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malfutka [58]
3 years ago
8

A 3.0-kilogram mass is traveling in a circle of

Physics
1 answer:
dmitriy555 [2]3 years ago
4 0

Answer:

a=20\ m/s^2

Explanation:

Given that,

The mass of an object, m = 3 kg

The radius of a circle, r = 0.2 m

The speed of the object, v = 2 m/s

We need to find the centripetal acceleration. Its formula is given by :

a=\dfrac{v^2}{r}\\\\a=\dfrac{(2)^2}{0.2}\\\\a=20\ m/s^2

So, the centripetal acceleration is 20\ m/s^2.

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Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but have a much smaller dia
Crank

Answer:

Wn = 9.14 x 10¹⁷ N

Explanation:

First we need to find our mass. For this purpose we use the following formula:

W = mg

m = W/g

where,

W = Weight = 675 N

g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²

m = Mass = ?

Therefore,

m = (675 N)/(9.8 m/s²)

m = 68.88 kg

Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:

gn = (G)(Mn)/(Rn)²

where,

gn = acceleration due to gravity on surface of neutron star = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg

Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m

Therefore,

gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)

gn = 13.27 x 10¹⁵ m/s²

Now, my weight on neutron star will be:

Wn = m(gn)

Wn = (68.88)(13.27 x 10¹⁵ m/s²)

<u>Wn = 9.14 x 10¹⁷ N</u>

3 0
3 years ago
The Venus flytrap is known for which of these behaviors?
tensa zangetsu [6.8K]
Snapping a leaf shut around an insect, I think.
8 0
2 years ago
Define rectilinear propagation of light.??​
Tju [1.3M]

Answer:

it relates to the light propensity to travel over one straight line without having any interference in its trajectory

Explanation:

5 0
3 years ago
Read 2 more answers
What are the directions of movement for the Sagittarius, frontal, transverse planes ?
Dima020 [189]

Answer:

Squats involve flexion (forward motion) and extension (backward on the way up), so would fit into the sagittal plane. Frontal plane motion would include leaning from left to right as in sidebends and lateral raises, or perhaps you might picture jumping jacks for a good image of movement along the frontal plane.

5 0
3 years ago
Can any kind soul help me​
Wittaler [7]

Answer:

2917.4 m/s

Explanation:

From the question given above, the following data were:

Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth

Radius (r) of the Moon = 1737 Km

Escape velocity (v) =?

Next, we shall determine the gravitational acceleration of the Moon. This can be obtained as follow:

Gravitational acceleration of the earth = 9.8 m/s²

Gravitational acceleration of the Moon (g) = 0.25 times the gravitational acceleration of the earth

= 0.25 × 9.8 = 2.45 m/s²

Next, we shall convert 1737 Km to metres (m). This can be obtained as follow:

1 Km = 1000 m

Therefore,

1737 Km = 1737 Km × 1000 m / 1 Km

1737 Km = 1737000 m

Thus, 1737 Km is equivalent to 1737000 m

Finally, we shall determine the escape velocity of the rocket as shown below:

Gravitational acceleration of the Moon (g) = 2.45 m/s²

Radius (r) of the moon = 1737000 m

Escape velocity (v) =?

v = √2gr

v = √(2 × 2.45 × 1737000)

v = √8511300

v = 2917.4 m/s

Thus, the escape velocity is 2917.4 m/s

4 0
3 years ago
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