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3 years ago

For typical rubber-on-concrete friction, what is the shortest time in which a car could accelerate from 0 to 50 mph

Physics

1 answer:

SVEN [57.7K]3 years ago

6 0

Answer:

2.5sec

Explanation:

for the typical rubber on concrete it will take 2.5seconds which is the shortest time.

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A 7-hp (shaft) pump is used to raise water to an elevation of 19 m. if the mechanical efficiency of the pump is 82 percent, dete

Hoochie [10]

Convert the units of power,W = 7 hp = 7 * 745.69 = 5219.83 WCalculate the power input to the pump using the efficiency of the pump equationn=Wpump/Wshaft
Substitute 0.82 for n and 5219.83 for Wshaft0.82=Wpump/5219.83 Wpump=0.82*5219.83=4280.26 WCalculate the mass flow ratein=Wpump/(gz_2 )Where g is the acceleration due to gravity, and z_2 is the elevation of water. Substitute 4280.26 for Wpump, 9.81 m/s^2 for g, and 19m for z_2in = 4280.26 / 9.81 * 19 = 22.9640 m^3/sCalculate the volume flow rate of waterV=m/ρWhere ρ is the density of water. Substitute 22.9640 m^3/s for in and 1000 m^3/kg for ρ, we get V = 22.9640 / 1000 = 0.0230 kg/sTherefore, the volume flow rate of water is 0.0230 kg/s

7 0

3 years ago

4- Protein foods?? pls answer these i really need these

vampirchik [111]

The #1 answer would be meats and eggs. But, as a veggan, I present you with meatless protien-rich food!

1 - Pea protein (used in Beyond Meat, which tastes just like meat btw!)

2 - Soy protein such as tofu

3 - Nuts

4 - Beans! oml so many beans have MORE protien than meat!! ikr!?!?

may I have Brainliest pls =)

8 0

3 years ago

Suppose the ring rotates once every 4.30 ss . If a rider's mass is 58.0 kgkg , with how much force does the ring push on her at

Stells [14]

Answer:

422.36 N

Explanation:

given,

time of rotation = 4.30 s

T = 4.30 s

Assuming the diameter of the ring equal to 16 m

radius, R = 8 m

$v = \dfrac{2\pi R}{T}$

$v = \dfrac{2\pi\times 8}{4.30}$

v = 11.69 m/s

now, Force does the ring push on her at the top

$- N - m g = \dfrac{-mv^2}{R}$

$N + m g = \dfrac{mv^2}{R}$

$N = \dfrac{mv^2}{R}- m g$

$N = m(\dfrac{v^2}{R}- g)$

$N = 58\times (\dfrac{11.69^2}{8}- 9.8)$

N = 422.36 N

The force exerted by the ring to push her is equal to 422.36 N.

6 0

3 years ago

Jason fought in the war in Iraq and witnessed some horrific sights. When he got home from the war, Jason didn't seem to remember

zimovet [89]

Answer:

Jason has repressed the memories

Explanation:

7 0

3 years ago

Read 2 more answers

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

5 0

3 years ago