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Inessa [10]
3 years ago
8

______ help keep plants from being eaten by animals.

Physics
2 answers:
yanalaym [24]3 years ago
8 0
Thorns help keep plants from being eaten by animals.


Vera_Pavlovna [14]3 years ago
7 0
Thorns and plants is what protects them
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What process is occurring in the picture?
Alex777 [14]

Two fat black arrows are swimming along together, when they see a single skinny black arrow coming toward them. They are afraid of strangers, and they know that the skinny one must be mean and tough if it's not afraid to travel alone. So they turn to the side and get out of the skinny arrow's way.

3 0
3 years ago
An egg rolls off a kitchen counter and breaks as it hits the floor. The counter is 1.0 m high, the mass of the egg is about 50 g
crimeas [40]
Use equations of motion to find the velocity just before it hits the floor: 
<span>Vf^2 = Vi^2 + 2gx </span>
<span>Final velocity = 4.42m/s </span>

<span>Impulse is change in momentum so: </span>
<span>m(Vf - Vi) = 0.05(0 - 4.42) </span>
<span>= - 0.221 kg.m/s

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

</span>
3 0
3 years ago
What is the kinetic energy of an object that has a mass of 12 kilograms and moves with a velocity of 10 m/s?
Alik [6]
1/2mv^2

1/2x12x10^2=600J

The kinetic energy is 600J
3 0
3 years ago
Read 2 more answers
You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an
a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, \mu=6\times 10^{-2}

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

KE=PE+W

\dfrac{1}{2}mv^2=mgh+W

W is the work done by the friction.

W=f\times d

f=\mu mg\ cos\theta

W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}

W=\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}

\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}

v = 8.19 m/s

So, the speed of the box is 8.19 m/s. Hence, this is the required solution.

8 0
3 years ago
Professional Application: A 30,000-kg freight car is coasting at 0.850 m/s with negligible friction under a hopper that dumps 11
Nataliya [291]

Answer:

0.182 m/s

Explanation:

m1 = 30,000 kg, m2 = 110,000 kg, u1 = 0.85 m/s

let the velocity of loaded freight car is v

Use the conservation of momentum

m1 x u1 + m2 x 0 = (m1 + m2) x v

30,000 x 0.85 = (30,000 + 110,000) x v

v = 0.182 m/s

5 0
3 years ago
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