The combination of all of the forces acting on an object is called the

A student wants to determine the speed of sound at an elevation of one mile. To do this the student performs an experiment to de

Lorico [155]

Answer:

The correct answer is a

Explanation:

The speed of a sound wave depends on the square root of the modulus of compressibility and the density of the medium.

For the same medium, the speed of sound depends on the temperature of the fora

v = $v_o \ \sqrt{1 + \frac{T}{273} }$

Therefore, the different results that are obtained are due to changes in temperature. The correct answer is a

since this way it has the values of the speed of sound for each temperature, for which it can compare with the results obtained from the trip.

3 0

2 years ago

A projectile is fired vertically from earth's surface with an initial speed of 5.3 km/s. neglecting air drag, how far above the

MaRussiya [10]

It will rise, and then fall back to the Earth’s surface.

6 0

3 years ago

When a 0.1-kilogram pendulum bob reaches the top of its swing, how much kinetic energy does it have?

zvonat [6]

When ANY pendulum bob reaches the top of its swing,
it stops momentarily and then begins to drop. At the instant
when it stops, its kinetic energy is zero.

7 0

3 years ago

A buoy is a floating device that can have many purposes, but often as a locator for ships. Collin constructs a hollow metal buoy

Yuri [45]

Answer:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water

B )angular frequency ( w ) = $\sqrt{\frac{3g}{h} }$

c ) h = 1.34 m ( 4 $\pi ^{2}$ / 3g )

Explanation:

A) since the density of the buoy is half the density of sea water when the buoy is at rest on an ocean half of the buoy will be submerged in water this is because the substances with lesser density floats when placed in a substance with a higher density

B ) when the Buoy is at rest ( t = 0 ) and is then pushed down calculate angular frequency ( w ) of small oscillations in terms of the given variables

volume of cone = hA /3.

h = height of cone, A = Base Area.

therefore the total volume of the Buoy above water level ( at rest )

= ( $\pi r^{2} * \frac{h}{3}$ ) * 2 = $\frac{\pi r^{2} h }{6}$

The part of the buoy immersed in water = x

then the net upward force will be

fb ( force of Buoy ) = fg ( force of gravity )

note force = Mass * Acceleration

force of buoy = $\frac{\alpha }{2} *$ $\frac{\pi r^{2} }{h}$ * a = ( force of gravity ) $\alpha * T\frac{r^{2} }{4} * x * g$

therefore a = $\frac{3g}{h} * x$

angular frequency ( w ) = $\sqrt{\frac{3g}{h} }$

C ) height of the each cone

$\frac{2\pi }{w} = 1$ therefore w = 2 $\pi$

back to the angular frequency : $\frac{3g}{h} = 4\pi ^{2}$

therefore h = 1.34 m ( 4 $\pi ^{2}$ / 3g )

4 0

3 years ago

In a closed system, the loss of momentum of one object ________ the gain in momentum of another object.

Virty [35]

In a closed system, the loss of momentum of one object is same as________ the gain in momentum of another object

according to law of conservation of momentum, total momentum before and after collision in a closed system in absence of any net external force, remains conserved . that is

total momentum before collision = total momentum after collision

P₁ + P₂ = P'₁ + P'₂

where P₁ and P₂ are momentum before collision for object 1 and object 2 respectively.

P'₁ - P₁ = - (P'₂ - P₂)

so clearly gain in momentum of one object is same as the loss of momentum of other object

8 0

3 years ago