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Scrat [10]
3 years ago
13

I need science fair ideas for 4th grade, But i need something that won't need a oven, sunlight, ETC. Also, i need something that

isn't Common (example: volcano) but still simple!
Physics
1 answer:
Lina20 [59]3 years ago
5 0
Something simple?
Hmm, how about a diagram of a fod chain? Its simple, if you dont have a printer to print the images of a food chain, or animals, then just try to draw it the best you can...
Make it into a poster, write some information about the animals, who eats who, are they a carnivor, herbivor, you know... Stuff like that :)
~.~.~.~.~.~.~.~.
<em>Hope this helped! :)</em>
You might be interested in
In an inelastic collision, as compared to an elastic collision, what is to be expected?.
motikmotik
The loss or conservation of kinetic energy is the difference between an elastic and an inelastic collision. Kinetic energy is not preserved in an inelastic collision, and it will change forms into sound, heat, radiation, or another form. The kinetic energy in an elastic collision is preserved and does not change forms.
3 0
2 years ago
A child whose weight is 287 N slides down a 7.20 m playground slide that makes an angle of 31.0° with the horizontal. The coeff
natulia [17]

Answer:

a

H  =212.6 \  J

b

v  =  7.647  \  m/s

Explanation:

From the question we are told that

   The child's weight is  W_c  =  287 \ N

    The length of the sliding surface of the playground is  L =  7.20 \  m

    The coefficient of friction is  \mu =  0.120

      The angle is \theta = 31.0 ^o

      The initial  speed is  u =  0.559 \  m/s

Generally the normal force acting on the child is mathematically represented as

=>    N  =  mg  *  cos \theta

Note  m *  g  =  W_c

Generally the frictional force between the slide and the child is    

         F_f  =  \mu *  mg  *  cos \theta

Generally the resultant force acting on the child due to her weight and the frictional  force is mathematically represented as

      F =m* g sin(\theta) - F_f

Here  F is the resultant force and it is represented as  F =  ma

=>   ma =   m* g sin(31.0)  - \mu *  mg  *  cos (31.0)

=>   a =  g sin(31.0)-  \mu *  g  *  cos (31.0)

=>  a =    9.8 *  sin(31.0) - 0.120 *  9.8  *  cos (31.0)

=>a =  4.039 \ m/s^2

So

   F_f  =  0.120  * 287  *  cos (31.0)

=> F_f  = 29.52 \  N

Generally the heat energy generated by the frictional  force which equivalent tot the workdone by the frictional force  is mathematically represented as

     H  =  F_f  * L

=>  H  = 29.52 *  7.2

=>  H  =212.6 \  J

Generally from kinematic equation we have that

    v^2  =  u^2  +  2as

=>  v^2  =  0.559^2  +  2 * 4.039 * 7.2

=>  v  =  \sqrt{0.559^2  +  2 * 4.039 * 7.2}

=>  v  =  7.647  \  m/s

   

6 0
3 years ago
Difference between displacement and vector quantity
daser333 [38]

The only 'difference' is that they are different categories.

It's like asking "What's the difference between Susie and girl ?"

Or "What's the difference between Cadillac and car ?"

Displacement <em>IS</em> a vector quantity.

5 0
3 years ago
What is the average velocity of atoms in 1.00 mol of neon (a monatomic gas) at 465 K? For m, use 0.0202 kg.
TiliK225 [7]

Answer: 757m/s

Explanation:

Given the following :

Mole of neon gas = 1.00 mol

Temperature = 465k

Mass = 0.0202kg

Using the ideal gas equation. For calculating the average kinetic energy molecule :

0.5(mv^2) = 3/2 nRt

Where ;

M = mass, V = volume. R = gas constant(8.31 jK-1 mol-1, t = temperature in Kelvin, n = number of moles

Plugging our values

0.5(0.0202 × v^2) = 3/2 (1 × 8.31 × 465)

0.0101 v^2 = 5796.225

v^2 = 5796.225 / 0.0101

v^2 = 573883.66

v = √573883.66

v = 757.55109m/s

v = 757m/s

5 0
3 years ago
The Moon orbits Earth in an average of p = 27.3 days at an average distance of a =384,000 kilometers. Using Newton’s version of
Korvikt [17]

Answer:

The mass of the earth, M=6.023\times 10^{24}\ kg

Explanation:

It is given that,

Time taken by the moon to orbit the earth, T=27.3\ days=2358720\ m

Distance between moon and the earth,r=384000\ km=384\times 10^6\ m

We need to find the mass of the Earth using Kepler's third law of motion as :

T^2=\dfrac{4\pi^2}{GM}r^3

M=\dfrac{4\pi^2r^3}{T^2G}

M=\dfrac{4\pi^2\times (384\times 10^6)^3}{(2358720)^2\times 6.67\times 10^{-11}}

M=6.023\times 10^{24}\ kg

So, the mass of the earth is 6.023\times 10^{24}\ kg. Hence, this is the required solution.

7 0
3 years ago
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