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2 years ago

You are riding in an elevator that is accelerating upward. Suppose you stand on a scale. The reading on the scale is __________.

Physics

2 answers:

enyata [817]2 years ago

6 0

Less than true weight

Darina [25.2K]2 years ago

5 0

Answer: A

Explanation: Your body gets pulled down, causing ur body to have more weight but not mass.

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4) A racing car undergoing constant acceleration covers 140 m in 3.6 s. (a) If it’s moving at 53 m/s at the end of this interval

Scorpion4ik [409]

Answer

given,

distance = 140 m

time, t = 3.6 s

moving speed = 53 m/s

a) distance = (average velocity) x time

$D = \dfrac{v_0 + v_1}{2}\times t$

$140 = \dfrac{v_0 + 53}{2}\times 3.6$

v₀ + 53 = 77.78

v₀ = 24.78 m/s or 25 m/s

b) $a = \dfrac{v-u}{t}$

$a = \dfrac{53-25}{3.6}$

a = 7.8 m/s²

using equation of motion

v₀² = v₁² + 2 a s

53² = 0²+ 2 x 7.8 x s

s = 180 m

3 0

3 years ago

A scientist notices that an oil slick floating on water when viewed from above has many different rainbow colors reflecting off

Semenov [28]

Answer:

Explanation:

In case of oil slick a thin layer of oil is formed on water . This thin layer creates a rainbow of colour . The phenomenon is due to interference of light waves , one reflected from the upper surface of oil and the other reflected from the lower surface of the oil.

For formation of bright colour

2 μ t = ( 2n + 1 ) λ / 2

μ is refractive index of oil , t is thickness of oil layer λ is wave length of light falling on the layer .

given μ = 1.2 , λ = 750 x 10⁻⁹ ,

2 x 1.2 t = ( 2n + 1 ) 750 x 10⁻⁹ / 2

For minimum thickness n = 0

2.4 t = 375 x 10⁻⁹

t = 156.25 n m

B ) If the refractive index of layer of medium below oil is less than that of oil , the condition of formation of colour changes

The new condition is

2 μ t = n λ

2 x 1.5 t = 750 nm , n = 1 for minimum wavelength .

t = 250 nm

C ) Light mostly transmitted means dark spot is formed at that point .

For that to be observed from water side , the condition is

2 μ t = ( 2n + 1 ) λ / 2

λ = 4μ t / ( 2n + 1 )

For maximum wavelength n = 0

λ = 4μ t

= 4 x 1.5 x 200 nm

= 1200 nm .

3 0

3 years ago

What happens when a proton is placed directly in the path of the proton cannon?

blondinia [14]

Answer:coherent light

Explanation:

5 0

3 years ago

One of your summer lunar space camp activities is to launch a 1130 kg1130 kg rocket from the surface of the Moon. You are a seri

maxonik [38]

Answer:

∆U = 2.296×10^10Joules

Explanation:

Gravitational potential energy is defined as the energy possessed by an object under the influence of gravity due to its virtue of position.

Potential energy U = Fr where;

F is the force of attraction between the masses of the moon and the rocket.

r is the radius or height of the object.

From Newton's law of universal gravitation, F = GMm/r²

Potential energy U = (-GMm/r²)×r

Potential energy U = -GMm/r

The force is negative because the objects act upward.

M is the mass of the rocket

m is the mass of the moon

Gravitational potential energy possessed by the rocket

U1 = -GMm/r1

r1 is the altitude covered by the rocket

Gravitational potential energy possessed by the Moon

U2 = -GMm/(r2+r1)

r2 is the radius of the moon

Change in gravitational potential energy ∆U = U2-U1

∆U = -GMm/(r2+r1)-(-GMm/r1)

∆U = -GMm/(r2+r1) + GMm/r1

∆U = -GMm{1/(r2+r1)-1/r1}

Given

G = 6.67×10^-11m³/kgs²

M = 1130kg

m = 7.36×10²²kg

r1 = 215km = 215,000m

r2 = 1740km = 1,740,000m

∆U = -6.67×10^-11× 7.36×10²² × 1130{1/(215,000+1,740,000)-1/215000}

∆U= -55.47×10¹⁴{1/1955000-1/215000}

∆U = -55.47×10¹⁴{5.12×10^-7 - 4.65×10^-6}

∆U = -284×10^7 + 257.94×10^8

∆U = 22,954,000,000Joules

∆U = 2.296×10^10Joules

8 0

3 years ago

A converging lens of focal length 20 cm is placed in contact with, and to the left of, a diverging lens of focal length 30 cm. I

scZoUnD [109]

Answer:

Magnification will be equal to 3

Explanation:

We have given focal length of the converging lens $F_1=20cm$

Focal length of the diverging lens $F_2=30cm$

Object is placed 40 cm to the length of the converging lens d = 40 cm

Combination of the focal length will be equal to

$\frac{1}{F}=\frac{1}{F_1}+\frac{1}{F_2}$

$\frac{1}{F}=\frac{1}{20}+\frac{1}{-30}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}$

F = 60 cm

So combination of the focal length will be 60 cm

Magnification is given by

$M=\frac{F}{F-d}=\frac{60}{60-40}=3$

So magnification will be equal to 3

3 0

3 years ago