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2 years ago

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When the skydiver descends to a certain height above the ground, she deploys her parachute to ensure a safe landing. Usually the

parachute is deployed when the skydiver reaches an altitude of about 900 m (3000 ft). Immediately after deploying the parachute, does the skydiver have a nonzero acceleration?

1 answer:

Ilya [14]2 years ago

6 0

Answer:

Yes

Explanation:

Yes, the skydiver cannot have zero acceleration as soon as the parachute is opened it takes some time to balance the rate of change in velocity with the drag force. This is in accordance with the Newton's second law of motion that the rate of change in momentum is directly proportional to the force applied.

As force is given as :

$F=\frac{d}{dt}(p)$

$F=m\times \frac{dv}{dt}$

$F=m.a$

here a=g

$F=m.g$

where:

$dp=$ change in momentum

$dt=$ time taken

$m=$ mass

$a=$ acceleration

$g=$ acceleration due to gravity

So there will be a non zero acceleration because the force acting on the body does not becomes zero immediately.

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Physics students use a spring scale to measure the weight of a piece of lead. The experiment was performed two times: once in th

Lady_Fox [76]

Answer:

The reading of the experiment made in air is 50 g more than the reading of the measurement made in water.

Explanation:

Knowing that the density of lead is $11,3 g/cm^{3}$ and the volume, we can calculate the true weight of the piece of lead:

$weight_{lead}=\rho _{lead}*V_{lead}=11,3 g/cm^{3} *50 cm^{3} = 565 g$

When the experiment is done in air, we can discard buoyancy force (due to different densities) made by air because it's negligible and the measured weight is approximately the same as the true weight.

When it is done in water, the effect of buoyancy force (force made by the displaced water) is no longer negligible, so we have to take it into account.

Knowing that the density of water is 1 g per cubic centimeter, and that the volume displaced is equal to the piece of lead (because of its much higher density, the piece of lead sinks), we can know that the buoyancy force made by water is 50 g, opposite to the weight of the lead.

$Weight_{measured}=weight_{lead}-weight_{water}=\frac{(565 g *9.8 m/s^{2} -50 g*9.8 m/s^{2})}{9,8m/s^{2} } = 515 g$

Now that we have the two measurements, we can calculate the difference:

$Difference= |Weight _{in water}- Weight _{in air}|=|515 g-565 g|=50 g$

The reading of the experiment made in air is 50 g more than the reading of the measurement made in water.

4 0

2 years ago

A simple generator has a square armature 6.0 cm on a side. The armature has 85 turns of 0.59-mm-diameter copper wire and rotates

FrozenT [24]

Answer:

f=15.5 Hz

Explanation:

Let's determine the internal resistance:

$R=\frac{(p*L)}{A}$

ρ = 1.68*10^-8 Ω m

$L=0.060m*4*60 = 14.4m$

$A=\pi*r^2\\A= \pi*(5.9x10^-4m/2)^2=2.734*10^-7m^2$

$R=(1.68*10^-8)*(14.4m)/(2.734*10^-7m^2)= 0.884$ Ω

Since the bulb is rated at 12.0 V and 25.0 W,

Current

$I=\frac{25W}{12.0v}=2.08 A$

Therefore, voltage drop inside generator =

$V=(2.08 A)*(0.88)=2.35v$

Actual EMF required is

$E_{mf}=12.0v+2.35v=14.35v$

Note that this is an RMS value.

The peak voltage is

$v_{peak}=14.15v*\sqrt{2} =20.29v$

For a generator, by Faraday's Law,

$E_{(max)}=N*B*A*w$

$20.29v=(60)*(0.650T)*(0.06m)^2$ *ω

ω $=144.5\frac{rad}{s}$

f=ω/(2π)=

f=144.5 rad/s/(2π)

f=23.001 Hz

6 0

2 years ago

a bicycle pump contains 20cm3 of air at a pressure of 100kpa the air is then pumped in a single stroke through a valve into a ty

riadik2000 [5.3K]

If we assume also that the temperature of the air does not change, we can use Boyle's Law:
p₁V₁ = p₂V₂

Now, we know:
p₁ = 100kPa
V₂ = 100cm³ (the volume of the tyre)
V₁ = 120cm³ (becuse the air is contained inside the tyre AND the pump)

We can solve for p₂:
p₂ = (p₁V₁)/V₂
= (100×120)/100
= 120kPa

Therefore your answer is: 120kPa

8 0

2 years ago

A curtain hangs straight down in front of an open window. A sudden gust of wind blows past the window; and the curtain is pulled

VikaD [51]

Answer:

option B.

Explanation:

The correct answer is option B.

The phenomenon of the curtains to pull out of the window can be explained using Bernoulli's equation.

According to Bernoulli's Principle when the speed of the moving fluid increases the pressure within the fluid decrease.

When wind flows in the outside window the pressure outside window decreases and pressure inside the room is more so, the curtain moves outside because of low pressure.

3 0

2 years ago

Is it possible for two pieces of the same metal to have different recrystallization temperatures? Is it possible for recrystalli

Butoxors [25]

Answer:

Explained

Explanation:

Two pieces of the same metal can have different recrystallization temperatures if the pieces have been cold worked to different amounts. The piece of work cold worked to greater extend will have more internal energy to drive the recrystalline process and lower recrystallization temperature.

Yes, its possible that recrystallization to take place in some regions of a part before it does in other regions of the same part if the work has been unevenly strained or if the part have different thickness at different sections.

6 0

2 years ago