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vampirchik [111]
3 years ago
13

A banked circular highway is designed for traffic moving

Physics
1 answer:
Nuetrik [128]3 years ago
3 0

Answer:

0.063

Explanation:

velocity of the car, v = 40 km/h = 11.11 m/s

radius, r = 200 m

Let the coefficient of friction is μ.

The coefficient of friction relates to the velocity on banked road is given by

\mu =\frac{v^{2}}{rg}

where, v is the velocity, r be the radius of the curve road and μ is coefficient of friction.

By substituting the values, we get

\mu =\frac{11.11^{2}}{200\times 9.8}

μ = 0.063

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Heat is added to a 2kg piece of ice at a rate of 793kW. How long will it take for ice to melt if it was initially 0?
Ede4ka [16]

Answer:

0.84 s

Explanation:

Step 1

Given information:

Mass of the ice (m) = 2.0 kg

Heat transfer rate (Q/T) = 793.0 kW

Latent heat of fusion of ice (Lf) = 334 kJ/kg

\frac{Q}{T}  =  \frac{mLf}{T}

Substituting the corresponding values we have:

793.0 kW=  \frac{2.0kg(334 kJ/kg)}{T} \\  T  =  \frac{2.0kg(334 kJ/kg)}{793.0kW}  =  \frac{668kJ}{793kW}   \\  = 0.84s

8 0
2 years ago
Why does sodium chloride and magnesium chloride conduct electricity?
mihalych1998 [28]
They only conduct when they are in solution form, because then their ions become mobile, and the ions conduct electricity.

Hope this helps!
5 0
3 years ago
A sailboat moves north for a distance of 15.00 km when blown by a wind from the exact southeast with a force of 3.00 x 10^-4 N.
Zolol [24]
These are actually 4 different exercises:

ex 1) The sailboat moves north, while the wind moves from southeast. This means the angle between the direction of the boat and the wind is 45^{\circ}.

Calling F the force of the wind, and d=15~km=15000~m the distance covered by the boat, the work done by the wind is:
W=Fdcos{\theta}=3\cdot10^{-4}~N \cdot 15000~m\cdot cos 45^{\circ}=3.18~J

The total time of the motion is t=1~h=3600~s and therefore the power of the wind is
P= \frac{W}{t} = \frac{3.18~J}{3600~s}=8.8\cdot10^{-4}~W

ex 2) First of all, let's calculate the length of the ramp. Given the two sizes 2.00 m and 6.00 m, we have
d= \sqrt{(2~m)^2+(6~m)^2}= 6.32~m

The mechanical advantage (MA) of the ramp is the ratio between the output load (W) and the input force (F). The output load is the weight of the load, mg, therefore:
MA= \frac{W}{F}= \frac{mg}{F}= \frac{195~Kg\cdot 9.81~m/s^2}{750~N}=2.55

Finally, the efficiency \epsilon of the ramp is the ratio between the output energy and the work done. The output energy is simply the potential energy (Ep) of the load, which is mgh, where h is the height of the ramp. The work done W is the product between the input force, F, and the displacement of the load, which is the length of the ramp: Fd. Therefore:
\epsilon =  \frac{E_p}{W}= \frac{mgh}{Fd}= \frac{195~Kg \cdot 9.81~m/s^2\cdot 2~m}{750~N\cdot6.32~m}=0.81

ex 3) the graph is missing

ex 4) We know that the power is the ratio between the work done W and the time t:
P= \frac{W}{t}
But we can rewrite the work as
W=Fdcos\theta
where F is the force applied, d the displacement of rock and \theta=60^{\circ] is the angle between the direction of the force and the displacement (3 m). 
Therefore we can rewrite the power as
P= \frac{W}{t} = \frac{F d cos\theta}{t}=F v cos\theta
where v=d/t=5~m/s is the velocity, Using the data of the exercise, we can then find the force, F:
F= \frac{P}{v cos\theta} =   \frac{250~W}{5~m/s \cdot cos 60^{\circ}}=100~N

and now we can also calculate the work, which is 
W=Fdcos 60^{\circ}=100~N\cdot 3~m \cos60^{\circ}=150~J
3 0
3 years ago
What could you do to increase the electric potential energy between two
SSSSS [86.1K]

Answer:

Increase the charge of one particle by a factor of 16

Explanation:

3 0
2 years ago
Read 2 more answers
One particular descent goes from 2100m to 1600m. Assuming work done against friction is 90% of the potential energy change of th
alisha [4.7K]

Answer:

1/2 M V^2 = .1 M g H       where 10% of PE goes into KE

V^2 = .2 g H = .2 * 9.8 * (2100 - 1600) = 980 m^2 / s^2

V = 31.1 m/s       increase in speed during descent

1 km / hr = 1000 m / 3600 sec = .278 m/s

V = 31.1 m/s / (.278 m/s / km /hr)= 112 km/hr

7 0
2 years ago
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