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alexira [117]
3 years ago
9

A student shoots a spitball with a perfectly horizontal velocity of 9.7 m/s from a height of 1.8 meters. How long will it take f

or the spitball to hit the ground?
(ignore air resistance) (include units and correct number of significant figures)
Physics
1 answer:
sergejj [24]3 years ago
3 0

Answer:A student shoots a spitball with a perfectly horizontal velocity of 9.7 m/s from a height of 1.8 meters. How long will it take for the spitball to hit the ground?

(ignore air resistance) (include units and correct number of significant figures)

Explanation:La respuesta es porque esa es la respuesta, la respuesta al número es 9.7 1.8 Divide =53.888

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A person wants to determine the spring constant of an exercise stretch cord. He pulls the cord with a force probe that exerts a
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Answer:

350 N/m

Explanation:

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The spring constant is 350 N/m

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What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 22 km/h and the coef
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First, let's put 22 km/h in m/s:

22 \frac{km}{h} \times  \frac{1000m}{1km}  \times  \frac{1h}{3600s}=6.11 \frac{m}{s}

Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by

F_{rad}=m \frac{v^2}{R}

The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:

F_{fric}= mg \mu_{s}

Notice we don't use the kinetic coefficient even though the bike is moving.  This is because when the tires meet the road they are momentarily stationary with the road surface.  Otherwise the bike is skidding.

Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve:

m\frac{v^2}{R} = mg \mu_{s} \\ \\ \frac{v^2}{R} = g \mu_{s} \\ \\R= \frac{v^2}{g \mu_{s}} \\ \\ R= \frac{6.11}{9.8 \times 0.37}=1.685m

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In the two-slit experiment, monochromatic light of frequency 5.00 × 1014 Hz passes through a pair of slits separated by 2.20 × 1
asambeis [7]

Explanation:

It is given that,

Frequency of monochromatic light, f=5\times 10^{14}\ Hz

Separation between slits, d=2.2\times 10^{-5}\ m

(a) The condition for maxima is given by :

d\ sin\theta=n\lambda

For third maxima,

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{n\lambda}{d})

\theta=sin^{-1}(\dfrac{nc}{fd})  

\theta=sin^{-1}(\dfrac{3\times 3\times 10^8\ m/s}{5\times 10^{14}\ Hz\times 2.2\times 10^{-5}\ m})  

\theta=4.69^{\circ}

(b) For second dark fringe, n = 2

d\ sin\theta=(n+1/2)\lambda

\theta=sin^{-1}(\dfrac{5\lambda}{2d})

\theta=sin^{-1}(\dfrac{5c}{2df})

\theta=sin^{-1}(\dfrac{5\times 3\times 10^8}{2\times 2.2\times 10^{-5}\times 5\times 10^{14}})

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Hence, this is the required solution.

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