What’s the weight and how high is the clif
Answer:28 m
Explanation:
Given
Direction is 
North of east i.e. 
with x axis
Also ball moved by 33 m
therefore its east component is 33cos58=17.48 m
Northward component 
<u>Answer:</u>
Both the objects A and B will have the same acceleration.
<u>Explanation
:</u>
The objects will have the same acceleration as both are under free fall condition. When objects are under the free fall condition, the only force that acts on the object is its weight.
Weight is the force acting on a body of some mass, and the formula for finding the weight of a body is- Weight = mass × acceleration due to gravity(g).
Therefore, here the different weight is due to the difference masses of both bodies, and not due to the different acceleration values.
The weightiness of the added
water displaced is equivalent to the joined weight of the two extra people who come
to be into the boat:
<span>m water g = 2 x 690 N</span>
<span> =
1,380 N</span>
<span>
</span>
The mass of the water displace
is then
<span>m water g = 1,380 N</span>
<span> = 1,380 N / 9.8 m/s^2</span>
<span> = 141 kg</span>
<span>
</span>
Compute the calculation for
density for the volume of water displace and practice this outcome for the mass
of the water displace to get the answer:
<span>p water = mass of water / volume of water</span>
<span>
</span>
<span>volume of water = mass of water / p water</span>
<span> = 141 kg / 1000 kg /m^3 eliminate
kilogram</span>
<span> = 0.14 m^3 the additional volume
of water that is displaced</span>
Answer:
P_2 = 1.62 atm
Explanation:
We know the formula for the rms speed of the ideal gas is given by
P= pressure of the surrounding
V= volume of the vessel
m= mass of the gas
Now, From this formula rms speed (v_rms) is directly proportional to square root is pressure.
Then
given that v_rsm,1= v0
and v_rsm,2=0.9v0
putting these values we get
P_2 = 1.62 atm
