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8090 [49]
3 years ago
10

If this speed is based on what would be safe in wet weather, estimate the radius of curvature for a curve marked 50 km/h . The c

oefficient of static friction of rubber on wet concrete is μs=0.7, the coefficient of kinetic friction of rubber on wet concrete is μk=0.5.
Physics
1 answer:
PtichkaEL [24]3 years ago
7 0

Answer:

Explanation:

v = 50 km / h

= 13.89 m /s

When a vehicle runs on a circular path , it is static friction which prevents it from getting overturned .

static friction = μs mg

centripetal force = m v² / R

m v² / R = μs mg

R = v² / μs x g

= 13.89² / .7 x 9.8

= 28.12 m .

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Answer:

dV/dt  = 9 cubic inches per second

Explanation:

Let the height of the cylinder is h

Diameter of cylinder = height of the cylinder = h

Radius of cylinder, r = h/2

dh/dt = 3 inches /s

Volume of cylinder is given by

V = \pi r^{2}h

put r = h/2 so,

V = \pi \frac{h^{3}}{4}

Differentiate both sides with respect to t.

\frac{dV}{dt}=\frac{3h^{2}}{4}\times \frac{dh}{dt}

Substitute the values, h = 2 inches, dh/dt = 3 inches / s

\frac{dV}{dt}=\frac{3\times 2\times 2}{4}\times 3

dV/dt  = 9 cubic inches per second

Thus, the volume of cylinder increases by the rate of 9 cubic inches per second.

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4 years ago
A 2.0-kilogram cantaloupe rolling north at 4.0 meters per second collides head on with a 1.0-kilogram orange rolling south at 8.
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when two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of
Ahat [919]

<h2>When two object P and Q are supplied with the same quantity of heat, the temperature change in P is observed to be twice that of Q. The mass of P is half that of Q. The ratio of the specific heat capacity of P to Q​</h2>

Explanation:

Specific heat capacity

It is defined as amount of heat required to raise the temperature of a substance by one degree celsius .

It is given as :

Heat absorbed = mass of substance x specific heat capacity x rise in temperature

or ,

Q= m x c x t

In above question , it is given :

For Q

mass of Q = m

Temperature changed =T₂/2

Heat supplied = x

Q= mc t

or

X=m x C₁ X T₁

or, X =m x C₁ x T₂/2

or, C₁=X x 2 /m x T₂                                 (equation 1 )

For another quantity : P

mass of P =m/2

Temperature= T₂

Heat supplied is same that is : X

so, X= m/2 x C₂ x T₂                            

or, C₂=2X/m. T₂                                   (equation 2 )

Now taking ratio of C₂ to c₁, We have

C₂/C₁= 2X /m.T₂  /2X  /m.T₂

so, C₂/C₁= 1/1

so, the ratio is 1: 1

8 0
4 years ago
Can an object have increasing speed while its acceleration is decreasing?
alexdok [17]
The best option is C. This is due to friction.
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