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Roman55 [17]
3 years ago
6

The earth exerts the necessary centripetal force on an orbiting satellite to keep it moving in a circle at constant speed. Which

one of the following statements best explains why the speed of the satellite does not change although there is a net force exerted on it?
A) The satellite is in equilibrium.
B) The acceleration of the satellite is zero m/s2 .
C) The centripetal force has magnitude mv2 /r.
D) The centripetal force is canceled by the reaction force.
E) The centripetal force is always perpendicular to the velocity.
Physics
1 answer:
jenyasd209 [6]3 years ago
8 0

Answer:

E) The centripetal force is always perpendicular to the velocity.

Explanation:

Due to gravity and inertia, the satellite follows a uniform circular motion. In this movement, the velocity is always tangent to the orbit and the centripetal force is directed towards the center. Therefore, there is no net acceleration in the same direction of velocity, which implies that it remains constant.

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If your brakes fail, look for something to rub against, try to pick something that will not give way.True or False?
Yanka [14]

Answer:

False

Explanation:

False

If brakes fail then we should not run against any object or wall to slow down the speed instead we should use friction forces slow down the car and shift gear to neutral to cease any acceleration.

By driving zig-zag motion friction force on the tires may stop the car after some distance.

Do not try to shut off the engine as it will lock the steering mechanism any prove fatal.

7 0
3 years ago
An MRI technician moves his hand from a region of very low magnetic field strength into an MRI scanner's 2.00 T field with his f
makkiz [27]

Answer:

2.62 A

Explanation:

B = 2 T, Diameter = 2.5 cm , radius, r = 0.0125 m, t = 0.45 s, r = 0.01 ohm

Induced emf, e = rate of change of magnetic flux

e = A x dB / dt = 3.14 x (0.0125)^2 x 2 / 0.45

e = 0.026 V

induced current, i = e / R = 0.026 / 0.01 = 2.62 A

7 0
2 years ago
What information is necessary to calculate a confidence interval?
Anni [7]
<span>Write down the phenomenon you'd like to test. ...Select a sample from your chosen population. ...Calculate your sample mean and sample standard deviation. ...Choose your desired confidence level. ...Calculate your margin of error. ...<span>State your confidence interval.</span></span>
6 0
3 years ago
Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are
Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

d_{12} = \sqrt{3^{2}+3^{2}  }  = \sqrt{18} m : distance from q₁ to q₂

(d₁₂)² = 18 m²

d_{13} =\sqrt{1^{2}+3^{2}  } = \sqrt{10} m  : distance from q₁ to q₃

(d₁₃)² = 10 m²

d_{14} =\sqrt{3^{2}+2^{2}  } = \sqrt{13} m  : distance from q₁ to q₄

(d₁₄)² = 13 m²

K=  8.98755 × 10⁹ N *m²/C²

q₁=  7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45°  ,  β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N  

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }

F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }

Fn₁= 23.95*10⁻⁹ N

8 0
3 years ago
An engineer has the task of producing an aluminum alloy with a density of 3.0 grams per cubic centimeter. She comes up with the
pochemuha

Answer:

The best option is for the following option m = 15 [g] and V = 5 [cm³]

Explanation:

We have that the density of a body is defined as the ratio of mass to volume.

Ro =m/V

where:

Ro = density = 3 [g/cm³]

Now we must determine the densities with each of the given values.

<u>For m = 7 [g] and V = 2.3 [cm³]</u>

Ro=7/2.3\\Ro=3.04 [g/cm^{3} ]

<u>For m = 10 [g] and V = 7 [cm³]</u>

<u />Ro=10/7\\Ro=1.42[g/cm^{3} ]\\<u />

<u>For m = 15 [g] and V = 5 [cm³]</u>

<u />Ro=15/5\\Ro=3[g/cm^{3} ]\\<u />

<u>For m = 21 [g] and V = 8 [cm³]</u>

<u />Ro=21/8\\Ro=2.625[g/cm^{3} ]\\<u />

5 0
2 years ago
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