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3 years ago

Determine the gradient and the co-ordinates of the x and y intercept of line whose equation is 2y + 3x = 1

Physics

1 answer:

Savatey [412]3 years ago

4 0

Answer:

The x - intercept is 1/3

The y - intercept is 1/2

The gradient is -3/2

Explanation:

To find the x - intercept of the equation 2y + 3x = 1, we find the value of x when y = 0. So,

2y + 3x = 1

2(0) + 3x = 1

0 + 3x = 1

3x = 1

x = 1/3

So, the x - intercept is 1/3

To find the y - intercept of the equation 2y + 3x = 1, we find the value of y when x = 0. So,

2y + 3x = 1

2y + 3(0) = 1

2y + 0 = 1

2y = 1

y = 1/2

So, the y - intercept is 1/2

To find the gradient of the equation 2y + 3x = 1, we re-write it in gradient intercept form by making y subject of the formula.

So, 2y + 3x = 1

2y = -3x + 1

y = -3x/2 + 1/2

The coefficient of x which equals -3/2 is the gradient.

The gradient is -3/2

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The correct answer is C) a motor.
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3 years ago

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3 years ago

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

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

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3 years ago

The resistance of a wire depends upon the material's resistivity and the length and cross‑sectional area of the wire. What will

andriy [413]

Answer:

R' = 4R

The resistance will become 4 times the initial value.

Explanation:

The resistance of a wire at room temperature, is given by the following formula:

R = ρL/A ----------- equation 1

where,

R = Resistance of wire

ρ = resistivity of the material

L = Length of wire

A = Cross-sectional area of wire

Now, if the length (L) is multiplied by 4, then resistance will become:

R' = ρ(4L)/A

R' = 4 (ρL/A)

using equation 1:

R' = 4R

The resistance will become 4 times the initial value.

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3 years ago

Determine the gradient and the co-ordinates of the x and y intercept of line whose equation is 2y + 3x = 1​

Determine the gradient and the co-ordinates of the x and y intercept of line whose equation is 2y + 3x = 1