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SVEN [57.7K]
3 years ago
13

John accidentally drops his keys off the balcony at his apartment. John's friend Tony just happens to walk by at that moment and

picks up the keys to throw them back up to John. If John is 5.33 meters above the sidewalk where Tony is, and Tony throws at 18 m/s at a 40 degree angle above the horizontal, will the keys make it back to John?
Physics
1 answer:
dalvyx [7]3 years ago
8 0

Answer:

the keys will make it back to John.

Explanation:

In order to find out if the keys will reach John or not, we can use the formula of projectile motion to find the maximum height reached by the keys:

H = V²Sin²θ/2g

where,

V = Launch Speed = 18 m/s

θ = Launch Angle = 40°

g = 9.8 m/s²

Therefore,

H = (18 m/s)²[Sin 40°]²/(2)(9.8 m/s²)

H = 6.83 m

Hence, the maximum height that can be reached by the projectile or the keys is greater than the height of John's Balcony(5.33 m).

<u>Therefore, the keys will make it back to John.</u>

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if you run around a circle at 4.5 m/s and the circle has a radius of 7.7 m, what is your centripetal acceleration?
madreJ [45]

Answer:

Centripetal acceleration,

a_{c} =2.63\ m/s^{2} }

Explanation:

Centripetal acceleration:

Centripetal acceleration is the idea that any object moving in a circle, in something called circular motion, will have an acceleration vector pointed towards the center of that circle.

Centripetal means towards the center.

Examples of centripetal acceleration (acceleration pointing towards the center of rotation) include such situations as cars moving on the cicular part of the road.

An acceleration is a change in velocity.

Formula for Centripetal acceleration:

a_{c} =\frac{(velocity)^{2} }{radius}

Given here,

Velocity = 4.5 m/s

radius = 7.7 m

To Find :

a_{c} = ?

Solution:

We have,

a_{c} =\frac{(velocity)^{2} }{radius}

Substituting  given value in it we get

a_{c} =\frac{(4.5)^{2}}{7.7} \\\\a_{c} =\frac{20.25}{7.7}\\\\a_{c} =2.629\ m/s^{2} \\\\\therefore a_{c} =2.63\ m/s^{2

Centripetal acceleration,

a_{c} =2.63\ m/s^{2} }

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A tennis rackets delivers 500 N on impact with a tennis ball. What is the force of the tennis ball on the racket?
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What is the acceleration of a proton moving with a speed of 7.7 m/s at right angles to a magnetic field of 1.9 T
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Answer:

The acceleration of the proton is 1.403 x 10⁹ m/s²

Explanation:

Given;

speed of proton, v = 7.7 m/s

magnitude of magnetic field, B = 1.9 T

Magnetic force of moving proton is given by;

F = qvBsinθ

Centripetal force on the moving proton is given by;

F = m(\frac{v^2}{r})\\\\F = m(a_c) \\\\a_c \ is \ the \ centripetal \ acceleration

qvBsin\theta = ma_c\\\\ac = \frac{qvBsin(90)}{m}

where;

q is charge of the proton = 1.602 x 10⁻¹⁹ C

m is mass of proton = 1.67 x 10⁻²⁷ kg

ac = \frac{(1.602*10^{-19})(7.7)(1.9)sin(90)}{1.67*10^{-27}}\\\\a_c = 1.403*10^{9} \ m/s^2

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