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solniwko [45]
3 years ago
14

Quick answer please

Physics
2 answers:
timama [110]3 years ago
6 0

Answer:

tq1=EoA/D

-2q1=-2EoAV/d

<u>Option C.</u>

Alekssandra [29.7K]3 years ago
6 0

Answer:

tq1=EoA/D

-2q1=-2EoAV/D

Explanation:

option C is correct

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A stationary shell is exploded in to three fragments A, B, C of masses in the ratio 1:2:3. A travels
spin [16.1K]

Answer:

20 m/s

Explanation:

If the mass of fragment A is m, then the mass of fragment B is 2m, and the mass of fragment C is 3m.

The velocity of A is 60 m/s at angle 0°.

The velocity of B is 30 m/s at angle 120°.

The velocity of C is v at angle θ.

In the x direction:

Momentum before = momentum after

(m + 2m + 3m) (0) = m (60 cos 0°) + 2m (30 cos 120°) + 3m (v cos θ)

0 = 60m − 30m + 3m v cos θ

0 = 30m + 3m v cos θ

-30m = 3m v cos θ

-10 = v cos θ

In the y direction:

Momentum before = momentum after

(m + 2m + 3m) (0) = m (60 sin 0°) + 2m (30 sin 120°) + 3m (v sin θ)

0 = 0 + 30√3 m + 3m v sin θ

-30√3m = 3m v sin θ

-10√3 = v sin θ

Square the two equations and add together:

(-10)² + (-10√3)² = (v cos θ)² + (v sin θ)²

100 + 300 = v² cos² θ + v² sin² θ

400 = v² (cos² θ + sin² θ)

400 = v²

v = 20

The speed of fragment C is 20 m/s.

7 0
3 years ago
Explain the relationship between air resistance and gravitational acceleration on a falling object
Allushta [10]
Air resistance, also called drag, acts upon a falling body by slowing the body down to thr point where it stops accelerating, and it falls at a constant speed, known as the terminal volocity of a falling object. Air resistance depends on the cross sectional area of the object, which is why the effect of air resistance on a large flat surfaced object is much greater than on a small, streamlined object.

 
7 0
3 years ago
The length of a rectangular sheet of metal decreases by 34.5 cm. Its width decreases proportionally. If the sheets original widt
devlian [24]
The original width was 94.71 cm 
<span>The area decreased 33.1% </span>

<span>The equation for the final size is </span>
<span>2X^2 = 1.2 m^2 </span>
<span>X^2 - 0.6 m^2 </span>
<span>X^2 = 10000 * .6 cm </span>
<span>X = 77.46 cm (this is the width) </span>

<span>The length is 2 * 77.46 = 154.92 cm </span>

<span>The original length was 154.92 + 34.5 = 189.42 cm </span>
<span>The original width was 189.42 / 2 = 94.71 cm </span>

<span>The original area was 94.71 * 189.92 = 17939.9 cm^2 </span>
<span>The new area is 79.46 * 154.92 = 12000.1 cm^2 </span>

<span>The difference between the original and current area is 17939.9 - 12000.1 = 5939.86 cm^2 </span>

<span>The percentage the area decreased is 5939.86 ' 17939.9 = 33.1%</span>
6 0
4 years ago
A plane mirror of circular shape with radius r=20cm is fixed to the ceiling. A bulb is to be placed on the axis of the mirror. A
KIM [24]

Answer:

0.75 m

Explanation:

Let's call the distance between the bulb and the mirror x.

The bulb and the length of the mirror form a triangle.  The mirror and the illuminated area on the floor form a trapezoid.  If we extend the lines from the mirror edge to the reflected image of the bulb, we turn that trapezoid into a large triangle.  This triangle and the small triangle are similar.  So we can say:

x / 0.4 = (3 + x) / 2

Solving for x:

2x = 0.4 (3 + x)

2x = 1.2 + 0.4 x

1.6 x = 1.2

x = 0.75

So the bulb should located no more than 0.75 m from the mirror.

5 0
3 years ago
Some neodymium glass lasers can provide 100TW of power in 1.0 ns pulses at a wavelength of 0.26 micrometers. how much energy is
Aleksandr-060686 [28]

Answer:

 E = 10⁵ J

Explanation:

given,

Power, P = 100 TW

           = 100 x 10¹² W

time, t = 1 ns

           = 1 x 10⁻⁹ s

The energy of a single pulse is:-

Energy = Power x time

 E = P t

 E = 100 x 10¹² x 1 x 10⁻⁹

 E = 10⁵ J

The energy contained in a single pulse is equal to 10⁵ J

7 0
4 years ago
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