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Vladimir79 [104]
1 year ago
8

A 3400 kg jet is flying at a constant speed of 170 m/s as it makes a vertical loop. At the top of the loop the pilot feels three

times his normal weight. What is the radius of the loop
Physics
1 answer:
velikii [3]1 year ago
4 0

Answer:

736.49 meters

Explanation:

Step one:

Given data

mass m= 3400kg

speed= 170 m/s

At the top of the loop the pilot feels three times his normal weight

hence we multiply mass by 3 in our solution

Centrepetal force = mv^2/r-mg=3mg

Thereofore, (his true weight) + (the weight he feels) = mg + 3mg

so

mv^2/r=4mg

Step two:

centripetal force = mv^2/r=4mg

mv^2=4mgr

v^2=4gr

substitue

centripetal force

170^2 = 4*3400*9.81

so

170^2 = 4*9.81*r  

28900 = 39.24*r  

28900 =39.24*r  

divide both sides by 39.24

r=28900 /39.24

r=736.49 meters

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Answer:

14.7 m/s

Explanation:

a = acceleration experienced by driver's head = 50 g = 50 x 9.8 m/s² = 490 m/s²

v₀ = initial speed of the driver = 0 m/s

v = final speed of the driver after 30 ms

t = time interval for which the acceleration is experienced = 30 ms = 0.030 s

Using the equation

v = v₀ + a t

Inserting the values

v = 0 + (490) (0.030)

v = 14.7 m/s

6 0
2 years ago
In a series circuit, a generator (1300 Hz, 12.0 V) is connected to a 14.0- resistor, a 4.40-μF capacitor, and a 6.00-mH inductor
klemol [59]

Answer:

(a) 2.8 V

(b) 5.6 V

(c) 9.8 V

Explanation:

Given:

Frequency of the generator (f) = 1300 Hz)

Terminal voltage (V) =12.0 V

Resistance of resistor (R) = 14.0 Ω

Capacitance of capacitor (C) = 4.40 μF = 4.40 × 10⁻⁶ F

Inductance of the inductor (L) = 6.00 mH = 6.00 × 10⁻³ H

In order to find the voltages across each, we first need to find the reactance and impedance.

Reactance of the inductor is given as:

X_L=2\pi f L\\\\X_L=2\times 3.14\times 1300\times 6.00\times 10^{-3}\\\\X_L=49\ \Omega

Reactance of the capacitor is given as:

X_C=\frac{1}{2\pi fC}\\\\X_C=\frac{1}{2\times 3.14\times 1300\times 4.40\times 10^{-6}}\\\\X_C =28\ \Omega

Now, impedance is given as:

Z=\sqrt{X_L^2+X_C^2}\\\\Z=\sqrt{(49)^2+(28)^2}\\\\Z=\sqrt{3185}=56.4\ \Omega

Current across the circuit is given as:

I=\frac{V}{Z}\\\\I=\frac{12}{56.4}=0.2\ A

As resistor, capacitor and inductor are connected in series, the current across each of them is same and equal to total current in the circuit.

(a)

Voltage across the resistor is given as:

V_R=IR\\\\V_R=0.2\times 14=2.8\ V

Therefore, the voltage across resistor is 2.8 V.

(b)

Voltage across the capacitor is given as:

V_C=IX_C\\\\V_C=0.2\times 28=5.6\ V

Therefore, the voltage across the capacitor is 5.6 V.

(c)

Voltage across the inductor is given as:

V_L=IX_L\\\\V_L=0.2\times 49=9.8\ V

Therefore, the voltage across the inductor is 9.8 V.

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2 years ago
Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e
Arlecino [84]

Answer:

The radius is r =  4.434 *10^{-5} \ m

Explanation:

From the question we are told that

    The magnetic field is  B =   90 mT =  90*10^{-3} \ T

     The electron kinetic energy is  KE  =  1.4 eV = 1.4 * (1.60*10^{-19})  =2.24*10^{-19} \ J

Generally for the collision to occur the centripetal force of the electron in it orbit is equal to the magnetic force applied  

   This is mathematically represented as

   \frac{mv^2}{r}  =  qvB

=>    r =  \frac{m* v}{q *  B}

Where  m is the mass of electron with values m  =  9.1 *10^{-31} \ kg  

             v is the escape velocity  which is mathematically represented as

                v  = \sqrt{\frac{2 * KE}{m} }

So  

       r =  \frac{m}{qB}  *  \sqrt{\frac{2 *  KE}{m} }

     apply indices

    r = \frac{\sqrt{2 * KE * m} }{qB}

substituting values

   

        r = \frac{\sqrt{2 * 2.24*10^{-19}* 9.1 *10^{-31}} }{ 1.60 *10^{-19}* 90*10^{-3}}

       r =  4.434 *10^{-5} \ m

     

6 0
1 year ago
A professor determined the relationship between the time spent studying (in hours) and performance on an exam.
lubasha [3.4K]
Lets write again formula for determening Ann's performance.

P = 70.443 + 4,885*t

where t is in hours. This is equation with P(t) which means that P only depends on variable t. If we express t=2.6 in formula we will find her expected performance.

P = 83.144

Now, since it says that she scored 16 points less than expected we need to find value of P-16

P - 16 = 67.144

After round we get that the answer is 67
3 0
2 years ago
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Answer:

The work flow required by the compressor = 100.67Kj/kg

Explanation:

The solution to this question is obtained from the energy balance where the initial and final specific internal energies and enthalpies are taken from A-17 table from the given temperatures using interpolation .

The work flow can be determined using the equation:

M1h1 + W = Mh2

U1 + P1alph1 + ◇U + Workflow = U2 + P2alpha2

Workflow = P2alpha2 - P1alpha1

Workflow = (h2 -U2) - (h1 - U1)

Workflow = ( 684.344 - 491.153) - ( 322.483 - 229.964)

Workflow = ( 193.191 - 92.519)Kj/kg

Workflow = 100.672Kj/kg

6 0
2 years ago
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