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4 years ago

What is the only organism feed off other living organisms? Please help! WILL GIVE 50 POINTS!!!

Physics

1 answer:

Bad White [126]4 years ago

8 0

A carnivore is the only animal that feeds off of other animals

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The mass of the sun is 2*10^30 kg and its radius is

Marta_Voda [28]

Explanation:

Distance d=1.5×108 km=1.5×1011 m

Mass of the sun, m=2×1030 kg

Mass of the earth, M=6×1024 kg

Force of gravitation, F=G×d2m×M

F=6.7×10−11×(1.5×1011)22×1030×6×1024=3.57×1022 N

4 0

3 years ago

A rock is suspended by a light string. When the rock is in air, the tension in the string is 51.9 N . When the rock is totally i

Luden [163]

Answer:

$\rho _{liquid}=1995.07kg/m^{3}$

Explanation:

When the rock is immersed in unknown liquid the forces that act on it are shown as under

1) Tension T by the string

2) Weight W of the rock

3) Force of buoyancy due to displaced liquid B

For equilibrium we have $T_{3}+B = W_{rock}$

$T_{3}+\rho _{Liquid}V_{rock}g$ = $W_{rock}.....(\alpha)$

When the rock is suspended in air for equilibrium we have

$T_{1}=W_{rock}....(\beta)$

When the rock is suspended in water for equilibrium we have

$T_{2}$ + $\rho _{water}V_{rock}g$ = $W_{rock}.....(\gamma)$

Using the given values of tension and solving α,β,γ simultaneously for $\rho _{Liquid}$ we get

$W_{rock}=51.9N\\31.6+1000\times V_{rock}\times g=51.9N\\\\11.4+\rho _{liquid}V_{rock}g=51.9N\\\\$

Solving for density of liquid we get

$\rho _{liquid}=\frac{51.9-11.4}{51.9-31.6}\times 1000$

$\rho _{liquid}=1995.07kg/m^{3}$

5 0

3 years ago

A 4.0×1010kg asteroid is heading directly toward the center of the earth at a steady 16 km/s. To save the planet, astronauts str

blondinia [14]

Answer:

a. t = 69.4 hr = 2.89 days

b. theta = 91.67*10^-3 degrees

c. deflection_angle = 0.134 degrees

Explanation:

a).

The asteroid impacts the earth in t

t = x/v = (4.0*10^6 km)/(16 km/sec)

t = 2.5 * 10^5 sec

t = 69.4 hr = 2.89 days

b).

tan(theta) = 6400 km/(4.0*10^6 km)

tan(theta) = 1.6*10^-3

theta = arctan(1.6*10^-3)

theta = 1.6*10^-3 radians (for small angles, tan(theta) ~= theta)

theta = 91.67*10^-3 degrees

c).

v_minimum = 6400 km/(2.5 * 10^5 sec)

v_minimum = 25.6 m/s

Using F = m*a, we can calculate the acceleration of the asteroid due to the rocket's thrust:

5.0*10^9 N = 4.0*10^10 kg * a

a = (5.0*10^9 N)/(4.0*10^10 kg)

a = 0.125 m/s^2

The transverse velocity after 300 seconds of this acceleration is:

v_transverse = a*t = 0.125 m/s^2 * 300 s

v_transverse = 37.5 m/s = 37.5*10^-3 km/s

tan(deflection_angle) = v_transverse/(20 km/s)

tan(deflection_angle) = (37.5*10^-3 km/s)/(16 km/s) = 2.34^-3

deflection_angle = arctan(2.34*10^-3)

deflection_angle = 2.34*10^-3 radians = 0.134 degrees

v_transverse/(16 km/s) > (6400km)/(5.0*10^6 km)

(note that the right hand side if this inequality is tan(theta) calculated above)

v_transverse > 23.704 m/

8 0

3 years ago

A student wanted to investigate whether or not the temperature of the water would affect the solubility of sugar. She had three

erastova [34]

ind variable ... temp

dep variable .... amount of sugar dissolved

5 0

3 years ago

A ray of light bends upon entering a new medium. What is this phenomenon called?

hoa [83]

Ii believe it's a refraction

4 0

4 years ago

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