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grigory [225]
3 years ago
11

At what speed will a car round a 52-m-radius curve, banked at a 45???? angle, if no friction is required between the road and ti

res to prevent the car from slipping? (g = 9.8 m/s2)a. 27 m/s
b. 17 m/s
c. 23 m/s
d. 35 m/s
Physics
1 answer:
Serhud [2]3 years ago
4 0

Answer:

c. 23 m/s

Explanation:

\theta = Angle of banking = 45°

r = Radius of turn = 52 m

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of car

As the car is tilted we have the relation

tan\theta=\dfrac{v^2}{rg}\\\Rightarrow v=\sqrt{tan\theta rg}\\\Rightarrow v=\sqrt{tan45\times 52\times 9.8}\\\Rightarrow v=22.57432\ m/s\approx 23\ m/s

The velocity of the car is 23 m/s if the car is not slipping

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Currents during lightning strikes can be up to 50000 A (or more!). We can model such a strike as a 47500 A vertical current perp
Rufina [12.5K]

Answer:

57 N

Explanation:

Force on a current carrying conductor in a magnetic field

B = 12 X 10⁻⁴ T

= Bil where B is magnetic field , i is current and l is length of conductor

force required = 12 x10⁻⁴ x 47500 x 1

= 57 N

6 0
2 years ago
Water enters the constant 130-mm inside-diameter tubes of a boiler at 7 MPa and 65°C and leaves the tubes at 6 MPa and 450°C wit
snow_lady [41]

The inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

Explanation:

When water entering the tube of constant diameter flows through the tube, it exhibits continuity of mass in the hydrostatics. So the mass of water moving from the inlet to the outlet tend to be same, but the velocity may differ.

As per mass flow equality which states that the rate of flow of mass in the inlet is equal to the product of area of the tube with the velocity of the water and the density of the tube.

Since, the inlet volume flow is measured as the product of velocity with the area.

Inlet volume flow=Inlet velocity*Area*time

And the mass flow rate is  

Mass flow rate in the inlet=density*area*inlet velocity*time

Mass flow rate in the outlet=density*area*outlet velocity*time

Since, the time and area is constant, the inlet and outlet will be same as

(Mass inlet)/(density*inlet velocity)=Area*Time

(Mass outlet)/(density*outlet velocity)=Area*Time

As the ratio of mass to density is termed as specific volume, then  

(Specific volume inlet)/(Inlet velocity)=(Specific volume outlet)/(Outlet velocity)

Inlet velocity=  (Specific volume inlet)/(Specific volume outlet)*Outlet velocity

As, the specific volume of water at inlet is 0.001017 m³/kg and at outlet is 0.05217 m³/kg and the outlet velocity is given as 72 m/s, the inlet velocity

is

Inlet velocity = \frac{0.001017}{0.05217}*72 =1.4035 m/s

So, the inlet velocity is 1.4035 m/s.

Then the inlet volume will be

Inlet volume = inlet velocity*area of circle=\pi  r^{2}*inlet velocity

As the diameter of tube is 130 mm, then the radius is 65 mm and inlet velocity is 1.4 m/s

Inlet volume = 1.4*3.14*65*65*10^{-6} =0.019 \frac{m^{3} }{s}

So, the inlet volume is 0.019 m³/s.

Thus, the inlet velocity is 1.4 m/s and inlet volume is 0.019 m³/s.

4 0
3 years ago
Which of the following could classify as a non-essential need according to the hierarchy of needs? A. healthy food B. human cont
sergejj [24]
D. According to the hierarchy of needs the body and mind must be taken care of first and foremost.

4 0
3 years ago
A horizontal force of magnitude 30.2 N pushes a block of mass 3.50 kg across a floor where the coefficient of kinetic friction i
marshall27 [118]

Answer:

A) 89.39 J

B) 30.39J

C) 23.8 J

Explanation:

We are given;

F = 30.2N

m = 3.5 kg

μ_k = 0.646

d = 2.96m

ΔEth (Block) = 35.2J

A) Work done by the applied force on the block-floor system is given as;

W = F•d

Thus, W = 30.2 x 2.96 = 89.39 J

B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;

ΔEth = μ_k•mgd

Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J

Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.

Thus,

ΔEth = ΔEth (Block) + ΔEth (floor)

Thus,

ΔEth (floor) = ΔEth - ΔEth (Block)

ΔEth (floor) = 65.59J - 35.2J = 30.39J

C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;

W = K + ΔEth

Therefore;

K = W - ΔEth

K = 89.39 - 65.59 = 23.8J

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2 years ago
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