     =C_D D*\frac{\rho v^2}{2}

Drag coffecient for smooth circular cylinder is 1.1

therefore Drag force is

f_D = 1.1*0.025 *\frac{1.048*15^2}{2}

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4 0
2 years ago
During a race the dirt bike was observed to leap up off the small hill at A at an angle of 60^o with the horizontal. If the poin
const2013 [10]

Answer:

Velocity is equal to 27.3 feet per second  and time is equal to 1.4668 seconds

Explanation:

Given

The horizontal distance traveled by  dirt bike before landing = 20 feet

Angle of flight = 60 degree

As we know that Horizontal distance (H) is equal to

= H_0 + V_0 * t\\

Where H_0 is the initial horizontal distance

V_0 is the velocity with which the bike is travelling in horizontal direction

and t is the time in seconds

Substituting the given values, we get -

H = H_0 + v*t\\20 = 0 + v * cos \theta * t\\20 = v * cos 60 * t\\t = \frac{20}{v * cos 60} \\t = \frac{40}{v}

Now distance traveled in vertical direction is equal to

Y = Y_0 + v_0 * t + \frac{1}{2} a * t^2

here acceleration will be equal to acceleration due to gravity which is equal to - 32.2 \frac{ft}{s^2}. It is negative as its is acting in upward direction

Thus,

Y = 0 + v * sin 60 + \frac{1}{2}  * (-32.2) * (\frac{40}{v} )^2\\0 = 0  + 0.866 v + \frac{-25760}{v^2} \\0.866 v = \frac{-25760}{v^2}\\v^3 = \frac{-25760}{0.866} \\v = 27.3

Velocity is equal to 27.3 feet per second  and time is equal to 1.4668 seconds

8 0
2 years ago
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