How many planets are there in the milky way galaxy?

Andre, whose mass is 77.1 kg, sits on a diving board above a dunk tank. If he is sitting 0.90 m above the water, what is his tot

Ganezh [65]

Answer:

680 J

Explanation:

Mechanical energy = potential energy + kinetic energy

ME = PE + KE

ME = mgh + ½ mv²

ME = (77.1 kg) (9.8 m/s²) (0.90 m) + ½ (77.1 kg) (0 m/s)²

ME = 680 J

8 0

3 years ago

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A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o

gulaghasi [49]

A. 0.77 A

Using the relationship:

$P=\frac{V^2}{R}$

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega$

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

$R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega$

The two light bulbs are connected in series, so their equivalent resistance is

$R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega$

The two light bulbs are connected to a voltage of

V = 240 V

So we can find the current through the two bulbs by using Ohm's law:

$I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A$

B. 142.3 W

The power dissipated in the first bulb is given by:

$P_1=I^2 R_1$

where

I = 0.77 A is the current

$R_1 = 240 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_1 = (0.77 A)^2 (240 \Omega)=142.3 W$

C. 42.7 W

The power dissipated in the second bulb is given by:

$P_2=I^2 R_2$

where

I = 0.77 A is the current

$R_2 = 72 \Omega$ is the resistance of the bulb

Substituting numbers, we get

$P_2 = (0.77 A)^2 (72 \Omega)=42.7 W$

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

$P=\frac{E}{t}$

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0

3 years ago

Suppose a soccer player kicks the ball from a distance 29 m toward the goal. find the initial speed of the ball if it just passe

Rudik [331]

The ball's vertical velocity at the time it just passes over the goal is 0 m/s. Its initial vertical velocity is unknown and we denote it by $v\sin39^\circ$ , where $v$ here is the ball's initial speed. Vertically, the only force acting on the ball is gravity, which attributes a downward acceleration of 9.8 m/s^2. We expect the maximum height achieved by the ball to be 2.4 m, so we can find the initial speed by solving

$\left(0\,\dfrac{\mathrm m}{\mathrm s}\right)^2-\left(v\sin39^\circ\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(2.4\,\mathrm m)$

$\implies v=11\,\dfrac{\mathrm m}{\mathrm s}$

6 0

3 years ago

If a galaxy has an apparent velocity of 2300 km/s, what is its distance if the Hubble constant is assumed to be 70 km/s/Mpc

prohojiy [21]

The distance of the galaxy is 32.86 Mpc.

Using the hubble law, v = H₀D where v = apparent velocity of galaxy = 2300 km/s, H = hubble constant = 70 km/s/Mpc and D = distance of galaxy.

Since we require the distance of the galaxy, we make D subject of the formula in the equation. So, we have

D = v/H₀

Substituting the values of the variables into the equation, we have

D = 2300 km/s ÷ 70 km/s/Mpc

D = 32.86 Mpc

So, the distance of the galaxy is 32.86 Mpc

Learn more about hubble law here:

brainly.com/question/18484687

4 0

2 years ago

1. El puente mas largo del mundo es el puente Akashi Kaikyo, en Japón. El puente mide 3910 m de largo y está construido con acer

bearhunter [10]

Answer:

It's 1.0000042 times longer in summer than in winter. It represents a 1.6 centimeters difference between seasons.

Explanation:

The linear coefficient of thermal expansion for steel is about $1.2*10^{-7}\°C^{-1}$ . From the equation of linear thermal expansion, we have:

$L_f=L_0(1+\alpha\Delta T)$

Taking the winter day as the initial, and the summer day as the final, we can take the relationship between them:

$L_{summer}=L_{winter}[1+(1.2*10^{-7}\°C^{-1})(30\°C+5\°C)]\\\\L_{summer}=(1.0000042)L_{winter}$

It means that the bridge is 1.0000042 times longer in summer than in winter. If we multiply it by the length of the bridge, we obtain that the difference is of about 1.6 centimeters between the two seasons.

8 0

3 years ago