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nikklg [1K]
3 years ago
9

Which of Newton’s laws of accounts for the following statement?

Physics
1 answer:
Neko [114]3 years ago
4 0
The answer is B


second law
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Một xe đang di chuyển với tốc độ 36km/h thì gia tốc và sau 2s xe lên tới tốc độ 54km/h. Tính gia tốc của xe trong 2s ?
yulyashka [42]

Answer:

which language is these

Explanation:

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please tell us that it is which language. From that we will slove this

3 0
3 years ago
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Nicole's office is 5 m wide and 3 m long. What is the area?
Sloan [31]
Area is calculated as length times width.
In this case, the width is 5 m and the length is 3 m.
5 • 3 = 15
So the area of Nicole's office is 15 square meters.
Square meters because only length and width is 2 dimensional.

Hope this helps!
6 0
3 years ago
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A force of 8,480 N is applied to a cart to accelerate it at a rate of 26.5 m/s2. What is the mass of the cart?
katrin2010 [14]

Answer:

<h2>320</h2>

Just use F =ma where m is mass and a is acc.

8 0
3 years ago
The Sun orbits the center of the Milky Way galaxy once each 2.60 × 108 years, with a roughly circular orbit averaging 3.00 × 104
Mamont248 [21]

To solve this problem it is necessary to apply the kinematic equations of linear and angular motion, as well as the given definitions of the period.

Centripetal acceleration can be found through the relationship

a_c = \frac{v^2}{R}

Where

v = Tangential Velocity

R = Radius

At the same time linear velocity can be expressed in terms of angular velocity as

v = R\omega

Where,

R = Radius

\omega = Angular Velocity

PART A) From this point on, we can use the values used for the period given in the exercise because the angular velocity by definition is described as

\omega = \frac{2\pi}{T}

T = Period

So replacing we have to

\omega = \frac{2\pi}{2.6*10^8years}\\\omega = 2.4166*10^{-8}rad/years\\\omega = 2.4166*10^{-8}rad/years(\frac{1years}{365days})(\frac{1day}{86400s})\\\omega = 7.663*10^{-16}rad/s

Since 1 Light year = 9.48*10^{15}m

Then the radius in meters would be

R = (3*10^4ly)(\frac{9.48*10^{15}m}{1ly})

R = 2.844*10^{20}m

Then the centripetal acceleration would be

a_c = \frac{v^2}{R}\\a_c = \frac{(R\omega)^2}{R}\\a_c = R\omega^2 \\a_c = 2.844*10^{20}(7.663*10^{-16})^2\\a_c = 1.67*10^{-10}m/s^2

From the result obtained, considering that it is an unimaginably low value of an order of less than 10^{-10} it is possible to conclude that it supports the assertion on the inertial reference frame.

8 0
3 years ago
What would happen if you took a really strong metal box and you started teleporting in hundreds and hundreds of small cotton bal
mojhsa [17]
It will not explode since the mass of the cotton balls is so low but rather will most likely break the lock and hinges and come out.
4 0
3 years ago
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