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Peter designed a road with a curve of radius 30 m that is banked so that a 950 kg car traveling at 40.0 km/h can round it even i

spayn [35]

Answer:

$v = 15.56 m/s$

$v = 56 km/h$

Explanation:

When coefficient of friction is approximately zero then we have

$F_ncos\theta = mg$

$F_n sin\theta = \frac{mv^2}{R}$

$tan\theta = \frac{v^2}{Rg}$

here we know that

$v = 40 km/h = 11.11 m/s$

R = 30 m

$tan\theta = \frac{11.11^2}{30\times 9.81}$

$\theta = 22.75 degree$

now when friction coefficient is 0.30 then we have

$F_n cos\theta = mg + F_f sin\theta$

$F_f cos\theta + F_n sin\theta = \frac{mv^2}{R}$

now we have

$v = \sqrt{Rg(\frac{\mu + tan\theta}{1 - \mu tan\theta})}$

$v = \sqrt{30(9.81)(\frac{0.30 + tan22.75}{1 - (0.30) tan22.75})}$

$v = 15.56 m/s$

$v = 56 km/h$

3 0

3 years ago

According to the rule of 72 and about how many years will $78 be worth $39 if the rate of inflation is 5.8%

Monica [59]

The answer is 12.4 years

3 0

3 years ago

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A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm

I am Lyosha [343]

Answer:

Rate of change of magnetic field is $3.466\times 10^3T/sec$

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop $r=\frac{0.13}{2}=0.065m$

Length of the circular loop $L=2\pi r=2\times 3.14\times 0.065=0.4082m$

Wire is made up of diameter of 2.6 mm

So radius $r=\frac{2.6}{2}=1.3mm=0.0013m$

Cross sectional area of wire $A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2$

Resistivity of wire $\rho =2.18\times 10^{-8}m$

Resistance of wire $R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm$

Current is given i = 11 A

So emf $e=11\times 1.67\times 10^{-3}=0.0183volt$

Emf induced in the coil is $e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}$

$0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}$

$\frac{dB}{dt}=3.466\times 10^3=T/sec$

8 0

3 years ago

A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500

defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

$M = \frac{\mu_0 N_1 N_2A_1}{l_1}$

Where,

$\mu =$ permeability of free space

$N_1 =$ Number of turns in solenoid 1

$N_2 =$ Number of turns in solenoid 2

$A_1=$ Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

$\mu_0 = 4\pi *10^{-7}H/m$

$N_1 = 500$

$N_2 = 40$

$A = 7.5*10^{-4}m^2$

$l = 0.5m$

Substituting,

$M = \frac{\mu_0 N_1 N_2A_2}{l_1}$

$M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}$

$M = 3.77*10^{-4}H$

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0

4 years ago

Which type of radiation can reach the furthest? A. Beta B. Gamma C. Alpha

Mashutka [201]

Answer:

C. Alpha

Explanation:

Gamma radiation, unlike alpha or beta, does not consist of any particles, instead consisting of a photon of energy being emitted from an unstable nucleus. Having no mass or charge, gamma radiation can travel much farther through air than alpha or beta, losing (on average) half its energy for every 500 feet.

7 0

3 years ago

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