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3 years ago

Which of the following is true about all electromagnetic waves in a vacuum

Physics

2 answers:

Yakvenalex [24]3 years ago

6 0

They all travel at the same speed!

Helen [10]3 years ago

4 0

Electro waves in a vacuum air is deals with this and electricity when the air and the electricity it makes electro magnets.

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The motion of a car on a position-time graph is represented with a horizontal line. What does this indicate about the car's moti

Art [367]

Answer:

Explanation:

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2 years ago

A procedure called______is used to find the volume of a rock

KengaRu [80]

Displacement is usually how the messure the rock

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3 years ago

Read 2 more answers

For a caffeinated drink with a caffeine mass percent of 0.65% and a density of 1.00 g/mL, how many mL of the drink would be requ

slava [35]

Explanation:

First we will convert the given mass from lb to kg as follows.

157 lb = $157 lb \times \frac{1 kg}{2.2046 lb}$

= 71.215 kg

Now, mass of caffeine required for a person of that mass at the LD50 is as follows.

$180 \frac{mg}{kg} \times 71.215 kg$

= 12818.7 mg

Convert the % of (w/w) into % (w/v) as follows.

0.65% (w/w) = $\frac{0.65 g}{100 g}$

= $\frac{0.65 g}{(\frac{100 g}{1.0 g/ml})}$

= $\frac{0.65 g}{100 ml}$

Therefore, calculate the volume which contains the amount of caffeine as follows.

12818.7 mg = 12.8187 g = $\frac{12.8187 g}{\frac{0.65 g}{100 ml}}$

= 1972 ml

Thus, we can conclude that 1972 ml of the drink would be required to reach an LD50 of 180 mg/kg body mass if the person weighed 157 lb.

5 0

2 years ago

An object moving at a constant speed of 25 m/s is making a turn with a radius of curvature of 7 m (this is the radius of the "ki

prisoha [69]

Answer:

- 278.34 kg m/s^2

Explanation:

The rate of the change of momentum is the same as the force.

The force that an object feels when moviming in a circular motion is given by:

F = -mrω^2

Where ω is the angular speed and r is the radius of the circumference

Aditionally, the tangential velocity of the body is given as:

v = rω

The question tells us that

v = 25 m/s

r = 7m

mv = 78 kg m/s

Therefore:

m = (78 kg m/s) / (25 m/s) = 3.12 kg

ω = (25 m/s) / (7 m) = 3.57 (1/s)

Now, we can calculate the force or rate of change of momentum:

F = - (3.12 kg) (7 m)(3.57 (1/s))^2

F = - 278.34 kg m/s^2

4 0

3 years ago

A steel wire of length 31.0 m and a copper wire of length 17.0 m, both with 1.00-mm diameters, are connected end to end and stre

Brut [27]

Answer:

The time taken is $t = 0.356 \ s$

Explanation:

From the question we are told that

The length of steel the wire is $l_1 = 31.0 \ m$

The length of the copper wire is $l_2 = 17.0 \ m$

The diameter of the wire is $d = 1.00 \ m = 1.0 *10^{-3} \ m$

The tension is $T = 122 \ N$

The time taken by the transverse wave to travel the length of the two wire is mathematically represented as

$t = t_s + t_c$

Where $t_s$ is the time taken to transverse the steel wire which is mathematically represented as

$t_s = l_1 * [ \sqrt{ \frac{\rho * \pi * d^2 }{ 4 * T} } ]$

here $\rho_s$ is the density of steel with a value $\rho_s = 8920 \ kg/m^3$

$t_s = 31 * [ \sqrt{ \frac{8920 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_s = 0.235 \ s$

And

$t_c$ is the time taken to transverse the copper wire which is mathematically represented as

$t_c = l_2 * [ \sqrt{ \frac{\rho_c * \pi * d^2 }{ 4 * T} } ]$

here $\rho_c$ is the density of steel with a value $\rho_s = 7860 \ kg/m^3$

$t_c = 17 * [ \sqrt{ \frac{7860 * 3.142* (1*10^{-3})^2 }{ 4 * 122} } ]$

$t_c =0.121$

$t = t_c + t_s$

$t = 0.121 + 0.235$

$t = 0.356 \ s$

4 0

2 years ago