Which is not a device for reproducing sound?

Complete the following statement. Our solar system includes the sun, _________, and their moons. galaxies stars planets universe

GuDViN [60]

The planets have (their) moons. So the answer would be planets.

5 0

3 years ago

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Mike is standing on the roof of a building looking at the roof of the neighboring building that is 15 meters away and 10 meters

amid [387]

Answer:

Part a)

$t = 1.65 s$

Part b)

$x = 40.4 m$

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

$v = 27.3 m/s$

direction of velocity is given as

$[tex]\theta = 26.35 degree$

Explanation:

Part a)

acceleration due to gravity on this planet is 3/4 times the gravity on earth

So the acceleration due to gravity on this new planet is given as

$a = \frac{3}{4}(9.81)$

$a = 7.36 m/s^2$

now the vertical displacement covered by the canister is given as

$y = 10 m$

now by kinematics we have

$y = \frac{1}{2}gt^2$

$10 = \frac{1}{2}(7.36)t^2$

$t = 1.65 s$

Part b)

Horizontal speed of the canister is given as

$v_x = 24.5 m/s$

now the distance moved by it

$x = v_x t$

$x = 24.5 (1.65)$

$x = 40.4 m$

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

Final velocity in X direction will remains the same

$v_x = 24.5 m/s$

final velocity in Y direction

$v_y = v_i + at$

$v_y = 0 + (7.36)(1.65)$

$v_y = 12.14 m/s$

now magnitude of velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{24.5^2 + 12.14^2}$

$v = 27.3 m/s$

direction of velocity is given as

$\theta = tan^{-1}\frac{v_y}{v_x}$

$\theta = tan^{-1}\frac{12.14}{24.5}$

$[tex]\theta = 26.35 degree$

6 0

3 years ago

What is work done aganist gravity?

Rudik [331]

Answer:

Work done against gravity in lifting an object becomes potential energy of the object-Earth system. The change in gravitational potential energy, ΔPEg, is ΔPEg = mgh, with h being the increase in height and g the acceleration due to gravity.

Explanation:

You're Welcome.

8 0

3 years ago

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An infinitely long straight wire has a uniform linear charge density of Derive the 4. equation for the electric field a distance

marshall27 [118]

Answer:

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

Explanation:

Let the linear charge density of the charged wire is given as

$\frac{q}{L} = \lambda$

here we can use Gauss law to find the electric field at a distance r from wire

so here we will assume a Gaussian surface of cylinder shape around the wire

so we have

$\int E. dA = \frac{q}{\epsilon_0}$

here we have

$E \int dA = \frac{\lambda L}{\epsilon_0}$

$E. 2\pi r L = \frac{\lambda L}{\epsilon_0}$

so we have

$E = \frac{\lambda}{2\pi \epsilon_0 r}$

4 0

3 years ago

A golf club hits a 0.04551 kg golf ball off a golf tee. The club is in contact with the ball for 0.020 s, and the force applied

Juliette [100K]

Answer:

v = 50.5 m/s

Explanation:

F = (m)(^v/^t)

115N = (0.04551kg)(v/(0.020s))

2,526.917161 m/s² = v/(0.020s)

v = 50.53834322 m/s

v = 50.5 m/s

8 0

3 years ago