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4 years ago

Which image illustrates refraction

Physics

2 answers:

photoshop1234 [79]4 years ago

4 0

Answer:

it loaded and it is C. buddy sorry about that :)

Alex777 [14]4 years ago

4 0

Answer:

Explanation:

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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

A pilot heads his jet due east. The jet has a speed of 425 mi/h relative to the air. The wind is blowing due north with a speed

Alja [10]

Answer:

(a)Velocity of wind = 35 j

(b)Velocity of jet relative to air = 425 i

(c)True velocity of jet = 425 i + 35 j

(d)True speed of jet = $\sqrt{425^{2}+35^{2}}$ = 426.88 mi/h

Direction of jet is, ∅ = $tan^{-1} \frac{40}{425}$ = 5.38°

Explanation:

We can represent east direction by i and north direction by j.

The jet has a relative speed of 425 mi/h relative to the air.

The wind is blowing due north with a speed of 35 mi/h = 35 j

425 mi/h is the relative speed with respect to wind that is

Velocity of jet wrt wind= $V_{jw}$ = $V_{j}-V_{w}$

425 i = $V_{j}$ - 35 j

$V_{j}$ = 425 i + 35 j

(a)Velocity of wind = 35 j

(b)Velocity of jet relative to air = 425 i

(c)True velocity of jet = 425 i + 35 j

(d)True speed of jet = $\sqrt{425^{2}+35^{2}}$ = 426.88 mi/h

Direction of jet is,

∅ = $tan^{-1} \frac{40}{425}$ = 5.38°

7 0

4 years ago

How do you find the number of neutrons

sammy [17]

the mass number minus the atomic number

8 0

3 years ago

How does the density of fluid affect the magnitude of buoyancy acting on an object immersed in it

Oksi-84 [34.3K]

Answer:

Explanation:

more the density , more will be the buoyant force acting on it , less the density less will be the buoyant force acting on it. This is why people float in dead sea and sink in other seas

5 0

3 years ago

2. Kevin works as a janitor, and he is pushing a fully-

dybincka [34]

The time taken for him to move the bin 6.5 m is 2.30 s.

The given parameters;

weight of the load, w = 557 N
force applied , F = 410 N
angle of force, = 15°
coefficient of kinetic friction = 0.46
distance moved, d = 6.5 m

The net horizontal force on the recycling bin is calculated as follows;

$Fcos\theta - F_k = ma$

where;

m is the mass of the recycling bin
$F_k$ is the frictional force

W = mg

$557 = 9.8m\\\\m = \frac{557}{9.8} \\\\m = 56.84 \ kg$

The net horizontal force on the recycling bin is calculated as;

$Fcos \theta - F_k = ma\\\\Fcos\theta - \mu_kF_n = ma\\\\410\times cos(15) \ - \ 0.46(557) = 56.84 a\\\\139.8 = 56.84a\\\\a = \frac{139.8}{56.84} \\\\a = 2.46 \ m/s^2$

The time taken for him to move the bin 6.5 m is calculated as follows;

$s = v_0t + \frac{1}{2} at^2\\\\6.5 = 0 + \frac{1}{2} \times 2.46\times t^2\\\\6.5 = 1.23 t^2\\\\t^2 = \frac{6.5 }{1.23} \\\\t^2 = 5.285\\\\t = \sqrt{5.285} \\\\t = 2.30 \ s$

Thus, the time taken for him to move the bin 6.5 m is 2.30 s.

Learn more here:brainly.com/question/21684583

7 0

3 years ago