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3 years ago

Which of the following do waves carry?

Physics

1 answer:

kumpel [21]3 years ago

8 0

I would go with energy, because they give alllooooot of energy, from the waves.
hope that this would help you! =)

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A bowling ball is launched from the top of a building at an angle of 35° above the horizontal with an initial speed of 15 m/s. T

Mamont248 [21]

Let $y_0$ be the height of the building and thus the initial height of the ball. The ball's altitude at time $t$ is given by

$y=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ\,t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity.

The ball reaches the ground when $y=0$ after $t=2.9\,\mathrm s$ . Solve for $y_0$ :

$0=y_0+\left(15\dfrac{\rm m}{\rm s}\right)\sin35^\circ(2.9\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(2.9\,\mathrm s)^2$

$\implies y_0\approx16.258\,\mathrm m$

so the building is about 16 m tall (keeping track of significant digits).

3 0

2 years ago

A concave mirror has a focal length of 13.5 cm. This mirror forms an image located 37.5 cm in front of the mirror. Find the magn

77julia77 [94]

Explanation:

It is given that,

Focal length of the concave mirror, f = -13.5 cm

Image distance, v = -37.5 cm (in front of mirror)

Let u is the object distance. It can be calculated using the mirror's formula as :

$\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{u}=\dfrac{1}{f}-\dfrac{1}{v}$

$\dfrac{1}{u}=\dfrac{1}{(-13.5)}-\dfrac{1}{(-37.5)}$

u = -21.09 cm

The magnification of the mirror is given by :

$m=\dfrac{-v}{u}$

$m=\dfrac{-(-37.5)}{(-21.09)}$

m = -1.77

So, the magnification produced by the mirror is (-1.77). Hence, this is the required solution.

7 0

2 years ago

The rate at wich energy is used in a circuit

Vaselesa [24]

Power is the rate at which energy is used in a circuit.

3 0

3 years ago

A blue car pulls away from a red stop-light just after it has turned green with a constant acceleration of 0.2 m/s2. A green car

jolli1 [7]

Answer:

After 15 seconds, the green car will catch up with the blue car

Explanation:

Let the time for the green car to catch up with the blue car be T

When the green car catches up to the blue car, the distances covered by each car after time T will be equal. Also, their velocities at that instant will be equal

Distance covered by blue car after time T is given by: s = ut + 0.5 at²

Where u = 0, a = 0.2 m/s², t = T

S = 0.5 × 0.2 × T² = 0.1 T²

Velocity of blue car, v = u+ at

v = 0.2T

Distance covered by green car at T is given as: S = Velocity × time

Where v = 0.2T, t = T - 7.5 (since the blue car started 7.5 seconds earlier)

S = 0.2T (T - 7.5)

S = 0.2 T² - 1.5T

Equating the distance covered by the two cars

0.2T² - 1.5T = 0.1T²

0.1T² - 1.5T = 0

T(0.1T - 1.5) = 0

T = 0 or

T = 1.5/0.1 = 15 secs

Therefore, after 15 seconds, the green car will catch up with the blue car

8 0

2 years ago

A container is filled with an ideal diatomic gas to a pressure and volume of P1 and V1, respectively. The gas is then warmed in

lilavasa [31]

Answer:

Explanation:

The change is as follows

P₁ V₁ to 3P₁, V₁ ( constt volume ) --- first process

3P₁,V₁ to 3P₁ , 5V₁ ( constt pressure ) ---- second process

In the first process Temperature must have been increased 3 times . So if initial temperature is T₁ then final temperature will be 3 T₁

P₁V₁ = n R T₁ , n is no of moles of gas enclosed.

nRT₁ = P₁V₁

Heat added at constant volume = n Cv ( 3T₁ - T₁)

= n x 5/3 R X 2T₁ ( for diatomic gas Cv = 5/3 R)

= 10/3 x nRT₁

= 10/3x P₁V₁

In the second process, Temperature must have been increased 5 times . So if initial temperature is 3T₁ then final temperature will be 15 T₁

Heat added at constant pressure in second case

= n Cp ( 15T₁ - 3T₁)

= n x 7/3 R X 12T₁ ( For diatomic gas Cp = 7/3 R)

= 28 x nRT₁

= 28 P₁V₁

6 0

3 years ago