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3 years ago

An aeroplane, flying in a straight line at a constant

Physics

1 answer:

dedylja [7]3 years ago

3 0

Answer:

200

Explanation:

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A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero

Rainbow [258]

Answer:

Explanation:

For destructive interference , the condition is

2μt = nλ

2μt = n x 678

For constructive interference , the condition is

2μt = (2n+1)λ₁ /2

n x 678 = (2n+1)λ₁ /2

λ₁ = 1356 n / ( 2n + 1 )

λ₁ = 1356 / ( 2 + 1/n )

For longest wavelength , denominator should be smallest or n should be largest . The longest value of n is infinity so

λ₁ = 1356 / 2

= 678 nm .

7 0

3 years ago

A ball thrown straight up climbs for 3.0 sec before falling. Neglecting air resistance, with what velocity was the ball thrown?

sesenic [268]

Answer:

Speed, u = 29.4 m/s

Explanation:

Given that, A ball thrown straight up climbs for 3.0 sec before falling, t = 3 s

Let u is speed with which the ball is thrown up. When the ball falls, v = 0

Using first equation of motion as :

v = u + at

Here, a = -g

So, u = g × t

$u=9.8\times 3$

u = 29.4 m/s

So, the speed with which the ball was thrown is 29.4 m/s. Hence, this is the required solution.

7 0

4 years ago

The pressure exerted by 15m of liquid is 1500pa.The acceleration due to gravity g=10m/s^2.Calculate the density liquid.

Natali [406]

Answer:

1500 divided by 150(15m x 10m/s^2) = 10

8 0

3 years ago

One billiard ball is shot east at 2.2m/s. A second, identical billiard ball is shot west at 1.2m/s. The balls have a glancing co

Kruka [31]

Answer:

$v_{1}=1.886 \frac{m}{s}$

β= 57.99 south of east

Explanation:

$v_{1}=2.2 \frac{m}{s} \\v_{2}=1.2 \frac{m}{s} \\m_{1}=m_{2}=m\\v_{fx}=1.6 \frac{m}{s} \\v_{fy}=$ ?

Velocity in axis x the two balls come one from east and west

$m_{1}*v_{1x}+m_{2}*v_{2x}=m_{1}*v_{fx1}+m_{2}*v_{fx2}\\m*(v_{1x}+v_{2x})=m*(v_{fx1}+v_{fx2})\\v_{fx2}=0\\v_{1x}+v_{2}=v_{f1}+0\\v_{fx1}=2.2 \frac{m}{s}+(1.2\frac{m}{s})\\ v_{fx1}=1 \frac{m}{s} \\$

Velocity in axis y initial is zero so:

$v_{y1}+v_{y2}=v_{y1f}+v_{y2f}\\v_{y1}=0\\v_{y2}=0\\v_{y1f}+v_{y2f}=0\\v_{y1f}=-v_{y2f}\\v_{y2f}=1.6\frac{m}{s}$

$v=\sqrt{v_{1fx}^{2}+v_{1fy}^{2}}\\ v=\sqrt{1^{2}+1.6^{2}}\\v=1.886 \frac{m}{s}$

Angle is find using:

tan(β)= $\frac{v_{fy}}{v_{fx}}$

$\beta =tan^{-1}*\frac{1.6}{1}=57.99$

5 0

3 years ago

Can someone help me :)

Mamont248 [21]

Answer:

mass= 60kg

velocity= 10m/s

momentum = mass × velocity

p=m×v

p= 60×10

p= 600kgm/s

5 0

3 years ago