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3 years ago

What is a robin is it a herbivores or carnivores or omnivores or decompers

1 answer:

8 0

A robin is a small carnivore bird which feed on insects, earthworms, spiders and bees. It needs about 60 calories per day from these organisms. This bird has a short lifespan of only 7 years. Hope this answers the question. Have a nice day.

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-09utydrbfshjklo iuohyfdgxfncj

adoni [48]

Answer:

thanks

Explanation:

5 0

2 years ago

A water line with an internal radius of 5.29 x 10-3 m is connected to a shower head that has 15 holes. The speed of the water in

fiasKO [112]

Answer:

(a) 3.44 x 10^-3 m^3/s

(b) 8.4 m/s

Explanation:

area of water line, A = 5.29 x 10^-3 m

number of holes, N = 15

Speed of water in line, V = 0.651 m/s

(a) Volume flow rate is given by

V = area of water line x speed of water in water line

V = 5.29 x 10^-3 x 0.651 = 3.44 x 10^-3 m^3/s

(b) area of one hole, a = 4.13 x 10^-4 m

Let v be the velocity of water in each hole

According to the equation of continuity

A x V = a x v

5.29 x 10^-3 x 0.651 = 4.1 x 10^-4 x v

v = 8.4 m/s

5 0

3 years ago

QUESTION 3

Alisiya [41]

The force of frictions is opposed to relative motion.

The acceleration of the crate is approximately 2.937 m/s².

Reason:

The given parameters are;

The mass of the wood, m = 100 kg

The force which can move the wood, F = 588 N

Wood on wood static friction, $\mu_s$ = 0.5

Wood on wood kinetic friction, $\mu_k$ = 0.3

Solution;

The force of friction, $F_f$ , acting when the crate is moving is given as

follows;

$F_f$ = m × g × $\mu_k$

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore, we have;

$F_f$ = 100 × 9.81 × 0.3 = 294.3

The force of friction, $F_f$ = 294.3 N

The force with which the crate moves, F = 588 - 294.3 = 293.7

The force with which the crate moves, F = 293.7 N

Force = Mass, m × Acceleration, a

$a = \dfrac{F}{m}$

Therefore;

$a = \dfrac{293.7 \ N}{100 \ kg} = 2.937$

The acceleration of the crate, a ≈ 2.937 m/s².

Learn more about friction here:

brainly.com/question/94428

8 0

2 years ago

hink about the friction activity. Return to this activity to see it again if you need to. Explain what happened in each of the t

ankoles [38]

from what i know the force of gravity and its effects on objects, regardless of their mass, fall to the ground at the same rate.

7 0

3 years ago

The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge

lesya [120]

Answer:

Loss, $\Delta E=-10.63\ J$

Explanation:

Given that,

Mass of particle 1, $m_1=m =0.66\ kg$

Mass of particle 2, $m_2=7.4m =4.884\ kg$

Speed of particle 1, $v_1=2v_o=2\times 6=12\ m/s$

Speed of particle 2, $v_2=v_o=6\ m/s$

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

$m_1v_1+m_2v_2=(m_1+m_2)V$

$V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}$

$V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}$

V = 6.71 m/s

Initial kinetic energy before the collision,

$K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)$

$K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)$

$K_i=135.43\ J$

Final kinetic energy after the collision,

$K_f=\dfrac{1}{2}(m_1+m_2)V^2$

$K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2$

$K_f=124.80\ J$

Lost in kinetic energy,

$\Delta K=K_f-K_i$

$\Delta K=124.80-135.43$

$\Delta E=-10.63\ J$

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0

3 years ago