Which best describes the speed of the ball as it is thrown straight up into the air and comes back down

1 answer:

4 0

Assuming that up is the positive direction, the velocity of the ball goes from it's initial and approaches zero, which it reaches at the apex of it's height. After peaking, the velocity is now reaching the same numbers it had on the way up, only with a negative value this time. If it had a velocity of 5m/s upon leaving the hand, it will have a velocity of -5m/s when it came back to the hand.

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A sound wave traveling at 343 m/s is emitted by the foghorn of a tugboat. an echo is heard 2.80 s later. how far away is the ref

Nostrana [21]

The echo is heard 2.80 s later, this means this is the time the sound takes to travel to the reflecting object and then back to us. So, during this time, the sound wave has covered the distance L between us and the object twice:
$S=2L$
The speed of the sound wave is: $v=343 m/s$ , and since it is moving by uniform motion, we can find the distance covered by the wave using
$S=vt=(343 m/s)(2.80 s)=960 m$
And we said this corresponds to twice the distance between us and the reflecting object, so:
$L= \frac{S}{2} = \frac{960 m}{2} =480 m$
so, the object is 480 meters away.

3 0

3 years ago

A satellite orbits earth with a mean altitude of 361 km. If the orbit is circular, what are the satellite's time period and spee

Advocard [28]

Answer:

v = 7.69 x 10³ m/s = 7690 m/s

T = 5500 s = 91.67 min = 1.53 h

Explanation:

In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:

$F_{gravitation}= F_{centripetal}\\\\\frac{GM_{s} M_{E}}{r^2} = \frac{M_{s} v^2}{r}\\\\\frac{GM_{E}}{r} = v^2\\\\v = \sqrt{\frac{GM_{E}}{r} } \\\\$

where,

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Me = Mass of Earth = 5.97 x 10²⁴ kg

r = distance between the center of Earth and Satellite = Radius of Earth + Altitude = 6.371 x 10⁶ m + 0.361 x 10⁶ m = 6.732 x 10⁶ m

v = orbital speed = ?

Therefore,

$v = \sqrt{\frac{(6.67 x 10^{-11}N.m^2/kg^2)(5.97 x 10^{24} kg)}{6.732 x 10^6 m} }\\\\$

For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.

So, its orbital speed can be given as:

$v = \frac{Circumference of Circle at Given Altitude}{T}\\\\v = \frac{2\pi r}{T}\\\\$

where,

T = Time Period of Satellite = ?

Therefore,

$T = \frac{2\pi r}{v}\\\\T = \frac{(2\pi )(6.732 x 10^6 m}{7.69 x 10^3 m/s}\\\\$

7 0

3 years ago

A wrench 0.500 m long is applied to a nut with a force of 80.0 N. Because of the cramped space, the force must be exerted upward

riadik2000 [5.3K]

Answer:

Torque, $\tau=34.6\ N.m$

Explanation:

It is given that,

Length of the wrench, l = 0.5 m

Force acting on the wrench, F = 80 N

The force is acting upward at an angle of 60.0° with respect to a line from the bolt through the end of the wrench. We need to find the torque is applied to the nut. We know that torque acting on an object is equal to the cross product of force and distance. It is given by :

$\tau=Fr\ sin\theta$