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zlopas [31]
3 years ago
9

"If an object is thrown up at an initial speed of 9.8 m/sec, what would be the final speed upon impact with the ground?"

Physics
1 answer:
swat323 years ago
5 0
Https://www.jiskha.com/search/index.cgi?query=If+an+object+is+thrown+up+at+an+initial+velocity+of+9.... this website gives you all the infromation to this question i cant go against policy so this will help
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Beryllium has a nucleus composed of protons and neutrons. Given the data, how many neutrons are in a typical Beryllium nucleus?
son4ous [18]
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GPS consists of three segments what are they?
WARRIOR [948]

Answer:

GPS consist of the space segment, the control segment, and the user segments

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Earth rotates on is axis approximately once every 24 hours, causing the day to night cycle and?
katovenus [111]
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6 0
3 years ago
A 9.0-V battery is connected to a resistor so that there is a 0.50-A current through the resistor.For how long should the batter
salantis [7]

Answer:

35.6 s

Explanation:

The power through the resistor is given by:

P=VI

where V=9.0 V is the voltage and I=0.50 A is the current. Substituting into the formula, we find

P=(9.0 V)(0.5 A)=4.5 W

The power is also equal to:

P=\frac{W}{t}

where W is the work done while t is the time taken. Since we know the work done, W=160 J, we can re-arrange the equation to find the time taken:

t=\frac{W}{P}=\frac{160 J}{4.5 W}=35.6 s

7 0
4 years ago
To take off from the ground, an airplane must reach a sufficiently high speed. The velocity required for the takeoff, the takeof
BlackZzzverrR [31]
<h2>Answer: 26,8 s</h2>

Explanation:

If we are talking about an acceleration at a constant rate , we are dealing with constant acceleration, hence we can use the following equations:

{V_{f}}^{2}={V_{o}}^{2}+2ad (1)

V_{f}=V_{o}+at (2)

Where:

V_{f} is the final velocity of the plane (the takeoff velocity in this case)

V_{o}=0 is the initial velocity of the plane (we know it is zero because it starts from rest)

a=5m/s^{2} is the constant acceleration of the plane to reach the takeoff velocity

d=1800m is the distance of the runway

t is the time

Knowing this, let's begin with (1):

{V_{f}}^{2}=0+2(5m/s^{2})(1800m) (3)

{V_{f}}^{2}=18000m^{2}/s^{2} (4)

V_{f}=134.164 m/s (5)

Substituting (5) in (2):

134.164 m/s=0+(5m/s^{2})t (6)

Finding t:

t=26.8 s This is the time needed to take off

6 0
3 years ago
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