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3 years ago

Does sunlight really take 8 minutes to reach your eyes?

Physics

2 answers:

Leona [35]3 years ago

4 0

Yes,it can if you look straight at the sun

Brums [2.3K]3 years ago

3 0

It takes sunlight 8 minutes to reach earth , so yes

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A roller coaster car rapidly picks up speed as it rolls down a slope. as it starts down the slope, its speed is 4 m/s. but 3 sec

Arisa [49]

Vi = 4 m/s
vf = 22 m/s
t = 3s
a = ?

vf = vi + a * t
vf - vi = a * t
a = (vf - vi) / t
a = (22 - 4) / 3
a = 6 m / s^2

8 0

3 years ago

A 100 kg marble slab falls off a skyscraper and falls 200 m to the ground without hitting anyone. Its fall stops within millisec

Radda [10]

Answer:

Δ T = 2.28°C

Explanation:

given,

mass of marble = 100 Kg

height of fall = 200 m

acceleration due to gravity = 9.8 m/s²

C_marble = 860 J/(kg °C)

using conservation of energy

Potential energy = heat energy

$m g h = m C_{marble}\Delta T$

$g h =C_{marble}\Delta T$

$\Delta T= \dfrac{g h}{C_{marble}}$

$\Delta T= \dfrac{9.8 \times 200}{860}$

Δ T = 2.28°C

7 0

4 years ago

Mass is a measure of weight. True False

8_murik_8 [283]

Answer: False

Explanation: Mass is a measure of the amount of matter in an object.

5 0

3 years ago

Read 2 more answers

A 1.00 kg object is attached to a horizontal spring. the spring is initially stretched by 0.500 m, and the object is released fr

valina [46]

The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is
$T=2 t = 2 \cdot 0.100 s = 0.200 s$
Which means that the frequency is
$f= \frac{1}{T}= \frac{1}{0.200 s}=5 Hz$
and the angular frequency is
$\omega=2 \pi f = 2 \pi (5 Hz)=31.4 rad/s$

In a spring-mass system, the maximum velocity of the object is given by
$v_{max} = A \omega$
where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
$v_{max} = A \omega = (0.500 m)(31.4 rad/s)= 15.7 m/s$

6 0

3 years ago

You stand on a merry-go-round which is spinning at f = 0:25 revolutions per second. You are R = 200 cm from the center. (a) Find

ivanzaharov [21]

Answer:

(a) ω = 1.57 rad/s

(b) ac = 4.92 m/s²

Explanation:

(a)

The angular speed of the merry go-round can be found as follows:

ω = 2πf

where,

ω = angular speed = ?

f = frequency = 0.25 rev/s

Therefore,

ω = (2π)(0.25 rev/s)

ω = 1.57 rad/s

(b)

The centripetal acceleration can be found as:

ac = v²/R

but,

v = Rω

Therefore,

ac = (Rω)²/R

ac = Rω²

therefore,

ac = (2 m)(1.57 rad/s)²

ac = 4.92 m/s²

(c)

In order to avoid slipping the centripetal force must not exceed the frictional force between shoes and floor:

Centripetal Force = Frictional Force

m*ac = μs*R = μs*W

m*ac = μs*mg

ac = μs*g

μs = ac/g

μs = (4.92 m/s²)/(9.8 m/s²)

μs = 0.5

7 0

3 years ago