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Pepsi [2]
3 years ago
14

A truck with 34-in.-diameter wheels is traveling at 55 mi/h. Find the angular speed of the wheels in rad/min, *hint convert mile

s to inches & hours to minutes:
Physics
1 answer:
julia-pushkina [17]3 years ago
7 0

Answer:

w =3416 rad/min  

Explanation:

The angular speed of the wheels in radians can be defined as follows:

w=\frac{s}{r}

where

w = angular speed

r = radius of the wheel

s =distance traveled per minute

in this case we have

r = 34/2 inches = 17 inches

s = (55 mi/hr) (5280 ft/mi) (12 in/ft) (1hr/60min) = 58,080 inches/min

Therefore, w = 58,080/ 17 = 3416 rad/min  

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The linear impulse delivered by the hit of a boxer is 202 N · s during the 0.244 s of contact. What is the magnitude of the aver
zlopas [31]

Answer: Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).

Explanation: Impulse is defined as the force acting on an object for a short period or interval of time.

Mathematically it is given by the relation:

Impulse = Force \times Time

According to the numerical values given in the question, I = 202 Ns and T = 0.244 s

So, Force F = \frac{Impulse}{Time} = \frac{202}{0.244} = 827.86 N

Magnitude of the average force exerted on the glove by the other boxer is 827.86 N (approximately 828 N).

7 0
3 years ago
uppose a wheel is initially rotating at 10.0 rad/s while undergoing constant angular acceleration reaching a speed of 30.0 rad/s
Paul [167]

Answer:

t = 22.2 s

Explanation:

angular distance covered in the 36.0 s is

θ = ω(avg)t = ½(10.0 + 30.0)36 = 720 radians

720/2 = 360 radians

α = Δω/t = (30 - 10)/36 = 5/9 rad/s²

θ = ω₀t + ½αt²

360 = 10.0t + ½(5/9)t²

   0 = (5/18)t² + 10.0t - 360

   0 = t² + 36t - 1296

t = (-36 ±√(36² - 4(1)(-1296))) / 2

t = (-36 ±√(6480)) / 2

t = -18 ±√1620

we ignore the negative time result as it occurs before we care.

t = -18 + √1620 = 22.249223... s

8 0
3 years ago
An output is a Push or pull _________________________ on the object
Ugo [173]

Answer:

The input force that you use on an inclined plane is the force with which you push or pull an object. The output force is the force that you would need to lift the object without the inclined plane. This force is equal to the weight of the object.

Explanation:

5 0
3 years ago
What is 1/12+7/9 hj​
Umnica [9.8K]

Explanation:

\frac{1}{12}  +  \frac{7}{9}  \\  \\  =  \frac{1 \times 3}{12 \times 3}  +  \frac{7 \times 4}{9 \times 4} \\  \\  =  \frac{3}{36}  +  \frac{28}{36}  \\  \\  =  \frac{3 + 28}{36}  \\  \\  =  \frac{31}{36}  \\  \\   \huge \purple{ \boxed{\therefore \:  \frac{1}{12}  +  \frac{7}{9}  =  \frac{31}{36} }} \\

6 0
3 years ago
Read 2 more answers
A 2 kg ball is moving 3 m/s when it starts rolling up a hill.
AURORKA [14]

Answer:

the height reached is = 0.458 [m]

Explanation:

We need to make a sketch of the ball and see the location of the reference point where the potential energy is zero. But the kinetic energy will be defined by the following expression:

Ek=\frac{1}{2} *m*v^{2} \\where:Ek= kinetic energy [J]\\m = mass of the ball [kg]\\v = velocity of the ball [m/s]

Replacing the values on the equation we have:

Ek=\frac{1}{2}*(2)*(3^{2} )\\ Ek=9[J]\\

This kinetic energy will be transformed in potential energy in the moment when the ball starts to rolling up. Therefore the maximum height reached by the ball depends of the initial velocity given to the ball.

Ek=Ep\\where\\Ep=potential energy [J]\\Ep=m*g*h\\where\\g=gravity = 9.81[m/s^2]\\h=height reached [m]\\

Now we have:

h=\frac{Ep}{m*g} \\h=\frac{9}{2*9.81} \\\\h=0.45 [m]

In that moment when the ball reach the 0.45 [m] the potencial energy will be maximum and equal to the kinetic energy when the ball has a velocity of 3[m/s]

6 0
3 years ago
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