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labwork [276]
3 years ago
9

Define average velocity ​

Physics
2 answers:
maxonik [38]3 years ago
7 0
Hello! mark me brainliest please

The average speed of an object is defined as the distance traveled divided by the time elapsed. Velocity is a vector quantity, and average velocity can be defined as the displacement divided by the time.
dolphi86 [110]3 years ago
4 0
HELLO,!! MARK ME AS EXPERT

THE AVERAGE VELOCITY OF AN OBJECT IS DEFINED AS THE DISTANCE TRAVELED DIVIDED BY THE TIM ELAPSED.VELOCITY IS A VECTOR QUANTITY,AND AVRAGE VELOCITY CAN BE DEFINED AS THE DISPLACEMENT DIVIDE BY THE TIME
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Yea, gonna need some help. Thanks
natulia [17]

Answer:

t = 3.48 s

Explanation:

The time for the maximum height can be calculated by taking the derivative of height function with respect to time and making it equal to zero:

h(t) = -16t^2+v_ot+h_o\\\\\frac{dh(t)}{dt}=0=-32t+v_o\\\\v_o = 32t

where,

v₀ = initial speed = 110 ft/s

Therefore,

110 = 32t\\\\t = \frac{110}{32}\\\\

<u>t = 3.48 s</u>

8 0
2 years ago
A package is dropped from an air balloon two times. In the first trial the distance between the balloon and the surface is Hand
enyata [817]

Answer:

<em>The final speed of the second package is twice as much as the final speed of the first package.</em>

Explanation:

<u>Free Fall Motion</u>

If an object is dropped in the air, it starts a vertical movement with an acceleration equal to g=9.8 m/s^2. The speed of the object after a time t is:

v=gt

And the distance traveled downwards is:

\displaystyle y=\frac{gt^2}{2}

If we know the height at which the object was dropped, we can calculate the time it takes to reach the ground by solving the last equation for t:

\displaystyle t=\sqrt{\frac{2y}{g}}

Replacing into the first equation:

\displaystyle v=g\sqrt{\frac{2y}{g}}

Rationalizing:

\displaystyle v=\sqrt{2gy}

Let's call v1 the final speed of the package dropped from a height H. Thus:

\displaystyle v_1=\sqrt{2gH}

Let v2 be the final speed of the package dropped from a height 4H. Thus:

\displaystyle v_2=\sqrt{2g(4H)}

Taking out the square root of 4:

\displaystyle v_2=2\sqrt{2gH}

Dividing v2/v1 we can compare the final speeds:

\displaystyle v_2/v_1=\frac{2\sqrt{2gH}}{\sqrt{2gH}}

Simplifying:

\displaystyle v_2/v_1=2

The final speed of the second package is twice as much as the final speed of the first package.

5 0
3 years ago
A 10.0-kg box starts at rest on a level floor. An external, horizontal force of 2.00 × 102 N is applied to the box for a distanc
Harman [31]

Answer:

vf = 11.2 m/s

Explanation:

m = 10 Kg

F = 2*10² N

x = 4.00 m

μ = 0.44

vi = 0 m/s

vf = ?

We can apply Newton's 2nd Law

∑ Fx = m*a   (→)

F - Ffriction = m*a  ⇒  F - (μ*N) = F - (μ*m*g) = m*a   ⇒  a = (F - μ*m*g)/m

⇒    a = (2*10² N - 0.44*10 Kg*9.81 m/s²)/10 Kg = 15.6836 m/s²

then , we use the equation

vf² = vi² + 2*a*x    ⇒    vf = √(vi² + 2*a*x)

⇒   vf = √((0)² + 2*(15.6836 m/s²)*(4.00m)) = 11.2 m/s

7 0
3 years ago
Ionic compounds _____.
aleksandr82 [10.1K]

Explanation:

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<em><u>i</u></em><em><u> </u></em><em><u>hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>

6 0
3 years ago
What are the differences between a magnetic and gravitational interaction
GREYUIT [131]
Classically, gravitational attraction is due to the masses of objects. It is inversely proportional to the square of the distances between the objects.

Magnetic attraction is due to the magnetism in objects which is due to uncompensated electron spins in certain atoms.  The force due to magnetism less easy to put into one equation than gravity since magnetic fields can have different shapes, but the simplest one (the dipolar field) is inversely proportional to the cube of the distance between the magnetic dipoles.
3 0
3 years ago
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