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Free_Kalibri [48]
3 years ago
11

How many moles of MgCl2 are there in 329 g of the compound? Your Answer:

Physics
2 answers:
Nookie1986 [14]3 years ago
6 0

Answer:3.46

Explanation:

mass=329

molar mass=71

no. of moles=329/71

=3.46

Debora [2.8K]3 years ago
5 0

Answer:3.46

Explanation:

mass=329

molar mass=71

no. of moles=329/71

=3.46

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25% part (c) assume that d is the distance the cheetah is away from the gazelle when it reaches full speed. Derive an expression
levacccp [35]

maximum speed of cheetah is

v_1 = v_{max}

speed of gazelle is given as

v_2 = v_{g}

Now the relative speed of Cheetah with respect to Gazelle

v_{12} = v_1 - v_2

v_{12} = v_{max} - v_g

now the relative distance between Cheetah and Gazelle is given initially as "d"

now the time taken by Cheetah to catch the Gazelle is given as

d = v_{12}* t

so by rearranging the terms we can say

t = \frac{d}{v_{12}}

t = \frac{d}{v_{max} - v_g}

so above is the relation between all given variable

6 0
3 years ago
For a transverse wave, what is a wavefront?
vichka [17]

Explanation:

A wavefront is the long edge that moves, for example, the crest or the trough. Each point on the wavefront emits a semicircular wave that moves at the propagation speed v. These are drawn at a time t later, so that they have moved a distance s = vt.

4 0
3 years ago
Read 2 more answers
A baseball is launched horizontally from a height of 1.8 m. The baseball travels 0.5 m before hitting the ground. How fast is th
zysi [14]

Answer:

0.83 m/s

Explanation:

FIrst of all, we have to find the time of flight, i.e. the time the baseball needs to reach the ground. This can be done by using the equation for the vertical motion:

h=ut+\frac{1}{2}gt^2

where

h is the initial height

u = 0 is the initial vertical velocity

g = 9.8 m/s^2 is the acceleration of gravity

t is the time

Substituting h = 1.8 m and solving for t,

t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2(1.8)}{9.8}}=0.61 s

We know that the horizontal distance travelled by the ball is

d = 0.5 m

Therefore, we can find the horizontal velocity (which is constant during the whole motion):

v= \frac{d}{t}=\frac{0.5}{0.60}=0.83 m/s

4 0
3 years ago
A 15.0 kg crate, initially at rest, slides down a ramp 2.0 m long and inclined at an angle of 20.0° with the horizontal. Using t
Elis [28]

The component of the crate's weight that is parallel to the ramp is the only force that acts in the direction of the crate's displacement. This component has a magnitude of

<em>F</em> = <em>mg</em> sin(20.0°) = (15.0 kg) (9.81 m/s^2) sin(20.0°) ≈ 50.3 N

Then the work done by this force on the crate as it slides down the ramp is

<em>W</em> = <em>F d</em> = (50.3 N) (2.0 m) ≈ 101 J

The work-energy theorem says that the total work done on the crate is equal to the change in its kinetic energy. Since it starts at rest, its initial kinetic energy is 0, so

<em>W</em> = <em>K</em> = 1/2 <em>mv</em> ^2

Solve for <em>v</em> :

<em>v</em> = √(2<em>W</em>/<em>m</em>) = √(2 (101 J) / (2.0 m)) ≈ 10.0 m/s

3 0
3 years ago
PLEASE ASAP ILL GIVE BRAINLIEST.
Sergeu [11.5K]

Answer:

law of gravity

Explanation:

cause the ball was still moving

6 0
3 years ago
Read 2 more answers
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