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PolarNik [594]
3 years ago
10

Gravity affects atmospheric pressure true or false

Physics
1 answer:
Burka [1]3 years ago
5 0

Answer:

false

Explanation:

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Which of the following are located on a plant's leaves?
valina [46]

i think its stomata....

4 0
2 years ago
In a Broadway performance, an 77.0-kg actor swings from a R = 3.65-m-long cable that is horizontal when he starts. At the bottom
krek1111 [17]

Answer: h =1.22 m

Explanation:

from the question we were given the following

mass of performer ( M1 ) = 77 kg

length of cable ( R ) = 3.65 m

mass of costar ( M2 ) = 55 kg

maximum height (h) = ?

acceleration due to gravity (g) = 9.8 m/s^2  (constant value)

We first have to find the velocity of the performer. From the work energy theorem work done = change in kinetic energy

work done = 1/2 x mass x ( (final velocity)^2 - (initial velocity)^2 )

initial velocity is zero in this case because the performer was at rest before swinging, therefore

work done = 1/2 x 77 x ( v^2 - 0)

work done = 38.5 x ( v^2 ) ......equation 1

work done is also equal to m x g x distance ( the distance in this case is the length of the rope), hence equating the two equations we have

m x g x R =  38.5 x ( v^2 )

77 x 9.8 x 3.65 =  38.5 x ( v^2 )

2754.29 = 38.5 x ( v^2 )

( v^2 ) =  71.54

v = 8.4 m/s  ( velocity of the performer)

After swinging, the performer picks up his costar and they move together, therefore we can apply the conservation of momentum formula which is

initial momentum of performer (P1) + initial momentum of costar (P2) = final momentum of costar and performer after pick up (Pf)  

momentum = mass x velocity therefore the equation above now becomes

(77 x 8.4) + (55 x 0) = (77 +55) x Vf  

take note the the initial velocity of the costar is 0 before pick up because he is at rest

651.3 = 132 x Vf

Vf = 4.9 m/s

the performer and his costar is 4.9 m/s after pickup

to finally get their height we can use the energy conservation equation for from after pickup to their maximum height. Take note that their velocity at maximum height is 0

initial Kinetic energy + Initial potential energy = Final potential energy + Final Kinetic energy

where

kinetic energy = 1/2 x m x v^2

potential energy  = m x g x h

after pickup they both will have kinetic energy and no potential energy, while at maximum height they will have potential energy and no kinetic energy. Therefore the equation now becomes

initial kinetic energy = final potential energy

(1/2 x (55 + 77) x 4.9^2) + 0 = ( (55 + 77) x 9.8 x h) + 0

1584.7 = 1293 x h

h =1.22 m

3 0
4 years ago
N april 1974, steve prefontaine completed a 10-km race in a time of 27 min, 43.6 s. suppose \"pre\" was at the 7.15-km mark at a
aliya0001 [1]

solution:speed is 7.99km ,u=\frac{7990}{(25\times60)}=5.33m/s\\
now let the final velocity after 60s from 7.99km is v.\\
v-u =at\\v=5.33+a\times60\\  =60s+5.33\\now distance travelled during this time will be\\
s_{2}=vt\\
v\times103.6\\
=(60a+5.33)\times103.6\\
=551.84+6216am\\
now total distance travelled  is \\  also\\
s_{1}=ut+(\frac{1}{2})at^{2}      =5.33\times60+(0.5)9\times3600\\
=1800a+319.6m\\
Now the remaining time to complete rest race was\\
t=(27\times60+43.6)60-60(25\times60)\\
=103.6s\\Now, distance travelled during this time will be\\
s_{2}=vt\\
v\times103.6\\
=(60a+5.33)\times103.6\\
=551.84+6216am\\
now total distance travelled  is \\
s_{1}+s_{2}+7990=10000\\
1800a+319.6+551.84+6216a=2010\\
8016a=2010-871.44\\
8016a=1138.56\\
a=0.142m/s^2

6 0
3 years ago
What is the weight of a 225kg space probe on the moon and the acceleration of gravity on the moon is 1.62
olga nikolaevna [1]
<span>364N should be your answer.. hope this helps

</span>
7 0
3 years ago
Read 2 more answers
A helicopter is ascending vertically witha speed of 5.40 m/s. At a height of 105 m above the earth a package is dropped from the
Mrac [35]

Answer: 5.21 s

Explanation:

Given

Helicopter ascends vertically with u=5.4\ m/s

Height of helicopter h=105\ m

When the package leaves the helicopter, it will have the same vertical velocity

Using equation of motion

\Rightarrow h=ut+\dfrac{1}{2}at^2\\\\\Rightarrow 105=-5.4t+0.5\times 9.8t^2\\\Rightarrow 4.9t^2-5.4t-105=0\\\\\Rightarrow t=\dfrac{5.4\pm \sqrt{5.4^2+4\times 4.9\times 105}}{2\times 4.9}\\\\\Rightarrow t=\dfrac{5.4\pm 45.68}{9.8}\\\\\Rightarrow t=5.21\ s\quad \text{Neglect negative value}

So, package will take 5.21 s to reach the ground

4 0
3 years ago
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