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3 years ago

"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?"

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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3 years ago

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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:

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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