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Nookie1986 [14]
2 years ago
6

Michael is doing an experiment. He wants to drop a bowling ball and a stuffed bear out the window of a tall building. Under what

conditions would they have the same acceleration?
Physics
1 answer:
Anon25 [30]2 years ago
6 0

Answer:

If there was no air resistance

Explanation:

We know that free fall is a unique motion in which gravity only works on one object. Objects that are said to be free-falling do not experience a significant force of air resistance; They come under the sole effect of gravity. Under such conditions, all objects fall under the same acceleration, regardless of their mass.

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For an object with a given mass on Earth, calculate the weight of the object with the mass equal in magnitude to the number repr
leonid [27]

<u>Answer:</u> The weight of the object is 29.4 N

<u>Explanation:</u>

To calculate the weight of the object, we use the equation:

W=m\times g

where,

m = mass of the object = 3 kg

g = acceleration due to gravity = 9.8m/s^2

Putting values in above equation, we get:

W=3kg\times 9.8m/s^2\\\\W=29.4N

Hence, the weight of the object is 29.4 N

6 0
3 years ago
Provide descriptions and examples of the following forces: friction, gravity and elastic.
arsen [322]

Answer:

Friction:-

The friction force is the force exerted by a surface as an object moves across it or makes an effort to move across it. There are at least two types of friction force - sliding and static friction. Though it is not always the case, the friction force often opposes the motion of an object. For example, if a book slides across the surface of a desk, then the desk exerts a friction force in the opposite direction of its motion. Friction results from the two surfaces being pressed together closely, causing inter molecular attractive forces between molecules of different surfaces. As such, friction depends upon the nature of the two surfaces and upon the degree to which they are pressed together. The maximum amount of friction force that a surface can exert upon an

EG:-

A coaster sliding against a table.

Gravity:-

The force of gravity is the force with which the earth, moon, or other massively large object attracts another object towards itself. By definition, this is the weight of the object. All objects upon earth experience a force of gravity that is directed "downward" towards the center of the earth. The force of gravity on earth is always equal to the weight of the object as found

EG:-

The force that causes a car to coast downhill even when you aren't stepping on the gas.

Elastic:-

Elasticity is the ability of a material to return to its original shape after being stretched or compressed. When an elastic material is stretched or compressed, it exerts elastic force. This force increases the more the material is stretched or compressed.

EG:-

An archer's stretched bow

5 0
3 years ago
A laser is placed at the point $(3,5)$. The laser beam travels in a straight line. Larry wants the beam to hit and bounce off th
Black_prince [1.1K]

Answer:

4units

Explanation:

To calculate the total distance the beam will travel along this path, you will use the formula for calculating the distance between two coordinates expressed as;

D = √(x2-x1)²+(y2-y1)²

Given the coordinate points

(3,5) and (7,5)

Substitute

D = √(7-3)²+(5-5)²

D = √(7-3)²+0²

D = √4²

D = √16

D = 4

Hence the total distance the beam will travel along this path is 4units

6 0
3 years ago
A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar
denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of \beta

Using formula of \beta

\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}

Put the value into the formula

\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}

\beta=8.86\times10^{9}

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}

Put the value into the formula

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}

T=5.45\times10^{-4}

(b). We need to calculate the tunnel probability for width 1.00 nm

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}

T=7.74\times10^{-8}

Hence, The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

6 0
3 years ago
A driver of a 1900 kg car traveling with 460,000 J of kinetic energy must slam on the brakes in order to avoid hitting a deer in
elena-s [515]
Uhhh about 40 miles probably so round that
7 0
2 years ago
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