Ask question

Ask question

All categories

anyanavicka [17]

3 years ago

7

The releasing of an object from certain height to ground due to gravity is called____ A, Free falling B, Projectile C, curved li

near D,para pola

1 answer:

MArishka [77]3 years ago

7 0

Answer:

projectile

Explanation:

when an object is released from certain height to ground due to gravity is called projectile

You might be interested in

Consider the following equilibrium at 979 K for the dissociation of molecular iodine into atoms of iodine. I2(g) equilibrium rea

GenaCL600 [577]

Answer: $I_2$ = 0.0050 M

$I$ = 0.0155 M

Explanation:

Initial moles of $I_2$ = 0.072 mole

Volume of container = 3.9 L

Initial concentration of $I_2=\frac{moles}{volume}=\frac{0.072moles}{3.9L}=0.018M$

The given balanced equilibrium reaction is,

$I_2(g)\rightleftharpoons 2I(g)$

Initial conc. 0.018 M 0

At eqm. conc. (0.018-x) M (2x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[I]^2}{[I_2]}$

$K_c=\frac{(2x)^2}{0.2-x}$

we are given : $K_c=1.60\times 10^{-3}$

Now put all the given values in this expression, we get :

$1.60\times 10^{-3}=\frac{(2x)^2}{(0.018-x)}$

$x=0.0025$

So, the concentrations for the components at equilibrium are:

$[I]=2\times x=2\times 0.0025=0.0050$

$[I_2]=0.018-x=0.018-0.0025=0.0155$

Hence, concentrations of $I_2$ and $I$ are 0.0050 M ad 0.0155 M respectively.

4 0

4 years ago

In the one pulley system when you move the Mass from The 20-centimeter Mark to the 15 centimeter mark it moves to my 5cm how far

hoa [83]

Answer:

5 cm

Explanation:

Remember that in the pulley system the rope moves the same distance in both ends of the pulley, what the pulley system does is creating a mechanical advantage which basically means that it takes less effort to pull than the actual effort it would take, so for example if you have to lift a box that is 800 N, with a pulley system and the mechanical advantage you'd have to pull with less force.

8 0

3 years ago

4. A driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch,

stira [4]

Answer:

The driver's average velocity is 82.35 km/h.

Explanation:

Given:

The motion of the driver can be divided into 3 parts:

i. Displacement of the driver in 1.5 hours = 135 km

ii. Rest for 45 minutes.

iii. Displacement in next 2 hours = 215 km

The direction of motion remains same (east).

Now, total displacement of the driver is, $D_{Total}=135+215=350$ km.

Rest time is 45 minutes. Converting it to hours, we need to use the conversion factor $1\textrm{ min} = \frac{1}{60}$ hour.

So, 45 minutes in hours is equal to $\frac{45}{60}=0.75$ hours.

Now, total time taken for the complete journey is, $\Delta t=1.5+\frac{45}{60}+2=1.5+0.75+2=4.25\textrm{ h}$

Average velocity is given as:

$v_{avg}=\frac{\textrm{Total displacement}}{Total time}=\frac{350}{4.25}=82.35\textrm{ km/h}$

Therefore, the driver's average velocity is 82.35 km/h

4 0

3 years ago

Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's

mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

$T^2\alpha R^3\\T^2=kR^3.......................(1)$

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

$\frac{T^2}{R^3}=k.......................(2)$

Let the orbital period of the earth be $T_e$ and its mean distance of from the sun be $R_e$ .

Also let the orbital period of the planet be $T_p$ and its mean distance from the sun be $R_p$ .

Equation (2) therefore implies the following;

$\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)$

We make the period of the planet $T_p$ the subject of formula as follows;

$T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)$

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

$R_p=16R_e...............(5)$

Substituting equation (5) into (4), we obtain the following;

$T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}$

$R_e^3$ cancels out and we are left with the following;

$T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)$

Recall that the orbital period of the earth is about 365.25 days, hence;

$T_p=64*365.25\\T_p=23376days$

4 0

3 years ago

a ball is thrown striaght up in the air and then falls back to earth. if the downward fall takes 2.2s, how fast is the ball trav

lapo4ka [179]

The velocity of the ball when it strikes the ground, given the data is 21.56 m/s

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

Time to reach ground from maximum height (t) = 2.2 s
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Final velocity (v) =?

<h3>How to determine the velocity when the ball strikes the ground</h3>

The velocity of the ball when it strikes the ground can be obtained as illustrated below:

v = u + gt

v = 0 + (9.8 × 2.2)

v = 0 + 21.56

v = 21.56 m/s

Thus, the velocity of the ball when it strikes the ground is 21.56 m/s

Learn more about motion under gravity:

brainly.com/question/22719691

#SPJ1

5 0

2 years ago