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2 years ago

Calculate the potential energy of 5.0g (0.005 kg) paper airplane 5.0 meters above the ground

Physics

1 answer:

Readme [11.4K]2 years ago

4 0

PE = (mass) (gravity) (height)

PE = (0.005 kg) (9.8 m/s²) (5 m)

<em>PE = 0.245 Joule</em>

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Answer the question with step.

Maru [420]

Answer:

f1/f2 =W1/W2 = 1/3

.0 f2 = 3f1

As ,

1/F= 1/f1 +1/f2

...1/40 = 1/f1 - 1/3f1

f1=> 80/3 cm

... f2 = 2f1 = 3 x 80/3 = 80 cm

7 0

3 years ago

Durante uma corrida de carros, o piloto da escuderia ganhadora percorreu a primeira volta da pista com velocidade media de 310 k

lesya692 [45]

A velocidade mínima é 0. A velocidade máxima é inferior ou igual a 620 km/h.
<span>buena suerte mi amigo</span>

3 0

3 years ago

Which statement below best describes the relationship between the voltage, resistance, and current?

aliina [53]

Answer:

As voltage increases, current increases and resistance stays the same .

Explanation:

Ohm's law gives the relationship between the voltage, resistance, and current. The mathematical form of Ohm's law is given by :

$V\propto I\\\\V=IR$

R is resistance

I is current

V is voltage

So, as voltage increases, current increases and resistance stays the same. The correct option is (A).

4 0

3 years ago

A large box slides across a frictionless surface with a velocity of 12 m/s and a mass of 4

denpristay [2]

Answer= 8m/s

Because total Momentum before= total momentum after

Momentum before (p=mu)
p=(4)(12)= 48
p=2(0)=0
So total momentum before=48

Momentum after (p=mu)
Masses combined —2+4=6kg
p=6u

Mb=Ma
48=6u
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3 0

2 years ago

Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both

Nuetrik [128]

Answer:

Part a)

Width of the slit is

$a = 580 nm$

Part b)

Ratio of intensity is given as

$\frac{I}{I_o} = 0.81$

Explanation:

Part a)

As we know by the formula of diffraction we will have

$a sin\theta = \lambda$

so we have

$\theta = 90$

$\lambda = 580 nm$

so we will have

$a sin90 = 580 nm$

$a = 580 nm$

Part b)

As we know that the intensity in diffraction pattern is given as

$I = I_o (\frac{sin\theta}{\theta})^2$

$\frac{I}{I_o} = (\frac{sin\theta}{\theta})^2$

so for angle 45 degree

$\frac{I}{I_o} = (\frac{sin45}{\pi/4})^2$

$\frac{I}{I_o} = 0.81$

7 0

3 years ago