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4 years ago

Ejdjsjdjskskdkdkdkdkdkdkdkdkdkdkdkdjdjdjdjdjdjdjdjd

Physics

2 answers:

Inga [223]4 years ago

6 0

Answer:Ejdjsjdjskskdkdkdkdkdkdkdkdkdkdkdkdjdjdjdjdjdjdjdjd!

Explanation:

riadik2000 [5.3K]4 years ago

5 0

DOEOUEPUROYEOYEKYDKDLYDYLDULDULDULDLUDULDLYDLUDLUDLUDLUDLUDULSOYDYODYODOYXYL I LUV MY DADDY MY SUPER HIRRO

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You are at the top of the Empire State Building on the 102nd floor, which is located 373 m above the ground, when your favorite

Liula [17]

Answer:

1.5 m

Explanation:

H = actual height of the superhero = ?

H₀ = height of the superhero as observed = 1.73 m

v = speed of the superhero = 0.50 c

Using the equation

$H = H_{o} \sqrt{1 - \left ( \frac{v}{c} \right )^{2}}$

Inserting the values

$H = 1.73 \sqrt{1 - \left ( \frac{0.50 c}{c} \right )^{2}}$

H = 1.5 m

3 0

4 years ago

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

A 69 g particle is moving to the left at 25 m/s . How much net work must be done on the particle to cause it to move to the righ

LekaFEV [45]

Answer:11.59 J

Explanation:

Given

mass of Particle $m=69 gm$

Initially Particle moves towards left $v_1=25 m/s$

Final velocity of Particle is towards Right $v_2=31 m/s$

According to Work Energy theorem

Work done by all the Forces=change in Kinetic Energy

Work done by Force $=\frac{mv_2^2}{2}-\frac{mv_1^2}{2}$

$W=\frac{69\times 10^{-3}}{2}\left [ 31^-25^2\right ]$

$W=\frac{69\times 10^{-3}}{2}\left [ 961-625\right ]$

$W=11.59 J$

5 0

3 years ago

1. What is the momentum of a 0.15 kg arrow that is traveling at 120 m/s?

Korvikt [17]

For this case we have that by definition, the momentum equation is given by:

$p = m * v$

Where:

m: It is the mass

v: It is the velocity

According to the data we have:

$m = 0.15 \ kg\\v = 120 \frac {m} {s}$

Substituting:

$p = 0.15 * 120\\p = 18 \frac {kg * m} {s}$

On the other hand, if we clear the variable "mass" we have:

$m = \frac {p} {v}$

According to the data we have:

$p = 250 \frac {kg * m} {s}\\v = 5 \frac {m} {s}\\m = \frac {250} {5}\\m = 50 \ kg$

Thus, the mass is $50 \ kg$

Answer:

$p = 18 \frac {kg * m} {s}\\m = 50 \ kg$

3 0

3 years ago

Two forces P and Q act on an object of mass 15.0 kg with Q being the larger of the two forces. When both forces are directed to

Rom4ik [11]

Answer:

P=6.25N and Q=16.25N

Explanation:

In order to solve this problem we must first draw a free body diagram for both situation, (see attached picture).

Now, we need to analyze the two free body diagrams. So let's analyze the first diagram. Since the body is accelerated, then the sum of forces is equal to mass times acceleration, so we get:

$\Sigma F=ma$

We can assume there will be only the two mentioned forces P and Q, so

the sum of forces will be:

P+Q=ma

$P+Q=(15kg)(1.50m/s^{2})$

P+Q=22.5N

We can do the same analysis for the second free body diagram:

$\Sigma F=ma$

$Q-P=(15kg)(0.7m/s^{2})$

Q-P=10.5N

so now we have a system of equations we can solve by elimination:

Q+P=22.5N

Q-P=10.5N

Now, we can add the two equations together so the P force is eliminated, so we get:

2Q=32.5N

now we can solve for Q:

$Q=\frac{32.5N}{2}$

Q=16.25N

Now we can use any of the equations to find P.

Q+P=22.5N

P=22.5N-Q

when substituting for Q we get:

P=22.5N-16.25N

P=6.25N

5 0

4 years ago