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3 years ago

HELP PLS WHAT IS Y COMPONENT

Physics

1 answer:

jeka943 years ago

6 0

Answer:

11.5

Explanation:

you do 12(sin 73.3) because its on the y you do sin

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A radio wave has a speed of 3.00 x10^8 and a frequency of 100 MHz. What is the wavelength?

dlinn [17]

Answer:

Wavelength of radio is wave is 3 m

Explanation:

Wavelength of radio is wave is

$\lambda=\frac{v}{f}$

where

$v=3 \times 10^8 m\s\\f=100 Mhz$

wavelength is

$\lambda=\frac{3\times 10^8}{100\times 10^6} \\\\\lambda=3 m\\$

5 0

3 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

5 0

3 years ago

Please help ill mark youas brainliest !!

USPshnik [31]

Answer:

can you put on a clearer image this one is hard to see

8 0

2 years ago

2. How is frequency described in mathematical terms?

alexdok [17]

Answer:

Explanation:

Frequency refers to the number of occurrences of a periodic event per time and is measured in cycles/second. In this case, there is 1 cycle per 2 seconds. So the frequency is 1 cycles/2 s = 0.5 Hz.

4 0

3 years ago

Read 2 more answers

The potential difference between the surface of a 3.0-cm-diameter power line and a point 1.0 m distant is 3.9 kV. Find the line

Korolek [52]

Answer:

The linear charge density is 5.19 X 10⁻⁶ C/m

Explanation:

The potential difference between two cylinders, is given as

V = (λ/2πε)ln(b/a)

where;

λ is the line charge density on the power line.

b is the distance between the power line = 1 m

a is the radius of the wire = 1.5 cm = 0.015 m

ε is the permittivity of free space = 8.9 X 10⁻¹² C

V*2πε = λ* ln(b/a)

3900 *(2π*8.9 x10⁻¹²)= λ *ln(1/0.015)

2.1812 X 10⁻⁷ = 4.1997* λ

λ = 5.19 X 10⁻⁶ C/m

Therefore, the linear charge density is 5.19 X 10⁻⁶ C/m

6 0

3 years ago