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Sati [7]
3 years ago
11

A boy sledding down a hill accelerates at 1.40 m/s². If

Physics
1 answer:
mrs_skeptik [129]3 years ago
5 0

Answer: He would reach 7m/s at the distance of 17,5m.

Explanation: In order for us to know how long it takes for the speed to be reached, we use the equation V= Vo + at, with V=7m/s, Vo= 0m/s since you start from rest and a=1.4m/s². T is the time in seconds that we want to find out.

7 = 0 + 1.4t

7 = 1.4t

t= 5s

Now, we want to know the distance this boy reaches in 5 seconds, in a 7m/s speed and accelarating in 1.4m/s². For finding this out, we use another equation, S = So + Vot + (at²)/2, S being the final distance, So = 0m because he started from rest, and the other variables used before.

S = 0 + 0*5 + [1.4*(5²)]/2

S = 0 + 0 + (1.4*25)/2

S = 35/2

S = 17,5m

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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
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tensa zangetsu [6.8K]

Hope this helps!!!!!!!!!!!!!

7 0
2 years ago
As capacitor was discharging, what did you observe about q on its plates and the motion of charges in the external circuit?
liubo4ka [24]

As capacitor was discharging, The charge on the plate got reversed and the motion of charge is opposite to the flow of current.

The charging contemporary asymptotically processes 0 as the capacitor becomes charged up to the battery voltage.

The capacitor is completely charged when the voltage of the electricity supply is equal to that at the capacitor terminals. that is referred to as capacitor charging; and the charging segment is over when modern-day stops flowing thru the electrical circuit.

A capacitor can be slowly charged to the important voltage and then discharged quick to provide the power wanted. it's far even viable to charge several capacitors to a positive voltage and then discharge them in any such way as to get extra voltage out of the gadget than became installed.

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8 0
1 year ago
The latitude of the city of Arlington is about 32.7357°. Calculate the following: (a) The angular speed of the Earth. (b) The li
julia-pushkina [17]

Answer:

Explanation:

a ) The earth rotates by 2π radian in 24 x 60 x 60 s

so angular speed ( w )  = 2π / (24 x 60 x 60)  = 7.268 x  10⁻⁵ rad / s

b ) Linear speed of city of Arlington ( v )  = w r = w R Cosλ where R is radius of the earth and λ is latitude .

v = 7.268 x 10⁻⁵ x  6.371 x 10⁶ cos 32.7357

389.5 m /s

acceleration = w² r = w² R Cos 32.7357

= (7.268 x 10⁻⁵ )² x 6.371 x 10⁶ x cos 32.7357

=283.08 x 10⁻⁴ m/s²

c) velocity ratio =

w r /w R =

R cos 32.73/ R  

= Cos 32.73

= 0.84 .

6 0
3 years ago
What is the only way to let your body recover from overtraining? A. Take vitamins B. Rest for several days C. Eat a lot of prote
Luba_88 [7]

Answer:

B) Rest for several days

Explanation:

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7 0
3 years ago
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