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4 years ago

What formula is needed to calculate the acceleration of object in motion?

Physics

2 answers:

Butoxors [25]4 years ago

6 0

Answer:

Acceleration = Force/Mass

Explanation:

mixer [17]4 years ago

3 0

Answer: A = v - u

——-

Explanation: acceleration is the rate of change of velocity with respect to time.

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What is the fundamental unit for measuring mass in the metric system ?

goldenfox [79]

Kilogram(kg)
It's not the SI unit of mass in the metric system however.

6 0

3 years ago

Read 2 more answers

A charge q is at the point x = 5.0 m , y = 0. Write expressions for the unit vectors you would use in Coulomb's law if you were

Flauer [41]

Answer:

A) $\^r = \^y$

B) $\^r = -\^x$

C) $\^r = -0.14\^x + 0.99\^y$

Explanation:

Coulomb's Law is

$\vec{F}_{ab} = K\frac{q_1q_2}{r^2}\^r$

where r is the distance between the two point charges.

The question clearly asks the unit vector expressions of the force that q exerts on the other charge. So, we do not need to find the force, but only the distance vector, r, between charges, and then we can derive the unit vector pointing only the direction of the force.

A) $\vec{r} = \vec{r}_b - \vec{r}_a = (5\^x + \^y) - (5\^x + 0) = \^y\\\^r = \frac{\vec{r}}{|\vec{r}|} = \frac{\^y}{1} = \^y$

B) $\vec{r} = \vec{r}_b - \vec{r}_a = (0 + 0) - (5\^x + 0) = -5\^x\\\^r = \frac{\vec{r}}{|\vec{r}|} = \frac{-5\^x}{5} = -\^x$

C) $\vec{r} = \vec{r}_b - \vec{r}_a = (4.5\^x + 3.5\^y) - (5\^x + 0) = -0.5\^x + 3.5\^y\\\^r = \frac{\vec{r}}{|\vec{r}|} = \frac{-0.5\^x + 3.5\^y}{\sqrt{(0.5)^2+(3.5)^2}} = \frac{-0.5\^x + 3.5\^y}{3.53} = -0.14\^x + 0.99\^y$

7 0

3 years ago

Read 2 more answers

A water pipe having a 4.00 cm inside diameter carries water into the basement of a house at a speed of 1.00 m/s and a pressure o

posledela

Answer:

$8.16\ \text{m/s}$

Explanation:

$d_1$ = Initial diameter = 4 cm

$v_1$ = Initial velocity = 1 m/s

$d_2$ = Final diameter = 7.8 m

$v_2$ = Final velocity

$A$ = Area = $\pi\dfrac{d^2}{4}$

From the continuity equation we get

$A_1v_1=A_2v_2\\\Rightarrow \pi\dfrac{d_1^2}{4}v_1=\pi\dfrac{d_2^2}{4}v_2\\\Rightarrow v_2=\dfrac{d_1^2}{d_2^2}v_1\\\Rightarrow v_2=\dfrac{4^2}{1.4^2}\times 1\\\Rightarrow v_2=8.16\ \text{m/s}$