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2 years ago

Does every light source emit only one type of light?

1 answer:

3 0

Yes it is possible. Spectrum of emitted light depends upon the chemical composition of the source. and the way of its excitation. a clear example to us is that of sun.

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A heavy log will float on water

posledela

Answer:

A it is less dense than water

Explanation:

The density, of a substance is its mass per unit volume

8 0

4 years ago

Read 2 more answers

You pull on a crate using a rope as in (Figure 1), except the rope is at an angle of 20.0 ∘ above the horizontal. The weight of

Nezavi [6.7K]

Hi there!

For an object on an incline with friction being pulled, the following forces are present.

Force due to Gravity
Force due to Friction
Force due to tension

The force due to friction opposes the force due to gravity which would cause the object to slide down. (The force due to friction acts up the incline). Additionally, the force due to the rope is also upward.

Let up the incline be positive, and down the incline be negative.

Doing a summation of forces:
$\Sigma F = F_T + F_f - F_g$

For the crate to be moving at a constant velocity, there is NO net force acting on the crate, so:
$0 = F_T + F_f - F_g$

Now, we can express each force as an equation.

Force due to tension:

Must be solved for.

Force due to gravity:

On an incline, this is equivalent to the SINE component of its weight. (Force of gravity is STILL THE WEIGHT, but on an incline, it contains a horizontal component that contributes to the net force)

This is expressed as:
$F_g = Mgsin\theta$

Force due to friction:

Equivalent to the normal force and coefficient of friction. The normal force is the VERTICAL component of the object's weight, so:

$N = Mgcos\theta$

$F_f = \mu Mgcos\theta$

Now, plug these expressions into the above equation.

$0 = F_T + \mu Mgcos\theta - Mgsin\theta\\\\Mgsin\theta - \mu Mgcos\theta = F_T$

Mg = 245 N (weight). Plug in all values:
$245sin(20) - 0.270(245)cos(20) = F_T\\\\F_T = \boxed{21.634 N}$

The normal force is equivalent to the vertical component (PERPENDICULAR TO THE INCLINE) of the weight (cosine), so:

$N = Mgcos\theta\\\\N = 245cos(20) = \boxed{230.225 N}$

5 0

2 years ago

How do Swati and Banks adjust their body position during a skydiving jump so they can fall at the same rate?

Wittaler [7]

Sky diving involves free fall under gravity along with the drag due to air molecules pushing against the body slowing the rate of fall of a body. This is actually a significant amount of force. The drag force depends on the contact surface area and weight of the body. More the surface area in contact, more would be the drag. The sitting position of the skydiver would experience less drag than the chest down position. This is because of the less contact surface area of the body with the air molecules while in the former case. Since no two persons have identical body shape and weight, the rate of fall can be made nearly equal but not exactly equal. This is would be possible when they are having same body position.

3 0

3 years ago

The back wall of a home aquarium is a mirror that is a distance of 46.0 cm away from the front wall. The walls of the tank are n

monitta

Answer:

i know the questin but i got to try and find it

Explanation:

5 0

3 years ago

The distance between two telephone poles is 50 m. A 1 kg bird lands on the telephone wire midway between the poles, and the wire

masha68 [24]

Answer:

the tension produce by the bird in the wire is 610.3 N.

Explanation:

given information:

mass of the bird, m = 1 kg

distance of two telephone pole, L = 50 m

wire sag, y = 0.2

the tension of the wire can be calculate by the following equation

tan θ = y/(L/2)

= 0.2/(50/2)

= 0.2/25

thus

θ = arc tan (0.2/25)

= 0.46°

according to Newton's first law

ΣF = 0, vertical direction

T sin θ + T sin θ - mg = 0

2 T sin θ = m g

T = mg/2 sinθ

= 1 x 9.8/(2 sin 0.46)

= 610.3 N

3 0

4 years ago