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Drupady [299]
4 years ago
7

How long will it take a cyclist with a forward acceleration of -0.50 m/s squared to bring a bicycle with an initial velocity of

forward of 13.5 m/s to a complete stop? show the formula and your work.
Physics
1 answer:
Liono4ka [1.6K]4 years ago
3 0
Using kinematic equation v=u+at......... 1.
v = final velocity = 0 m/s. u = initial velocity = 13.5 m/s.
a = -0.5 m/s². 
Plug all the known values in 1 to find t
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This graph shows a supply curve. A graph titled Supply Curve has Quantity Supplied on the x-axis, from 0 to 50 in increments of
Otrada [13]

Answer:

A just took the quiz on edge

Explanation:

7 0
3 years ago
When tightening a bolt, you push perpendicularly on a wrench with a force of 170 N at a distance of 0.155 m from the center of t
Thepotemich [5.8K]

Answer:

Explanation:

T=fd

f=170N

d=0.155m

T=170*0.155

T=26.35Nm⁻

i don't know how to convert to pound foot

the answer is correct in newton meters anyway(Nm⁻¹)

4 0
3 years ago
In an RC series circuit, ℰ = 17.0 V, R = 1.60 M, and C = 1.80 µF.
LiRa [457]

Answer:

(a) τ=2.88 s, (b) q₀= 30.6μC and (c) t=1.434s            

Explanation:

A RC circuit is an resistor(R)-capacitor(C) electric circuit.    

(a) In a resistor-capacitor circuit, the time constant (τ) can be calculated by:    

\tau = RC          

<em>where R: is the resistence and C: the capacitance of the capacitor</em>

\tau = (1.60\cdot 10^{6} \cdot 1.80\cdot 10^{-6} = 2.88 s  

 

(b) The maximum charge (q₀) is giving by:    

q_{0} = \epsilon \cdot C  

<em>where ε: is the voltage across the capacitor</em>

q_{0} = 17.0 V \cdot 1.80 \cdot 10^{-6} F = 3.06 \cdot 10^{-5} C = 30.6 \mu C                  

(c) The time (t) that take the charge (q) to build up to 12 μC can be calculated from the next equation:  

q = q_{0}(1 - e(\frac{-t}{RC}))  

t = RC Ln( \frac{q_0}{q_0 - q}) = \tau Ln( \frac{q_0}{q_0 - q})

t = 2.88 \cdot Ln (\frac{30.6 \cdot 10^{-6}}{(30.6 \cdot 10^{-6} - 12.0 \cdot 10^{-6}})

t = 1. 434 s  

Have a nice day!

5 0
3 years ago
Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above
poizon [28]

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

U_{g,1} = U_{g,2} + W_{dis}

Where:

U_{g,1}, U_{g,2} - Initial and final gravitational potential energy, measured in joules.

W_{dis} - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

U = m \cdot g \cdot y

Where:

m - Mass, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

y - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

W_{dis} = f \cdot \Delta s

Where:

f - Friction force, measured in newtons.

\Delta s - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s

f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}

If m = 1000\,kg, g = 9.807\,\frac{m}{s^{2}}, y_{1} = 40\,m, y_{2} = 25\,m and \Delta s = 400\,m, then:

f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}

f = 367.763\,N

The magnitude of the frictional force between the car and the track is 367.763 N.

7 0
4 years ago
Find the resistance at 50°c of copper wire 2mm in diameter and 3m long
mr Goodwill [35]
0.0179 ohms for copper.

0.0184 ohms for annealed copper



Ď = R (A/l) where

Ď = electrical resistivity

R = electrical resistance of a uniform specimen

A = cross sectional area

l = length



Solve for R by multiplying both sides by l/A

R = Ď(l/A)



The cross section of the wire is pi * 1^2 mm = 3.14159 square mm = 3.14159e-6 square meters.

The length is 3 meters. So l/A = 3/3.14159e-6 = 9.5493e5



Ď for copper is 1.68e-8 so 1.68e-8 * 9.5493e5 = 1.60e-2 ohms at 20 C

But copper has a temperature coefficient (α) of 0.00386 per degree C.

So the resistance value needs to be adjusted based upon how far from 20 C the temperature is.

50 - 20 = 30 C

So 0.00386 * 30 = 0.1158 meaning that the actual resistance at 50 C will be 11.58% higher.

So 1.1158 * 0.016 = 0.0179 ohms.



If you're using annealed copper, the values for Ď and the temperature coefficient change.

Ď = 1.72e-8

α = 0.00393



Doing the math, you get

1.72e-8 * 9.5493e5 * (1 + 30 * 0.00393) = 0.0184 ohms
8 0
3 years ago
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