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victus00 [196]
3 years ago
13

What is the efficiency of the machine in Figure 7.2 if the weight of the box is 200 N and required 98 N of effort Force?

Physics
1 answer:
zepelin [54]3 years ago
3 0
this question is tricky but the answer is C
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The Sun is expected to undergo hydrogen fusion for a total of _____ years. a million 10 million a billion 10 billion 100 billion
Elanso [62]
The correct answer is 10 billion years. The Sun is expected to undergo hydrogen fusion for a total of 10 billion years. The Sun generates its energy by nuclear fusion of hydrogen and produces helium nucleus. It fuses 620 million metric tons every second.
3 0
3 years ago
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A proton moves with a velocity of v with arrow = (4î − 6ĵ + k) m/s in a region in which the magnetic field is B with arrow = (î
nalin [4]

Answer:

F = [(6.4 × 10⁻¹⁹)î + (8.0 × 10⁻¹⁹)ĵ + (22.4 × 10⁻¹⁹)k] N

Magnitude of F = (2.466 × 10⁻¹⁸) N

Explanation:

The magnetic force, F, on a given charge, q, moving with velocity, v, in a magnetic field, B, is given as the vector product

F = qv × B

where v = (4î − 6ĵ + k) m/s

B = (î + 2ĵ − k) T

The particle is a proton, hence,

q = (1.602 × 10⁻¹⁹) C

F = qv × B = q (v × B)

(v × B) is given as (4î − 6ĵ + k) × (î + 2ĵ − k)

The cross product is evaluated as a determinant of

| î ĵ k |

|4 -6 1 |

|1 2 -1 |

î [(-6)(-1) - (2)(1)] - ĵ [(4)(-1) - (1)(1)] + k [(4)(2) - (-6)(1)]

î (6 - 2) - ĵ (-4 - 1) + k (8 + 6) = (4î + 5ĵ + 14k)

(v × B) = (4î + 5ĵ + 14k)

F = q (v × B) = (1.6 × 10⁻¹⁹) (4î + 5ĵ + 14k)

F = [(6.408 × 10⁻¹⁹)î + (8.01 × 10⁻¹⁹)ĵ + (22.428 × 10⁻¹⁹)k] N

Magnitude of F =

√[(6.408 × 10⁻¹⁹)² + (8.01 × 10⁻¹⁹)² + (22.428 × 10⁻¹⁹)²]

Magnitude of F = (2.466 × 10⁻¹⁸) N

Hope this Helps!!!

4 0
3 years ago
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A resistor and an inductor are connected in series to a battery. The battery is suddenly removed from the circuit. The time cons
Vladimir79 [104]
E, 63% of the value. I forget the rationale behind it but I learnt that in engineering. 90% confident for that answer.
3 0
3 years ago
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A concave spherical mirror has a radius of curvature of magnitude 20.0 cm. (a) Find the location of the image for object distanc
Tems11 [23]

Answer:

Answered

Explanation:

The radius of curvature of the mirror R = 20 cm

then the focal length f = R/2 = 10 cm

(a) From mirror formula

 1/f = 1/di + /1do

then the image distance

    di = fd_o / d_o - f

  = (10)(40) / 40-10

 = 30.76 cm

since the image distance is positive so the image is real

ii) when the object distance d_0=20 cm

 di = 10×20/ 20-10

= 20

Hence, the image must be  real

iii)when the object distance d_0 = 10

di = 10×10 / 10-10 =  ∞ (infinite)

the image will be formed at ∞

here also image will be real but diminished.

7 0
3 years ago
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The apparent height of a building 10.5 km away is 0.02 radians. What is the approximate height of the building to the nearest me
Ksenya-84 [330]

Answer:

Approximate height of the building is 23213 meters.

Explanation:

Let the height of the building be represented by h.

0.02 radians = 0.02 × \frac{180^{o} }{\pi }

                     = 0.02 x (180/\frac{22}{7})

0.02 radians  = 1.146°

10.5 km = 10500 m

Applying the trigonometric function, we have;

Tan θ = \frac{opposite}{adjacent}

So that,

Tan 1.146° = \frac{h}{10500}

⇒ h = Tan 1.146° x 10500

      = 2.21074 x 10500

      = 23212.77

h = 23213 m

The approximate height of the building is 23213 m.

8 0
2 years ago
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