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Nataly_w [17]
3 years ago
13

The relationship between sharks and remoras is characterizedTRUE    ORFALSE

Physics
1 answer:
Damm [24]3 years ago
8 0
False, sharks and remoras have a symbiotic relationship. The remora removes parasites from the sharks scales, and the shark provides protection for the remora
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The smallest unit which maintains the physical properties of a compound is a(n) _____.
Elodia [21]
A molecule, would be the correct answer
8 0
3 years ago
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A 1900kg airplane is flying at an altitude of 510 above the ground. What is the gravitational potential energy in Joules?
Natalija [7]

Answer: 9496200 joules

Explanation:

Gravitational potential energy, GPE is the energy possessed by the moving plane since it moves against gravity.

Thus, GPE = Mass m x Acceleration due to gravity g x Height h

Since Mass = 1900kg

g = 9.8m/s^2

h = 510 metres (units of height is metres)

Thus, GPE = 1900kg x 9.8m/s^2 x 510m

GPE = 9496200 joules

Thus, the gravitational potential energy of the airplane is 9496200 joules

8 0
3 years ago
A 50 kg student climbs 3 m to the top of a set of stairs.
VashaNatasha [74]
1) PE=mgh
mass 50 kg; height 3; g 9.81
50×3×9.81=1471.5 J

2. The student is opposing gravity force so 
Fgrav=m×g
50 × 9.81= 490N
Work=force×displacement
490×3=1470J(this should always be the same as the potential energy)

3.Power=work×time
4410 W=1470×3

4. Greater than: power is dependent on work, work is dependent on force, force is dependent on weight, 

3 0
3 years ago
Which one is Transverse Wave?
nekit [7.7K]

Answer:a

Explanation:

7 0
3 years ago
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In a physics lab experiment, a compressed spring launches a 32 g metal ball at a 35 degree angle compressing the spring 18 cm ca
Free_Kalibri [48]

Ball hits the floor at 1.6 m below the initial position

So here we can say that

\Delta y = -1.6 m

also it will hit at horizontal distance of 5.2 m

\Delta x = 5.2 m

let say its velocity in x and y directions are given as

v_x ,v_y

now we can say

v_x* t = 5.2

-1.6 = v_y * t - \frac{1}{2}gt^2

from above two equations

-1.6 = v_y* \frac{5.2}{v_x} - 4.9t^2

as we know that it is projected at an angle of 35 degree

so we know that

v_y = v_x tan35

-1.6 = 5.2 tan35 - 4.9 t^2

4.9 t^2 = 3.64 + 1.6 = 5.24

t = 1.03 s

now we have

v_x * 1.03 = 5.2

v_x = 5.03 m/s

also we can find vertical component of velocity as

v_y = v_x tan35 = 3.52 m/s

now we will find net velocity as

v^2 = v_x^2 + v_y^2

v^2 = 5.03^2 + 3.52^2

v = 6.14 m/s

now by energy conservation we can say

\frac{1}{2}kx^2 = \frac{1}{2}mv^2

k*(0.18)^2= 0.032*(6.14)^2

k = 37.2 N/m

so spring constant is 37.2 N/m

5 0
3 years ago
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