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3 years ago

an object moving with a speed of 5 m/s has a kinetic energy of 100 J. what is the mass of the object?

Physics

1 answer:

elixir [45]3 years ago

5 0

Answer:40kg

Explanation:

Velocity(v)=5m/s

Kinetic energy(ke)=100j

mass=?

Mass=(2xke) ➗ v^2

Mass=(2x500) ➗ 5^2

Mass=1000 ➗ 5x5

Mass=1000 ➗ 25

Mass=40kg

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Light from a helium-neon laser (λ=633nm) passes through a circular aperture and is observed on a screen 4.0 m behind the apertur

devlian [24]

Answer:

d = 0.247 mm

Explanation:

given,

λ = 633 nm

distance from the hole to the screen = L = 4 m

width of the central maximum = 2.5 cm

2 y = 0.025 m

y = 0.0125 m

For circular aperture

$sin \theta = 1.22\dfrac{\lambda}{d}$

using small angle approximation

$\theta = \dfrac{y}{D}$

now,

$\dfrac{y}{D} = 1.22\dfrac{\lambda}{d}$

$y = 1.22\dfrac{\lambda\ D}{d}$

$d = 1.22\dfrac{\lambda\ D}{y}$

$d = 1.22\dfrac{633\times 10^{-9}\times 4}{0.0125}$

d =0.247 x 10⁻³ m

d = 0.247 mm

the diameter of the hole is equal to 0.247 mm

5 0

3 years ago

What is the process called in which water vapor and carbon dioxide retain heat?

saw5 [17]

The process in which water vapour and carbon dioxide traps heat is called the “greenhouse effect”.

The greenhouse effect is a natural phenomenon which occurs every day. To illustrate an example of this natural phenomenon, during the day the Sun shines through the atmosphere. Earth's surface warms up because of the sunlight. Meanwhile at night in the absence of the sunlight, Earth's surface cools back and releasing the heat back into the air. However some of the heat is retained by the greenhouse gases (such as carbon dioxide and water vapour) in the atmosphere. This process what keeps our planet Earth warm and cozy at an average temperature of 16°C.

Answer:

greenhouse effect

6 0

3 years ago

irius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Ano

Norma-Jean [14]

The actual distance of Regulus from Earth is 23.81 parsecs.

Given:

Parallax of Regulus, p = 0.042 arc seconds

Calculation:

When an observer changes their position, an apparent change in the object's position takes place. This change can be calculated using the angle ( or semi-angle) made by the observer and object i.e. the angle made between the two lines of observation from the object to the observer.

Thus from the relation of parallax of a celestial body we get:

S = 1/ tan p ≈ 1 / p

where S is the actual distance between the object and the observer

p is the parallax angle observed

Here for Regulus, we get:

S = 1 / p

= 1 / (0.042) [ 1 parsecs = 1 arcseconds ]

= 23.81 parsecs

We know that,

1 parsecs = 3.26 light-years = 206,000 AU

Converting the actual distance into light years we get:

23.81 parsecs = 23.81 × (3.26 light yrs) = 77.658 light-years

Therefore, the actual distance of Regulus from Earth is 23.81 parsecs which is 77.658 in light years.

Learn more about astronomical units here:

brainly.com/question/16471213

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6 0

1 year ago

Awdfwkjbfgkjsenfkjnsekjfgnesklnslkenges

forsale [732]

Answer:

yes

Explanation:

3 0

3 years ago

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A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi

slega [8]

We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is
$dI =r^2*dm$
where $dm =M(b*dr)/(ab)$
Now we find the moment of inertia by integrating from $-a/2$ to $a/2$
The moment of inertia is
$I= \int\limits^{-a/2}_{a/2} {r^2*dm} = M \int\limits^{-a/2}_{a/2} r^2(b*dr)/(ab)=(M/a)(r^3/3)$ (from (-a/2) to $I=(M/3a)(a^3/8 +a^3/8)=(Ma^2)/12$ (a/2))

4 0

3 years ago