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3 years ago

What is the sprinters output at 2.0 s, 4.0 s and 6.0 s?

Physics

1 answer:

kari74 [83]3 years ago

5 0

A 59 kg sprinter, starting from rest, runs 47 m in 7.0 s at constant acceleration.?

What is the sprinter's power output at 2.0 s, 4.0 s, and 6.0 s?

Instantaneous Power is the force times velocity

P = Fv

Because the acceleration is constant, the force will be constant as well

F = ma

P = mav

for constant acceleration, the velocity at each time is found using

v = at

P = ma(at) = ma²t

find the acceleration using kinematic equation

s = ½at²

a = 2s/t²

a = 2(47) / 7.0²

a = 1.918 m/s²

P(2.0) = 59(1.918²)2.0 = 434.25 W = 0.43 kW

P(4.0) = 59(1.918²)4.0 = 868.51 W = 0.87 kW

P(6.0) = 59(1.918²)6.0 = 1302.76 W = 1.3 kW

I hope this helped.

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A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

Natasha2012 [34]

Answer:

a) $t=2s$

b) $h_{max}=1024ft$

c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$0=(64ft/s)^2-2(32ft/s^2)(y-960ft)$

$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

$a=-16\\b=64\\c=-64$

$t=2s$

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s$

Since the projectile is going down the velocity at the instant it reaches the ground is:

$v=-256ft/s$

5 0

3 years ago

Three resistors are wired in parallel with a battery. Two of the resistors have resistances of 38.7 Q/ and 89.5 Q. The current i

Lina20 [59]

Answer:

$214.9 \Omega$

Explanation:

The three resistors are connected in parallel: this means that the potential difference across each resistor is the same as the voltage of the battery. This can be calculated using the information about the $38.7 \Omega$ resistor: in fact, since we know its resistance and the current flowing through it (0.155 A), we can find the potential difference across this resistor, which is equal to the voltage of the battery:

$V=IR=(0.155 A)(38.7 \Omega)=6.0 V$

We also know the total current in the circuit, 0.250 A. This means that we can find the total resistance of the circuit, using Ohm's law:

$R_{eq}=\frac{V}{I}=\frac{6.0 V}{0.250 A}=24 \Omega$

So now we now the total resistance and the resistance of two of the 3 resistors; therefore, we can find the resistance of the 3rd resistor:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}\\\frac{1}{R_3}=\frac{1}{R_{eq}}-\frac{1}{R_1}-\frac{1}{R_2}=\frac{1}{24 \Omega}-\frac{1}{38.7\Omega}-\frac{1}{89.5\Omega}=0.00465 \Omega^{-1}\\R_3=\frac{1}{0.00465 \Omega^{-1}}=214.9 \Omega$

4 0

3 years ago

A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el

GuDViN [60]

Answer:

$7.2\cdot 10^{-19} J$

Explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

$\Delta U = q \Delta V$

where

q is the magnitude of the charge of the particle

$\Delta V$ is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

$q=3e=3\cdot 1.6\cdot 10^{-19} J=4.8\cdot 10^{-19}J$

- the potential difference is

$\Delta V=4.50 V$

So, the change in electrical potential energy is

$\Delta U=(1.6\cdot 10^{-19}C)(4.50 V)=7.2\cdot 10^{-19} J$

7 0

3 years ago

Which of the following would decrease the resistance in a wire?

padilas [110]

<u>Increase the thickness of the wire</u> would decrease the resistance in a wire

Explanation:

Thicker wires have a larger cross-section that increases the surface area with which electrons can flow unimpeded. The thicker the wire, therefore, the lower the resistance.

Thin wires have very high resistance the reason the thin tungsten in a bulb glows because it is heated from the high resistance of many electrons trying to pass through a very small cross-section.

6 0

3 years ago

ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself o

shtirl [24]

Q: ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself off in the opposite direction at 60 m/s. What is the magnitude of the average force the wall exerts on Ken during this rapid maneuver

Answer:

-6461.54 N

Explanation:

From Newton's Fundamental equation,

F = m(v-u)/t.................... Equation 1

Where F = Force exerted in sonic, m = mass of ken, v = final velocity, u = initial velocity, t = time.

Given: m = 0.75 kg, v = - 60 m/s (opposite direction), u = 52 m/s, t = 13 ms = 0.013 s

Substitute into equation 1

F = 0.75(-60-52)/0.013

F = 0.75(-112)/0.013

F = -84/0.013

F = -6461.54 N

Note: The negative sign tells that the force act in opposite direction to the initial motion of ken.

Hence the magnitude of the average force of the wall = -6461.54 N

4 0

3 years ago