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schepotkina [342]
2 years ago
14

Please help with this

Physics
1 answer:
sertanlavr [38]2 years ago
6 0

Answer:I have to say 56

Explanation: because it is going up by 8

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A thin uniform rod (length = 1.2 m, mass = 2.0 kg) is pivoted about a horizontal, frictionless pin through one end of the rod. (
Anika [276]

Answer:

a=9.8 rad/s^{2}

Explanation:

Torque, \tau is given by

\tau=Fr where F is force and r is perpendicular distance

R=0.5Lcos\theta where \theta is the angle of inclination

Torque, \tau can also be found by

\tau=Ia where I is moment of inertia and a is angular acceleration

Therefore, Fr=Ia and F=mg where m is mass and g is acceleration due to gravity

Making a the subject, a=\frac {Fr}{I}=\frac {mgr}{I} and already I is given as  

I=\frac {mL^{2}}{3} and r is 0.5Lcos\theta hence  

a=\frac {0.5mgLcos\theta}{1/3 mL^{2}}

a=\frac {3gcos\theta}{2L}

Taking g as 9.81, \theta is given as 37 and L is 1.2

a=\frac {3*9.81cos37}{2*1.2}=9.7932679419

a=9.8 rad/s^{2}

4 0
3 years ago
John is traveling north at 20 meters/second and his friend Betty is traveling south at 20 meters/second. If north is the positiv
Tpy6a [65]
I think it's just 20/20
3 0
3 years ago
Read 2 more answers
A circular parallel plate capacitor is constructed with a radius of 0.52 mm, a plate separation of 0.013 mm, and filled with an
olya-2409 [2.1K]

Answer:

289282

Explanation:

r = Radius of plate = 0.52 mm

d = Plate separation = 0.013 mm

A = Area = \pi r^2

V = Potential applied = 2 mV

k = Dielectric constant = 40

\epsilon_0 = Electric constant = 8.854\times 10^{-12}\ \text{F/m}

Capacitance is given by

C=\dfrac{k\epsilon_0A}{d}

Charge is given by

Q=CV\\\Rightarrow Q=\dfrac{k\epsilon_0AV}{d}\\\Rightarrow Q=\dfrac{40\times 8.854\times 10^{-12}\times\pi \times (0.52\times 10^{-3})^2\times 2\times 10^{-3}}{0.013\times 10^{-3}}\\\Rightarrow Q=4.6285\times 10^{-14}\ \text{C}

Number of electron is given by

n=\dfrac{Q}{e}\\\Rightarrow n=\dfrac{4.6285\times 10^{-14}}{1.6\times10^{-19}}\\\Rightarrow n=289281.25\ \text{electrons}

The number of charge carriers that will accumulate on this capacitor is approximately 289282.

6 0
2 years ago
A motorcycle that is slowing down uniformly. The motorcycle covers 1 ????m=1000 m in 80 sec⁡. The motorcycle then covers the nex
Lynna [10]

Answer:

Part a)

acceleration = -0.042 m/s/s

Part b)

initial speed = 14.17 m/s

final speed = 5.77 m/s

Explanation:

Part a)

Let the initial velocity of the motorcycle is

v_i = v_o

now at the end of 80 s let the speed is

v_f = v_1

after another 120 s let the speed will be

v_f' = v_2

now we know that

d = \frac{v_i + v_f}{2} (t)

d = \frac{v_o + v_1}{2}(80)

1000 = 40(v_o + v_1)

also we know that

v_1 - v_o = a(80)

also we have

1000 = \frac{v_1 + v_2}{2}(120)

1000 = 60(v_1 + v_2)

now we can say

(v_2 + v_1) - (v_o + v_1) = \frac{50}{3} - \frac{50}{2}

also we know

v_2 - v_o = a(120 + 80)

-8.33 = 200 a

a = -0.042 m/s^2

Part b)

now we have

v_1 + v_o = 25

v_1 - v_o = (-0.042)(80)

v_1 = 10.83 m/s

so the starting velocity of the trip is

v_o = 25 - 10.83 = 14.17 m/s

now speed after t = 200 s is given as

v_2 = v_o + at

v_2 = 14.17 - (0.042)(200)

v_2 = 5.77 m/s

5 0
3 years ago
You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and 5/8 inch inner
omeli [17]

To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

Q_1 = Q_2

V_1A_1=V_2A_2

Our values are given as,

A_1=\frac{1}{2}^2*\pi=0.785 in^2

A_2=\frac{5}{8}^2*\pi=1.227 in^2

Re-arrange the equation to find the first ratio of rates we have:

\frac{V_1}{V_2}=\frac{A_2}{A_1}

\frac{V_1}{V_2}=\frac{1.227}{0.785}

\frac{V_1}{V_2}=1.56

The second ratio of rates is

\frac{V2}{V1}=\frac{A_1}{A2}

\frac{V2}{V1}=\frac{0.785}{1.227}

\frac{V2}{V1}=0.640

3 0
3 years ago
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