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Rus_ich [418]
2 years ago
15

A plane starting at rest at one end of a runway undergoes a uniform acceleration of 4.8 m/s

Physics
1 answer:
yKpoI14uk [10]2 years ago
8 0
  • initial velocity=0m/s=u
  • Acceleration=a=4.8m/s^2
  • Time=t=15s

Final velocity be v

\\ \sf\longmapsto v=u+at

\\ \sf\longmapsto v=0+4.8(15)

\\ \sf\longmapsto v=72m/s

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What would happen if The DNA code in a daughter cell did not match the code in the parent cell
Vladimir [108]
A mutation would occur if the DNA code in a daughter cell did not match the code in the parent cell.
8 0
3 years ago
Read 2 more answers
When the input voltages of a difference amplifier are 5.1 v and 6.4 v, the output voltage is 64.7 v. The inputs are changed to 4
daser333 [38]

Answer:

a) Differential mode gain = 48

b) Common mode gain = 0.4

c) CMRR = 120

Explanation:

The output of a difference amplifier is related to the input by the equation:

V_{0} = A_{1} V_{1} + A_{2} V_{2} \\

When V₁ = 6.4 V, V₂ = 5.1 V and V₀ = 64.7 V, the equation becomes

6.4 A₁ + 5.1 A₂ = 64.7.....................(1)

When V₁ = 5.6 V, V₂ = 4.9 V and V₀ = 35.7 V, the equation becomes

5.6 A₁ + 4.9 A₂ = 35.7.....................(2)

Multiply equation (1) by 5.6  and (2) by 6.4

35.84 A₁ + 28.56A₂ = 362.32.....................(3)

35.84 A₁ + 31.36 A₂ = 228.48....................................(4)

Subtract equation (3) from (4)

2.8 A₂ = -133.84

A₂ = -133.84/2.8

A₂ = -47.8

Put the value of  A₂ into equation (1)

6.4 A₁ + 5.1 (-47.8) = 64.7

6.4 A₁ = 64.7 + 243.78

A₁ = 308.48/6.4

A₁ = 48.2

a) Common mode gain = A₁ + A₂ = 48.2 + (-47.8)

Common mode gain = 0.4

b) Differential mode gain = (A₁ -A₂)/2

Differential mode gain = (48.2 - (-47.8))/2

Differential mode gain = 96/2

Differential mode gain = 48

c) Common Mode Rejection Ratio (CMRR)

CMRR = |\frac{Differential Mode Gain}{Common Mode Gain} |

CMRR = |\frac{48}{0.4} |\\CMRR = 120

4 0
3 years ago
It is known that birds can detect the earth's magnetic field, but the mechanism of how they do this is not known. It has been su
Lubov Fominskaja [6]

Answer:

A) 0.50 mV

Explanation:

In this problem, we can think the wings of the bird as a metal rod moving across a magnetic field. So, and emf will be induced into the wings of the bird, according to the formula:

\epsilon = BvL sin \theta

where

B=5\cdot 10^{-5} T is the strength of the magnetic field

v = 13 m/s is the speed of the bird

L = 1.2 m is the wingspan of the bird

\theta=40^{\circ} is the angle between the direction of motion and the direction of the magnetic field

Substituting numbers into the formula, we find

\epsilon = (5.0\cdot 10^{-5} T)(13 m/s)(1.2 m) sin 40^{\circ}=0.00050 V = 0.50 mV

8 0
3 years ago
Which observational tool helped astronomers Arno Penzias and Robert Wilson discover the
iren2701 [21]

Answer:

A radio telescope helped the astronomers discover the CMB.

Explanation:

  • Penzias and Wilson while experimenting with a radio telescope in 1964, accidentally discovered the radiation that exists universally also known as the CMB.
  • This was used to support the "Big Bang Theory" and not the "Steady State Theory"
  • CMB is the faint cosmic radiation that fills up the universe. It provides important data for understanding early universe.
  • This data tells us about the composition of the universe and its age which raises new questions about the universe.
3 0
3 years ago
A wedge with an inclination of angle θ rests next to a wall. A block of mass m is sliding down the plane. There is no friction b
Softa [21]

Answer:

  The net force on the block  F(net)  = mgsinθ).

   Fw =mg(cosθ)(sinθ)

Explanation:

(a)

Here, m is the mass of the block, n is the normal force, \thetaθ is the wedge angle, and Fw  is the force exerted by the wall on the wedge.

Since the block sliding down, the net force on the block is along the plane of the wedge that is equal to horizontal component of weight of the block.

                    F(net)  = mgsinθ

The net force on the block  F(net)  = mgsinθ).

The direction of motion of the block is along the direction of net force acting on the block. Since there is no frictional force between the wedge and block, the only force acting on the block along the direction of motion is mgsinθ.

(b)

From the free body diagram, the normal force n is equal to mgcosθ .

                           n=mgcosθ

The horizontal component of normal force on the block is equal to force

                           Fw=n*sin(θ) that exerted by the wall on the wedge.

Substitute mgcosθ for n in the above equation;

                           Fw =mg(cosθ)(sinθ)

Since, there is no friction between the wedge and the wall, there is component force acting on the wall to restrict the motion of the wedge on the surface and that force is arises from the horizontal component for normal force on the block.

6 0
3 years ago
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