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3 years ago

Problem #1- In the first scene in which Miss Clark appears, she is supposed to be wearing a BLACK dress.

Physics

2 answers:

ASHA 777 [7]3 years ago

7 0

Answer:

You would have to shine green Light on the actress to make the dress appear to be red.

Explanation:

The reason it would appear to be red is because green reflects the red colour and absorbs the rest. So, when green light is shone on the dress, it absorbs all of the green light and not reflecting anything.The dress then appears to beblack.

katrin [286]3 years ago

6 0

Answer:

Blue Lighting

Explanation:

In order to make red look black, you must use blue light. The blue would be absorbed and there would be no red light to reflect.

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A 50 kg pitcher throws a baseball with a mass of 0.15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is

Andreas93 [3]

The velocity of the pitcher is <u>0.105 m/s</u> in a direction opposite to the velocity of the ball.

When no external force acts on a system, the total momentum of the system is conserved. The total initial momentum of the system is equal to the total final momentum of the system.

The pitcher and the ball are initially at rest, therefore, the total initial momentum of the system is zero.

Since no external forces act on the system comprising of pitcher and the ball, the total final momentum of the system is also equal to zero.

If the mass of the pitcher is mp and its speed is vp, the mass of the ball is mb and the ball's speed is vb, then the final momentum of the system of pitcher and the ball is given by,

$p=m_pv_p+m_bv_b=0$

Therefore,

$v_p=-\frac{m_b}{m_p} v_p$

Substituet 0.15 kg for mb, 50 kg for mp and 35 m/s for vb.

$v_p=-\frac{m_b}{m_p} v_p=-\frac{0.15 kg}{50 kg} (35m/s)=-0.105 m/s$

The pitcher has a velocity <u> 0.105 m/s</u> opposite to the direction of the velocity of the ball.

8 0

2 years ago

What are non examples of light years?

abruzzese [7]

Hi , here are some examples: An astronomical unit A parsec A meter

5 0

2 years ago

Read 2 more answers

Classify the properties as extensive or intensive: mass density; color volume; total energy; temperature; melting point

m_a_m_a [10]

Answer:

Intensive properties

Density

Color

temperature

Melting point

Extensive properties

Mass

Volume

Total Energy

Explanation:

Intensive properties: In Physics, Intensive properties which are not depend of the amount of matter in a sample, It only depends of the type of matter, some examples of intensive properties are:

1. Density: It is a intensive property. It can explain better with a example: the water density is 1000 kg/m3, So if we have 1 liter or 1000 liters of water the density will be the same for the two samples.

2. Color: Solid sodium chloride is white. If you have 2 samples the first recipient with 2 kilograms of NaCl and the second with 10 kilograms of NaCl. The color of the substance does not depend on the amount of the substance.

As was mentioned before the same theory is applied to temperature and melting point concepts.

On the other hand,

Extensive properties are properties of the matter which depend on the amount of matter that is present in the system or sample. some examples are:

1. Mass: It is a property that measures the amount of matter that an object contains. For example, 10 kilograms of solid Copper contains a higher mass than 2 kilograms of the same metal.

2. Volume: It is a property which measures the space occupied by an object or a substance. For example, the space occupied by a glass of milk is lower than the space occupied by a bottle of milk, Then the volume of the glass of milk is lower than the volume of the bottle of milk.

3. Finally the total energy is contained in molecules and atoms that constituted systems so, if the amount of matter increases the number of molecules too, then the total energy will increase.

I hope it helps you.

6 0

2 years ago

An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f

konstantin123 [22]

A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force $F=m \frac{v^2}{r}$ , directed toward the center of the circle (so, horizontally)
- the weight of the plane: $W=mg$ , downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is $r= \frac{260 m}{2} = 130 m$ , so the centripetal force acting on the plane is
$F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N$
On the vertical axis, we have two forces: the weight
$W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N$
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to $a=12 m/s^2$ , we can find the magnitude of this other force F by using Newton's second law:
$F+mg=ma$
$F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N$

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
$R= \sqrt{(F_c^2+(W+F)^2}=$
$= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2} =2.23 \cdot 10^6 N$

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
$\tan \theta = \frac{-(W+F)}{F_c}= -0.5$
And so, the angle is
$\theta = \arctan (-0.5)=-26.8 ^{\circ}$

7 0

2 years ago

Assume that the position vector of A is r=i+j+k . Determine the moment about the origin O if the force F=(1)i+(0)j+(5)k . The mo

ddd [48]

Answer:

M₀ = 5i - 4j - k

Explanation:

Using the cross product method, the moment vector(M₀) of a force (F) is about a given point is equal to cross product of the vector A from the point (r) to anywhere on the line of action of the force itself. i.e

M₀ = r x F

From the question,

r = i + j + k

F = 1i + 0j + 5k

Therefore,

M₀ = (i + j + k) x (1i + 0j + 5k)

M₀ = $\left[\begin{array}{ccc}i&j&k\\1&1&1\\1&0&5\end{array}\right]$

M₀ = i(5 - 0) -j(5 - 1) + k(0 - 1)

M₀ = i(5) - j(4) + k(-1)

M₀ = 5i - 4j - k

Therefore, the moment about the origin O of the force F is

M₀ = 5i - 4j - k

3 0

2 years ago