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3 years ago

Can someone help me with accerleration

Physics

2 answers:

dangina [55]3 years ago

6 0

Answer:

acceleration is the rate of change of the velocity of an object with respect to time. Accelerations are vector quantities. The orientation of an object's acceleration is given by the orientation of the net force acting on that object.

SI unit: m/s2, m·s−2, m s−2

Georgia [21]3 years ago

4 0

Answer:

Explanation:

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A car speed off around a bend at a constant 10m/s explain why it's velocity is not constant

VARVARA [1.3K]

It's velocity is not constant as direction is changing.

We know, velocity is speed with direction, so if direction is changing, velocity can't be constant, doesn't matter that speed is constant.

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3 years ago

When electrons are removed from the outermost shell of a calcium atom, the atom becomesA. an anion that has a larger radius than

Free_Kalibri [48]

Answer:

D. a cation that has a smaller radius than the atom.

Explanation:

When electrons are removed from the outermost shell of a calcium atom, the atom becomes a cation that has a smaller radius than the atom.

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3 years ago

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15. A locomotive moved 18.0 m [W] in a time of 6.00 s and stopped. After stopping, the

mariarad [96]

Answer:

Distance = 30m

Displacement = 6m W

Explanation:

Given the following:

Movement 1 = 18m W

Movement 2 = 12m E

Diatance is a scalar quantity with only magnitude and no direction. That is, in Calculating the distance moved by the locomotive, the direction of travel or movement of the object is not considered. It only measures the total amount of movement made during the Time of motion.

Therefore, total distance traveled equals :

Movement 1 + movement 2

18m + 12m = 30m

B) Displacement also measures the movement made by an object. However, Displacement is a vector quantity and therefore, considers both magnitude and direction of travel of the object. Therefore, it measures the overall change in position of the object from its starting position.

Therefore, Displacement of the locomotive equals:

18m W - 12m E = 6m E

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3 years ago

Charges q3 and q5 are now replaced by two charges, q2 and q6, having equal magnitude and sign (-3.4 μC). Charge q2 is located at

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3 years ago

Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the

Evgen [1.6K]

Answer:

g(h) = g ( 1 - 2(h/R) )

<em>*At first order on h/R*</em>

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

In order to do this, we must expand the newton's law of universal gravitation around r=R

Remember that this law is:

$F = G \frac{m_1m_2}{r^2}$

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

$F = m_2a$

Therefore, we can see that

$a(r) = G \frac{m_1}{r^2}$

With a the acceleration due to the earth's mass.

Now, the taylor series is going to be (at first order in h/R):

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}$

a(R) is actually the constant acceleration at sea level

and

$a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}$

Therefore:

$a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})$

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}$

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3 years ago

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