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hjlf
3 years ago
14

A rabbit runs 4.4 m across a lawn, stops, then runs 2.2 m back in the opposite direction. What is the rabbit’s displacement from

its starting point?
Physics
1 answer:
Alla [95]3 years ago
6 0

Answer:

it is 2.2 m

Explanation:

because he goes back 2.2 m so 4.4 minus 2.2 equals 2.2

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Which are causes of mechanical weathering? (check all that apply)
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The correct answers are:

B. plant growth;

C. animal actions;

The mechanical weathering is a type of weathering where physical force is included into the breaking up of the rocks. The plants and the animals are both causing this type of weathering with their actions. The plants can cause mechanical weathering with their roots, as they grow and surround a rock, they are able to create such a pressure that they can break the rock apart. Also, as their trunks are getting bigger, if there's rocks right next to them, the pressure from the growing of the trunk will crack the rocks. The animals are able to move the rocks, as well as pushing them, or even deliberately throwing them, so they manage to break up parts of them and cause mechanical weathering.

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4 years ago
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A 0.6 kg ball is initially at rest on a frictionless, horizontal surface. It is struck by a 0.4 kg ball initially moving with a
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Answer:

140

Explanation:

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3 years ago
What is a major contributor of greenhouse gases in the atmosphere?
ASHA 777 [7]

Answer:

a. burning of fossil fuel

Explanation:

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A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra
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Explanation:

Terminal velocity is given by:

v_t=\sqrt{\frac{2mg}{\rho C_dA}}

Here, m is the mass of the falling object, g is the gravitational acceleration,  C_d is the drag coefficient, \rho is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg

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v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}

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3 years ago
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