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3 years ago

White light falls on a yellow filter,if:

Physics

1 answer:

riadik2000 [5.3K]3 years ago

6 0

Answer:

all colours are absorbed except for the colour of the filter.

Explanation:

When white light passes through a coloured filter, all colours are absorbed except for the colour of the filter. For example, an orange filter transmits orange light but absorbs all the other colours. If white light is shone on an orange filter, only the orange wavelengths will be observed by the human eye.

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Consider the collision of a 1500-kg car traveling east at 20.0 m/s (44.7 mph) with a 2000-kg truck traveling north at 25 m/s (55

masha68 [24]

Answer:

1. $X_{cm}=-8.57m$

2. $Y_{cm}=-14.29m$

3. $V_{cm} = 16.66m/s$

4. α = 59.05°

5. $V_{cm} = 16.66m/s$

6. α = 59.05°

Explanation:

The position of the center of mass 1s before the collision is:

$X_{cm}=\frac{X_c*mc+X_t*m_t}{m_c+m_t}$

where

$X_c=-20m$ ; $m_c=1500kg$ ;

$X_t=0m$ ; $m_t=2000kg$ ;

Replacing these values:

$X_{cm}=-8.57m$

$Y_{cm}=\frac{Y_c*mc+Y_t*m_t}{m_c+m_t}$

where

$Y_c=0m$ ; $m_c=1500kg$ ;

$Y_t=-25m$ ; $m_t=2000kg$ ;

Replacing these values:

$Y_{cm}=-14.29m$

The velocity of their center of mass is:

$V_{cm-x}=\frac{V_{c-x}*mc+V_{t-x}*m_t}{m_c+m_t}$

where

$V_{c-x}=20m/s$ ; $m_c=1500kg$ ;

$V_{t-x}=0m/s$ ; $m_t=2000kg$ ;

Replacing these values:

$V_{cm-x}=8.57m/s$

$V_{cm-y}=\frac{V_{c-y}*mc+V_{t-y}*m_t}{m_c+m_t}$

where

$V_{c-y}=0m$ ; $m_c=1500kg$ ;

$V_{t-y}=-25m$ ; $m_t=2000kg$ ;

Replacing these values:

$V_{cm-y}=-14.29m$

So, the magnitude of the velocity is:

$V_{cm}=\sqrt{V_{cm-x}^2+V_{cm-y}^2}$

$V_{cm}=16.66m/s$

The angle of the velocity is:

$\alpha =atan(V_{cm-y}/V_{cm-x})$

$\alpha=59.05\°$

Since on any collision, the velocity of the center of mass is preserved, then the velocity after the collision is the same as the previously calculated value of 16.66m/s at 59.0° due north of east

4 0

3 years ago

4) The mass of Pluto is 1.31 x 1022 kg and its radius is 1.15 x 106 m. What is the acceleration of

Elina [12.6K]

Answer:

0.661 m/s²

Explanation:

g = MG / r²

g = (1.31×10²² kg) (6.67×10⁻¹¹ m³/kg/s²) / (1.15×10⁶ m)²

g = 6.61×10⁻¹ m/s²

g = 0.661 m/s²

3 0

3 years ago

PHYSICS

quester [9]

Answer:

c) 8.3 x 10^24 molecules

Explanation:

First of all, we convert the volume of the glass of water from mL to $cm^3$ :

$V=250 mL = 250 cm^3$

The relationship between mass and volume is:

$d=\frac{M}{V}$

where

$d=1.00 g/cm^3$ is the density of water

M is the total mass of the glass of water

$V=250 cm^3$ is the volume of the water in the glass

Solving for M,

$M=dV=(1.00)(250)=250 g = 0.250 kg$

Now we know that the mass of a single molecule of water is

$m=3.0\cdot 10^{-26}kg$

The total mass of the water in the glass can be written as

$M=Nm$

where

N is the number of molecules in the glass of water

Solving for N, we find:

$N=\frac{M}{m}=\frac{0.250}{3.0\cdot 10^{-26}}=8.3\cdot 10^{24}$

7 0

3 years ago

A snail at position 3 cm moves to position 20 cm in 8 seconds.

natulia [17]

Answer: 17cm.

Explanation:

The equation you're using is:

Δd = df - di

Which means the change in position is equal to the final position minus the starting position. In this case that works out to 20cm - 3cm = 17cm. We're only interested in how much the snail moved, not how long it took to move, so even though they give a time it actually doesn't matter for this question.

4 0

3 years ago

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.220 rev/s . The magnitude

matrenka [14]

Answer:

1) The fan's angular velocity after 0.208 seconds is approximately 2.585 rad/s

2) The number of revolutions the blade has travelled in 0.208 s is approximately 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is approximately 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds is approximately 2.312 m/s²

Explanation:

The given parameters are;

The initial velocity of the fan, n = 0.220 rev/s

The magnitude of the angular acceleration = 0.920 rev/s²

The direction of the angular acceleration and the angular velocity = Clockwise

The diameter of the circle formed by the electric ceiling fan blades, D = 0.800 m

1) The initial angular velocity of the fan, ω₀ = 2·π × n = 2·π × 0.220 rev/s = 1.38230076758 rad/s

The angular acceleration of the fan, α = 2·π×0.920 rad/s² = 5.78053048261 rad/s²

The fan's angular velocity, 'ω', after a time t = 0.208 seconds has passed is given as follows;

ω = ω₀ + α·t

From which we have;

ω = 1.38230076758 rad/s + 5.78053048261 rad/s × 0.208 s = 2.58465110796 rad/s

The fan's angular velocity after 0.208 seconds is ω ≈ 2.585 rad/s

2) The number of revolutions the blade has travelled in the given time interval is given from the angle turned, 'θ', in the given time as follows;

θ = ω₀·t + 1/2·α·t²

θ = 1.38230076758 × 0.208 + 1/2 × 5.78053048261 × 0.208² = 0.41256299505 radians

2·π radians = 1 revolution

∴ 0.41256299505 radians = 0.41256299505 radian× 1 revolution/(2·π radian) = 0.06566144 revolution

The number of revolutions the blade has travelled in 0.208 s ≈ 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is given as follows;

The tangential speed, $v_t$ = ω × r = ω × D/2

At t = 0.208 s, ω = 2.58465110796 rad/s, therefore, we have;

$v_t$ = ω × D/2 = 2.58465110796 × 0.800/2 = 1.0338604413

The tangential speed, $v_t$ = 1.0338604413 m/s

The tangential speed ≈ 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds, 'a' is given as follows;

a = α × r = α × D/2

a = 5.78053048261 × 0.800/2 = 2.31221219304

The tangential acceleration, a ≈ 2.312 m/s²

4 0

3 years ago