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sdas [7]
3 years ago
11

A car drives on a circular road of radius R. The distance driven by the car is given by = + [where a and b are constants, and t

in seconds will give d in meters]. In terms of a, b, and R, and when t = 3 seconds, find an expression for the magnitudes of (i) the tangential acceleration aTAN, and (ii) the radial acceleration aRAD3
Physics
1 answer:
pishuonlain [190]3 years ago
4 0

Answer:

The question is incomplete, the complete question is "A car drives on a circular road of radius R. The distance driven by the car is given by d(t)= at^3+bt [where a and b are constants, and t in seconds will give d in meters]. In terms of a, b, and R, and when t = 3 seconds, find an expression for the magnitudes of (i) the tangential acceleration aTAN, and (ii) the radial acceleration aRAD3"

answers:

a.18a m/s^{2}

b. a_{rad}=\frac{(27a +b)^{2}}{R}

Explanation:

First let state the mathematical expression for the tangential acceleration and the radial acceleration.

a. tangential acceleration is express as

a_{tan}=\frac{d|v|}{dt} \\

since the distance is expressed as

d=at^{3}+bt

the derivative is the velocity, hence

V(t)=\frac{dd(t)}{dt}\\V(t)=3at^{2}+b\\

hence when we take the drivative of the velocity we arrive at

a_{tan}=\frac{dv(t)}{dt}\\ a_{tan}=6at\\t=3 \\we have \\a_{tan}=18a m/s^2

b. the expression for the radial acceleration is expressed as

a_{rad}=\frac{v^{2}}{r}\\a_{rad}=\frac{(3at^{2} +b)^{2}}{R}\\t=3\\a_{rad}=\frac{(27a +b)^{2}}{R}

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Two electrons are separated by 1.70 nm. What is the magnitude of the electric force each electron exerts on the other?
lidiya [134]

Answer:

F=7.96*10^{-11}N

Explanation:

According to Coulomb's law, the magnitude of the electric force between two equals charges (q) is given by:

F=\frac{kq^2}{d^2}

Here k is the coulomb constant and d is the distance between the charges. For two electrons we have:

F=\frac{ke^2}{d^2}\\F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-1.6*10^{-19}C)^2}{(1.7*10^{-9}m)^2}\\F=7.96*10^{-11}N

8 0
3 years ago
a steam engine work on its vicinity 285 k heat is released with the help of 225 c energy absorbed to the system what is the effi
-Dominant- [34]

The efficiency of the steam engine is 78.9% because the rest of the work input is used to overcome friction.

<h3>What is efficiency of machines?</h3>

Efficiency of a machine expresses the useful work done by a a machine as a percentage.

  • Efficiency of a machine = work output/work input × 100 %%

The efficiency of machines are always less than 100% percent due to energy losses due to friction and heat.

For the steam engine:

Work output = 225 J

Work input = 285 J

Efficiency = 225/285 × 100% = 78.9 %

Therefore, the efficiency of the steam engine is 78.9% because the rest of the work input is used to overcome friction.

Learn more about efficiency of machines at: brainly.com/question/7536036

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6 0
2 years ago
A series circuit that is connected to a 50 V, 60 Hz source is made up of 25 ohm resistor, capacite wieh X= 18 ohms, and inductor
Allisa [31]

Answer:

the impedance of the circuit is 25.7 ohms.

Explanation:

It is given that,

Voltage, V = 50 volts

Frequency, f = 60 Hz

Resistance, R = 25 ohms

Capacitive resistance, X_C=18\ ohms

Inductive resistance, X_L=24\ ohms

We need to find the impedance of the circuit. It is given by :

Z=\sqrt{R^2+(X_L-X_C)^2}

Z=\sqrt{25^2+(24-18)^2}

Z = 25.7 ohms

So, the impedance of the circuit is 25.7 ohms. Hence, this is the required solution.

6 0
3 years ago
A circular swimming pool has a diameter of 12 m. The circular side of the pool is 3 m high, and the depth of the water is 2.5 m.
Anestetic [448]

Answer:

Explanation:

Diameter of pool = 12 m

radius of pool, r = 6 m

Total height raised, h = 3 + 2.5 = 5.5 m

density of water, d = 1000 kg/m³

Mass of water, m = Volume of water x density

m = πr²h x d

m = 3.14 x 6 x 6 x 5.5 x 1000

m = 113040 kg

Work = m x g x h

W = 113040 x 9.8 x 5.5

W = 6092856 J

7 0
3 years ago
You apply a horizontal force of 25N to push a shopping cart across the parking lot at a constant velocity. a) what is the net fo
AlekseyPX

(a) The net force on the shopping cart is zero.

(b) The the force of friction on the shopping cart is 25 N.

(c) When same force is applied to the shopping cart on a wet surface, it will move faster.

<h3>Net force on the shopping cart</h3>

The net force on the shopping cart is calculated as follows;

F(net) = F - Ff

where;

  • F is the applied force
  • Ff is the frictional force

ma = F - Ff

where;

  • a is acceleration of the cart
  • m is mass of the cart

at a constant velocity, a = 0

0 = F - Ff

F(net)  = 0

F = Ff = 25 N

Net force is zero, and frictional force is equal to applied force.

<h3>On wet surface</h3>

Coefficient of kinetic friction of solid surface is greater than that of wet surface.

Since frictional force limit motion, when the frictional force is smaller, the object tends to move faster.

Thus, the cart will move faster on a wet surface due to decrease in friction.

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4 0
2 years ago
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