A visible light emits light with a wavelength of 4.00x10^-7m.

Three resistors of 100 W, 3900 W, and 1000 W are connected in series across a 200-V battery. What is the voltage drop across the

Gala2k [10]

Answer:

40 V

Explanation:

I will assume that the resistors are

100 and 3900 and 1000 OHMS <=====(NOT W)

In series , the resistances add together 100 + 3900 + 1000 = 5000 ohms total

V = IR

I = V / R so the total current will be 200 v / 5000 ohms = .04 amps

this is the current through all of the resistors

so for the 1000 ohm resistor V = IR .04 (1000) = 40 V

7 0

2 years ago

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 11-m-hi

True [87]

Answer:

m = maximum mass of the coaster = 410 kg

d = maximum spring compression = 2.3 m

h = maximum height of the track = 11 m

H = maximum difference in height of the track = 19 m

g = acceleration by gravity = 9.8 m/s²

k = spring constant (without safety margin) = ?

K = spring constant (with safety margin) = ?

V = maximum speed of the coaster = ?

The gravitational potential energy of the coaster on the top of the 11 m high hill (relative to its initial starting point) is:

PEg = m g h

PEg = (410 kg) (9.8 m/s²) (11 m)

PEg = 44198 J

To reach that height, the elastic potential energy stored in the spring must be the same, so:

PEg = PEe = k d² / 2

(44198 J) = k (2.3 m)² / 2

k = 16710 N/m

Adding 14% to that value, you get:

K = 1.14 (16710 N/m)

K = 19045 N/m - answer spring constant

When fully compressed, the elastic potential energy stored in the spring is:

PEe = K d² / 2

PEe = (19045 N/m) (2.3m)² / 2

PEe = 51326 J

The difference in height between the starting point and the lowest point of the track is:

Δh = H - h

Δh = (19 m) - (11 m)

Δh = 8 m

So the initial gravitational potential energy of 330 kg coaster, relative to the lowest point, is

PEg = m g Δh

PEg = (340 kg) (9.8 m/s) (8 m)

PEg = 26656 J

The total energy of the coaster at its starting point (again, relative to the lowest point) is:

TE = PEe + PEg

TE = (51326J) + (26656 J)

TE = 77982J

At the lowest point of the track, all that energy is converted to kinetic energy, so the speed at that point will be:

TE = KE = m V² / 2

(77982 J) = (340kg) V² / 2

V = 21.46 m/s - answer maximum speed

4 0

3 years ago

Amir pitches a baseball at an initial height of 6 feet with a velocity of 73 feet per second. this can be represented by the fun

Romashka [77]

The values of t are 4.643 second for the function $H(t)=-16t^2+73t+6$

What is batter misses?

An out in baseball happens when the umpire declares a batter or baserunner out. A hitter or runner who is out is no longer able to score runs and must go back to the dugout until their subsequent turn at bat. The batting team's turn is over after three outs are recorded in a half-inning.

In order to signal an out, umpires typically make a fist with one hand and then flex that arm, either upward on pop flies or forward on regular plays at first base. To indicate a called strikeout, home plate umpires frequently use a "punch-out" action.When a batter is struck by a pitched ball without making a swing at it, it is referred to as a hit-by-pitch. He consequently gets first base.

We have been given that

s = 6 feet

v = 73 feet per second

Substituting these values in the formula $H(t)=-16t^2+vt+s$

$H(t)=-16t^2+73t+6$

When the ball hits the ground, the height becomes zero. Thus, H(t)=0

$-16t^2+73t+6=0$

We solve the equation using quadratic formula $x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Substituting the values a=-16, b= 73, c=6

$t_{1,2}=\frac{-73 \pm \sqrt{(73)^2-4(-16)(6)}}{2(-16)}\\\Rightarrow t_{1,2}=\frac{-73 \pm \sqrt{5713}}{2(-16)}\\\Rightarrow t_{1,2}=-0.081, 4.643$

Learn more about the batter misses with the help of the given link:

brainly.com/question/19475098

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8 0

2 years ago

when hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results

Nikitich [7]

When hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results deposition of sediment.

On the upcurrent side of the barrier, sediment is deposited as the longshore current slows.

What is Hard stabilization?

Hard stabilization is the prevention of erosion through the use of artificial barriers.
Other hard stabilization structures, such as breakwaters and seawalls, are built parallel to the beach to protect the coast from the force of waves.
Hard stabilization structures, such as groins, are built at right angles to the shore to prevent the movement of sand down the coast and maintain the beach.
These constructions are made to last for many years, but because they detract from the visual splendor of the beach, they are not always the ideal answer.
Additionally, they affect the habitats and breeding sites of native shoreline species, interfering with the ecosystem's natural processes.

Learn more about the Hard stabilization with the help of the given link:

brainly.com/question/16022736

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4 0

2 years ago

Suppose that diameters of a new species of apple have a bell-shaped distribution with a mean of 7.2cm and a standard deviation o

Svet_ta [14]

Answer:

97.5%

Explanation:

By the empirical rule (68-95-99.7),

68% of data are within μ - σ and μ + σ
95% of data are within μ - 2σ and μ + 2σ
99.7% of data are within μ - 3σ and μ + 2σ

σ and μ are the standard deviation and the mean respectively.

From the question,

μ = 7.2 cm

σ = 0.38 cm

7.96 = 7.2 + (n × 0.38)

n = 2

Hence, 7.96 represents μ + 2σ.

P(X < μ + 2σ) = P(X < μ) + P(μ < X < μ + 2σ)

P(X < μ) is the percentage less than the mean = 50%.

P(μ < X < μ + 2σ) is half of P(μ - 2σ < X < μ + 2σ) = 95% ÷ 2 = 47.5%.

Considering this, for apples that are no more than 7.96 cm,

P(X < 7.96) = P(X < 7.2) + P(7.2 < X < 7.96) = 50% + 47.5% = 97.5%

5 0

3 years ago