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rusak2 [61]
3 years ago
14

In an action movie, a stuntman leaps from the top of one building to the top of another building 5.2 m away. After a running sta

rt, he leaps at a velocity of 6.0 m/s at an angle of 15° above horizontal. Will he make it to the other roof, which is 2.9 m shorter than the building he jumps from?
Physics
1 answer:
Ksenya-84 [330]3 years ago
8 0
<span>stuntman's vertical velocity is vy= 6 sin 15=1.55m/s. Height that he goes up is, vertical kinetic energy = mgh 1/2 v² = gh, h=v²/(2g)=0.12 m. time wanted to go up= vy/9.8=0.16s, Time to fall through a height 0.12+2.9=3.02m is t=sqrt(3.02*2/9.8)=0.78 s Total time needed to go up and down is 0.78+0.16=0.94 s. to calculate the horizontal range, Horizontal velocity = 6 cos 15=5.8 m/s. Distance which he can cover is 5.8*0.94=5.44 m. If the distance between the two building is less than 5.44 m then he will be safe and he can jump that distance.</span>
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Two particles with charges of 5.00 μ C and -3.00 μC are placed 0.250 m apart. Where can a third charge be placed so that the net
MAXImum [283]

Answer:

0.86 m

Explanation:

q₁ = magnitude of positive charge = 5 x 10⁻⁶ C

q₂ = magnitude of negative charge = 3 x 10⁻⁶ C

r = distance between the two charges = 0.250 m

d = distance of the location of third charge from negative charge

q = magnitude of charge on third charge

Using equilibrium of electric force on third charge

\frac{kq_{2}q}{d^{2}} = \frac{kq_{1}q}{(r+d)^{2}}

\frac{q_{2}}{d^{2}} = \frac{q_{1}}{(r+d)^{2}}

\frac{(5\times 10^{-6})}{(0.250+d)^{2}} = \frac{(3\times 10^{-6})}{(d^{2}}

d = 0.86 m

4 0
3 years ago
Infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on Earth
liberstina [14]

Answer:

λ = 1.86 x 10⁻⁴ m = 186 μm

Explanation:

The relationship between the wavelength and the frequency of a wave is given by the following equation:

c = f\lambda\\\\\lambda = \frac{c}{f}

where,

λ = wavelength of infrared radiation = ?

c = speed of infrared radiation = speed of light = 3 x 10⁸ m/s

f = frequency of infrared radiation = 1.61 THz = 1.61 x 10¹² Hz

Therefore,

\lambda = \frac{3\ x\ 10^8\ m/s}{1.61\ x\ 10^{12}\ Hz}

<u>λ = 1.86 x 10⁻⁴ m = 186 μm</u>

7 0
3 years ago
A car’s velocity as a function of time is given by Vx (t) = α.t + β.t 2 , where α= 3m/s and β= 0.1m/s 3 . Calculate the average
s344n2d4d5 [400]

The definition of average acceleration allows to find the result for the average acceleration in the given time interval is:

          a_{average}= 1.5  \ \frac{m}{s^2}

Instantaneous acceleration is defined as the derivative of velocity with respect to time.

           a =   \frac{dv}{dt}

Where a is the acceleration, v the velocity and t the time.

They indicate that the speed of the car is given by the relation.

          v = α t + β t²

With α = 3 m / s and β = 0.1 m / s³

Let's make  the derivative.

           a = α + 2β t

Let's substitute

            a = 3 + 2 0.1 t

Average acceleration is the change in velocity in the time interval.  

          a_{average} = \frac{\Delta v}{\Delta t }

Let's find the velocity at the indicated time.

For t = 5 s

         v₅ = 3 + 0.1 5²

         v₅ = 5.5 m / s

For t = 10 s

          v₁₀ = 3 + 0.1 10²

          v₁₀ = 13 m / s

Let's calculate the average acceleration.

           a_{average} = \frac{13 - 5.5 }{ 10 - 5 }\\

           a_{average}= 1.5 \  m/s^2

In conclusion using the definition of mean acceleration we can find the result for the mean acceleration in the given time interval is:

           a_{average} =  1.5 m / s²

Learn more here: brainly.com/question/20057878

7 0
3 years ago
Two infinite nonconducting sheets of charge are parallel to each other, with sheet A in the x = -2.15 plane and sheet B in the x
GalinKa [24]

Answer:

a) (-367231.63i ,  367231.63i, 0) N/C

b) (0 , 0  , 367231.63i ) N/C

Explanation:

a)

Case x < -2.15

E =E_{+} + E_{+} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = - 367231.63 i

Case x > 2.15

E =E_{+} + E_{+} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = +367231.63 i

Case -2.15 < x <+2.15

E =E_{+} - E_{+} \\\\E = 0

b)

Case x < -2.15

E =E_{+} - E_{+} \\\\E = 0

Case x > 2.15

E =E_{+} - E_{+} \\\\E = 0

Case -2.15 < x <+2.15

E =E_{+} + E_{-} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = +367231.63 i

7 0
3 years ago
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