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ikadub [295]
2 years ago
9

A force of 6.00 newtons on a car's bumper makes an angle of 28.5 degrees with the ground. What is the vertical component of this

force vector?
Physics
1 answer:
Zielflug [23.3K]2 years ago
7 0
Your answer by using the following calculation"

6sin(28.5) = 2.862
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A frog is at the bottom of a 17-foot well. Each time the frog leaps, it moves up 3 feet. If the frog has not reached the top of
docker41 [41]

Answer:

The frog takes 8 jumps to reach top of well

Explanation:

Given data

Frog at bottom=17 foot

Each time frog leaps 3 feet

Frog has not reached the top of the well, then the frog slides back 1 foot

To Find

Total number of leaps the frog needed to escape from well

Solution

in 1 jump distance jumped=3+(-1)

                                           =2 feet

                                           =2×1 feet

The "-1" is because the frog goes back

Now After 2 jumps the distance jumped as:

                     Distance Jumped=2+2

                     Distance Jumped=2*2

                                                   =4 feet

Similarly after 7 jumps

                    Distance Jumped=2+2+......+2

                    Distance Jumped=2*7

                                                 =14 feet

Now after 8th jump the frog climbs but doesnot slide back as it is reached to the top of well.

So

              Distance Jumped=(Distance Jumped after 7 jumps)+3

                                           =14+3

                                           =17 feet

The frog takes 8 jumps to reach top of well                

7 0
3 years ago
a bubble of air of volume 1cm^3 is released by a deep sea diver at a depth where the pressure is 4.0 atmospheres. assuming its t
lys-0071 [83]

Answer:

hope this helps!

Explanation:

Volume of the air bubble, V1=1.0cm3=1.0×10−6m3

Bubble rises to height, d=40m

Temperature at a depth of 40 m, T1=12oC=285K

Temperature at the surface of the lake, T2=35oC=308K

The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa 

The pressure at the depth of 40 m: P1=1atm+dρg

Where,

ρ is the density of water =103kg/m3

g is the acceleration due to gravity =9.8m/s2

∴P1=1.103×105+40×103×9.8=493300Pa

We have T1P1V1=T2P2V2

Where, V2 is the volume of the air bubble when it reaches the surface.

V2=

8 0
2 years ago
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