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Yuri [45]
3 years ago
10

A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area.

Physics
1 answer:
babunello [35]3 years ago
8 0

Answer:

True

Explanation:

Pressure is defined as:

p=\frac{F}{A}

where

F is the magnitude of the force perpendicular to the surface

A is the surface

Therefore, pressure is inversely proportional to the area of the surface:

p\propto \frac{1}{A}

this means that, assuming that the forces in the two situations (which have same magnitude) are both applied perpendicular to the surface, the force exerted over the smaller area will exert a greater pressure. Hence, the statement"

<em>"A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area"</em>

is true.

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NASA is designing a Mars-lander that will enter the Martian atmosphere at high speed. To land safely it must slow to a constant
Viktor [21]

Answer:

a) maximum mass of the Mars lander to ensure it can land safely is 200 kg

b) area of the parachute required is 480 m² which is larger than 400 m²

c) area of the parachute should be 12.68 m²

Explanation:

Given the data in the question;

V = 20 m/s

A = 200 m²

drag co-efficient CD = 1.855

g = 3.71 m/s²

density of the atmospheric pressure β = 0.01 kg/m³

a. Calculate the maximum mass of the Mars lander to ensure it can land safely?

Drag force FD = 1/2 × CD × β × A × V²

we substitute

FD = 1/2 × 1.855 × 0.01 kg/m × 200 m² × ( 20 m/s )²

FD = 742 N

we know that;

FD = Fg

Fg = gravity force

Fg = mg

so

FD = mg

m = FD/g

we substitute

m = 742 N / 3.71 m/s²

m = 200 kg

Therefore, the maximum mass of the Mars lander to ensure it can land safely is 200 kg

b. The mission designers consider a larger lander with a mass of 480 kg. Show that the parachute required would be larger than 400 m²;

Given that;

M = 480 kg

Show that the parachute required would be larger than 400 m²

we know that;

FD = Fg = Mg = 480 kg × 3.71 m/s²

FD = 1780.8 N

Now, FD = 1/2 × CD × β × A × V², we solve for A

A = FD / 0.5 × CD × β × V²

we substitute

A = 1780.8  / 0.5 × 1.855 × 0.1 × (20)²

A = 1780.8 / 3.71

A = 480 m²

Therefore, area of the parachute required 480 m² which is larger than 400 m²

c. To test the lander before launching it to Mars, it is tested on Earth where g = 9.8 m/s^2 and the atmospheric density is 1.0 kg m-3. How big should the parachute be for the terminal speed to be 20 m/s, if the mass of the lander is 480 kg?

Given that;

g = 9.8 m/s²,

β" = 1 kg/m³

v" = 20 m/s

M" = 480 kg

we know that;

FD = Fg = M"g

FD = 480 kg × 9.8 m/s² = 4704 N

from the expression; FD = 1/2 × CD × β × A × V²

A = FD / 0.5 × CD × β" × V"²

we substitute

A = 4704 / 0.5 × 1.855 × 1 × (20)²

A = 4704 / 371

A = 12.68 m²

Therefore area of the parachute should be 12.68 m²

3 0
3 years ago
A child in danger of drowning in a river is being carried down-stream by a current that flows uniformly with a speed of 2.20 m/s
zmey [24]

Answer:

The angle is 65.6°.

Explanation:

Given that,

Speed = 2.20 m/s

Distance from the shore= 500 m

Distance from the bottom= 1100 m

Speed of boat = 7.30 m/s

According to figure,

We need to calculate the angle with shore

Using formula of angle

\tan\theta=\dfrac{y}{x}

Put the value into the formula

\tan\theta=\dfrac{500}{1100}

\theta=\tan^{-1}(\dfrac{500}{1100})

\theta=24.4^{\circ}

We need to calculate the angle

\alpha=90-\theta

Put the value into the formula

\alpha=90-24.4

\alpha=65.6^{\circ}

Hence, The angle is 65.6°.

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Assuming that the bridge segments are free to pivot at each intersection point, what is the tension T in the horizontal segment
Sergio039 [100]
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Answer in terms of T and L :
Tt = (TL.sqrt 3) / 2
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4 0
4 years ago
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