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mafiozo [28]
1 year ago
7

A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, take

n as the xy plane. The 4.20-kg puck has a velocity of 1.60î m/s at one instant. Eight seconds later, its velocity is (6.00î + 4.0ĵ) m/s. (a) Assuming the rocket engine exerts a constant horizontal force, find the components of the force.

Physics
1 answer:
adell [148]1 year ago
6 0

Answer:

F_{x}=2.31N to the right.

F_{y}=2.1N to in the upwards direction.

Explanation:

In order to solve this problem, we must first start by drawing a diagram of the situation. (See attached diagram).

So, remember that a force is determined by multiplying the mass of the parcticle by its acceleration:

F=ma

so in order to find the components of the force, we need to start by finding its acceleration.

Acceleration is found by using the following formula:

a=\frac{V_{f}-V{0}}{t}

so we can subtract the two vectors, like this:

a=\frac{(6.00i+4.0j)m/s-1.60i m/s}{8s}

which yields:

a=\frac{(4.4i+4.0j)m/s}{8s}

or:

a=(0.55i + 0.5j) m/s^{2}

so now I can find the components of the force:

F=(4.2kg)(0.55i + 0.5j) m/s^{2}

which yields:

F=(2.31i+2.1j)N

so the components of the force are:

F_{x}=2.31N to the right.

F_{y}=2.1N to in the upwards direction.

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At low temperatures, between 0 and 10 degrees Celsius – the enzymes that carry out photosynthesis do not work efficiently, and this decreases the photosynthetic rate.

At medium temperature above 10 degrees Celsius to below 40 degrees Celsius (e.g 27 degrees Celsius), the photosynthetic enzymes work at their optimum levels, so photosynthesis proceeds.

At a temperature above 40 degrees Celsius, the enzymes that carry out photosynthesis lose their shape and functionality, and the photosynthetic rate declines rapidly.

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6CO2 + 6H20 + (energy) = C6H12O6 + 6O2

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A force of 10.0 newtons acts at an angle 20.0 degrees from vertical. What are the horizontal and vertical components of the forc
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Instructions:Drag the tiles to the correct boxes to complete the pairs. Match each term with its definition. Tiles conductor rad
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Below is an attachment containing the solution.

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2 years ago
The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temper
meriva

Answer:

(a) 1.87 x 10⁻⁴ m/s

(b) 0.013V

Explanation:

(a) Drift speed, v_{d} , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;

v_{d} = \frac{I}{qnA}             ----------------(i)

Where;

q = amount of charge

n = free charge density

A = cross-sectional area of the wire

But current density, J, is the electric current per unit cross-section area. This  is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e

J = \frac{I}{A} = \frac{E}{p}

Equation (i) can then be written as follows;

v_{d} = \frac{J}{qn} = \frac{E}{qnp}

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From the question;

E = 0.0520N/C

p = 1.72 x 10⁻⁸ Ωm

n = 8.5 x 10²⁸ electrons/m³

c = charge on electron = 1.9 x 10⁻¹⁹C

Substitute these values into equation (ii) as follows;

v_{d} = \frac{0.0520}{1.9*10^{-19} * 8.5*10^{28} * 1.72*10^{-8}}

v_{d} = 1.87 x 10⁻⁴ m/s

(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e

V = E x d        [where d = 25.0cm = 0.25m]

V = 0.0520 x 0.25

V = 0.013V

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