Answer:
direct current
Explanation:
it has a direct path to go down to reach the specific point
Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.
Explanation:
- Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s
- Let
be the speed of the river's current given as 1.00 m/s.
- Note that this speed is the magnitude of the velocity which is a vector quantity.
- The direction of the swimmer is upstream.
Hence the resultant velocity is given as, 
= S — S 0
= 1.25 — 1
= 0.25 m/s.
Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.
Answer:
1000N,4.48m/s
Explanation:
Height(h) = 1m
P. E = 1000J
P. E = mass x acceleration due to gravity x height
Acceleration due to gravity = 10m/s
P. E = mgh
Where weight = mg
P. E = weight x height
Weight = P. E / height
Weight = 1000 / 1
Weight. 1000N
b) as he leaves the ground P. E = K. E
P. E = 1/2 mv2
P. E = 1000 , mass = weight / 10
Mass =1000/10 = 100kg
P. E = 1/2 mv2
1000 = 1/2 x 100 x v2
Multiply both sides by 2
1000 x 2 = 100 x v2
V2 = 2000/ 100
V2 =20
Square root both sides
V = square root of 20
V = 4.48m/s
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Answer:
300000.01008 Pa
123.76237 m/s²
Explanation:
= Density of liquid nitrogen = 808 kg/m³
h = Depth
g = Acceleration due to gravity
P = Atmospheric pressure
Absolute Pressure is given by
Below 2 m from surface
Below 5 m from surface
Subtracting the above equations we get
The acceleration due to gravity on the planet is 123.76237 m/s²
Equating the value of g in the first equation
The atmospheric pressure on the planet is 300000.01008 Pa
If an icy surface means no friction, then Newton's second law tells us the net forces on either block are
• <em>m</em> = 1 kg:
∑ <em>F</em> (parallel) = <em>mg</em> sin(45°) - <em>T</em> = <em>ma</em> … … … [1]
∑ <em>F</em> (perpendicular) = <em>n</em> - <em>mg</em> cos(45°) = 0
Notice that we're taking down-the-slope to be positive direction parallel to the surface.
• <em>m</em> = 0.4 kg:
∑ <em>F</em> (vertical) = <em>T</em> - <em>mg</em> = <em>ma</em> … … … [2]
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Adding equations [1] and [2] eliminates <em>T</em>, so that
((1 kg) <em>g</em> sin(45°) - <em>T </em>) + (<em>T</em> - (0.4 kg) <em>g</em>) = (1 kg + 0.4 kg) <em>a</em>
(1 kg) <em>g</em> sin(45°) - (0.4 kg) <em>g</em> = (1.4 kg) <em>a</em>
==> <em>a</em> ≈ 2.15 m/s²
The fact that <em>a</em> is positive indicates that the 1-kg block is moving down the slope. We already found the acceleration is <em>a</em> ≈ 2.15 m/s², which means the net force on the block would be ∑ <em>F</em> = <em>ma</em> ≈ (1 kg) (2.15 m/s²) = 2.15 N directed down the slope.
