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MAXImum [283]
2 years ago
14

Scientists plan to release a space probe that will enter the atmosphere of a gaseous planet. The temperature of the gaseous plan

et increases linearly with the height of the atmosphere as measured from the top of a visible boundary layer, defined as 0 kilometers in altitude. The instruments on board can withstand a temperature of 601 K. At what altitude will the probe's instruments fail?

Physics
1 answer:
ZanzabumX [31]2 years ago
5 0
From my research, the image supports the question. From the graph given, we can construct the equation of the line using the two-point formula. Using the given value of 601 K, we can solve for the missing value of altitude.

y - y1 = [(y2 - y1)/(x2 - x1)](x- x1)
y - 147.52 = [ (567 - 147.54)/(78.11 - 18.4) ](x - 18.4)

Substituting y = 601 to solve for x:
601 - 147.52 = [ (567 - 147.54)/(78.11 - 18.4) ](x - 18.4<span>)
</span>x = 83

Therefore, the probe's instruments will fail at 83 kilometers.

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A ball is thrown horizontally at a height of 2.2 meters at a velocity of 65m/s off a cliff. Assume no air resistance. How long u
slega [8]

The horizontal motion has no effect on the vertical drop.

From a drop, the distance the ball falls in 'T' seconds is

D = 4.9 T^2

so

2.2 = 4.9 T^2

T^2 = 2.2/4.9

T^2 = 0.449 sec^2

T = 0.67 second

8 0
2 years ago
A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turn
Cerrena [4.2K]

Answer:

e. Not enough information to determine

Explanation:

This question is incomplete. Here is the complete question with my solution afterwards;

A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing.

What is the first event that will occur?(Assume non-zero frictional force and the same coefficients of friction for both bugs.)

a. The ladybug begins to slide

b. The gentleman bug begins to slide

c. Both bugs begin to slide at the same time

d. Nothing ever happens

e. Not enough information to determine

The centripetal force acting on a rotating body or bugs can be written as,

F=mrw^2

m= mass of the corresponding bugs

r= corresponding radial distance of each bug

w= angular speed of the turntable

The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable

F ∝ mr

Since the radial distance from the axis of rotation of the turntable for each bug is given, but the mass is not given, the given information is therefore not enough to determine which bugs will slide first.

Option "e" is correct.

7 0
2 years ago
Read 2 more answers
3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-di
irinina [24]

Answer:

The initial velocity of the water from the tank is 5.42 m/s

Explanation:

By applying Bernoulli equation between  point 1 and 2

\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L

At the point 1

P₁=0  ( Gauge pressure)

V₁= 0 m/s

Z₁=3 m

At point 2

P₂=0  ( Gauge pressure)

Z₂= 0 m/s

h_L=1.5\ m

Now by putting the values

\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L

Z_1-h_L=\dfrac{V_2^2}{2g}

3-1.5=\dfrac{V_2^2}{2\times 9.81}

V_2=\sqrt{2\times 1.5\times 9.81}\ m/s

V₂= 5.42 m/s

The initial velocity of the water from the tank is 5.42 m/s

3 0
2 years ago
A parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates. A 1.0 g plastic bead, with a charg
marissa [1.9K]

Answer:

Please find the answer in the explanation

Explanation:

Given that A 1.0 g plastic bead, with a charge of -6.0 nC, is suspended between the two plates by the force of the electric field between them.

Since it is suspended, it must have been repelled by the bottom negative plate and trying to be attracted to the top plate.

We can therefore conclude that the upper plate, is positively charged

B.) The charge on the positive plate of parallel-plate capacitor is constructed of two horizontal 12.0-cm-diameter circular plates must be less than 6.0 nC

3 0
2 years ago
gaseous h2 and br2 are added to an evacuated 1.15L container kept at 298K. The intial partial pressurre of H2(g) is 0.782 atm an
Nastasia [14]

The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

<u>Explanation:</u>

H₂ + Br₂ ⇒ 2HBr

PH₂ = 0.782atm

PBr₂ = 0.493atm

Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹

At equilibrium:

Let 2x = pressure of HBr

PH₂ = 0.782 -x

PBr₂ = 0.493 - x

Kp = (2x)^2 / (0.782-x)(0.493-x)

Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.

Then,

Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)

x = 1.2X10⁻¹¹

PHBr = 2x = 2.4 X 10⁻¹¹ atm

Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm

3 0
2 years ago
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