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Reptile [31]
3 years ago
13

What homemade things could I use for wheels on a small balloon powered car?

Physics
2 answers:
Fantom [35]3 years ago
8 0

You can use mostly anything as long as it is circular. Depending on how big it is, you could use sturdy paper plates and use a stick/rod and tape to hold it together, or you could use bottle caps if the car you are trying to make is really small.

Brilliant_brown [7]3 years ago
8 0
CD disks can be used too.
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Look at the figure, which shows the motion of three balls. The curved paths followed by balls B and C are examples of _____. pro
Natali [406]

Answer:did you get the answer?

Explanation:

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3 years ago
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When a current flows in an aluminum wire of diameter 2.91 mm 2.91 mm , the drift speed of the conduction electrons is 0.000191 m
charle [14.2K]

Answer:

Number of electrons are flowing per second is 2.42 x 10¹⁹

Explanation:

The electric current flows through a wire is given by the relation :

I=envA   ....(1)

Here I is current, e is electronic charge, v is drift velocity of electrons and A is the Area of the wire.

But electric current is also define as rate of electrons passing through junction times their charge, i.e. ,

I=Ne      ....(2)

Here N is the rate of electrons passing through junction.

From equation (1) and (2).

eN = envA

N=nvA

But area of wire, A=\pi \frac{d^{2} }{4}

Here d is diameter of wire.

So, N = nv\pi \frac{d^{2} }{4}

Substitute 2.91 x 10⁻³ m for d, 0.000191 m/s for v and 6 x 10²⁸ m⁻³ for n in the above equation.

N = 6\times10^{28}\times 0.000191\times\pi \frac{(2.91\times10^{-3} )^{2} }{4}

N = 2.42 x 10¹⁹ s⁻¹  

8 0
2 years ago
Jane walked 5.00 meters on a road that inclines 13.0 degrees. How much distance did she cover horizontally?
Paladinen [302]
Any options available?
8 0
3 years ago
What activities should you avoid to prevent static discharge while working on a computer?
nordsb [41]
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6 0
3 years ago
A boy pulls a sled of mass 5.0 kg with a rope that makes a 60.0° angle with respect to the horizontal surface of a frozen pond.
Oduvanchick [21]

Answer:

μk = 0.124

Explanation:

Known data

m=5.0 kg : mass of the sled

T= 10 N   : force with which the boy pulls the rope

θ =60.0°  :angle of the rope with respect to the horizontal direction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law to the sled :

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

Forces acting on the sled

W: Weight of the sled : In vertical and downward direction

N : Normal force : In vertical and upwards direction

f : Friction force: parallel to the movement of the sled and in the opposite direction to the movement

T:Rope tension : forming angle 60.0° of  of the rope with respect to the horizontal direction

Calculated of the W  of the sled

W= m*g

W=  5.0 kg* 9.8 m/s² = 49 N

x-y  components  of the tension of the rope  T

Tx= 10*cos60°= 5 N

Ty=  10*sin60° = 8.66 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay    ay = 0

N+Ty -W = 0

N = 49 N  -  8.66 N

N = 40.34 N

Calculated of the f

f = μk* N

f = μk* 40.34 Equation (1)

We apply the formula (1) to calculated f

∑Fx = m*ax  the sled moves with constant velocity, then ax=0

∑Fx = 0

Tx-f = 0

5 - f = 0

f =  5N

We replace f in the equation (1)

5 = μk* 40.34

μk = 5 / 40.34

μk = 0.124

5 0
3 years ago
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