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3 years ago

8

Indicate all ways in which the graph used in question 4 would change if the Normal Force applied to the object was increased. Be

specific. (What happens to slopes? Y-values? X-values? Etc.)

1 answer:

Maksim231197 [3]3 years ago

5 0

Answer:

Answer:

PLEASE BRAINLESS ME INEED

# CARRY ON LEARNINGAnswer:

PLEASE BRAINLESS ME INEED

# CARRY ON LEARNING

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A goose with a mass of 2.0 kg strikes a commercial airliner with a mass of 160,000 kg head-on. Before the collision, the goose w

Goryan [66]

Answer:

The change in momentum of the goose during this interaction is 33.334 m/s

Explanation:

Given;

mass of goose, m₁ = 2.0 kg

mass of commercial airliner, m₂ = 160,000 kg

initial velocity of the bird, u₁ = 60 km/hr = 16.667 m/s

initial velocity of the airliner, u₂ = 870 km/hr = 241.667 m/s

Change in momentum is given as;

ΔP = mv - mu

where;

u is the initial velocity of the bird

v is the final velocity of the bird

Apply the principle of conservation of linear momentum;

Total momentum before collision = Total momentum after collision

m₁u₁ + m₂u₂ = v(m₁ + m₂)

where;

v is the final velocity of bird and airliner after collision;

(2 x 16.667) + (160,000 x 241.667) = v (2 + 160,000)

38,666,753.334 = 160,002v

v = 38,666,753.334 / 160,002

v = 241.664 m/s

Thus, the final velocity of the bird is negligible compared to final velocity of the airliner.

ΔP = mv - mu

ΔP = m(v - u)

ΔP = 2(0 - 16.667)

ΔP = -33.334 m/s

The negative sign implies a deceleration of the bird after the impact.

Therefore, the change in momentum of the goose during this interaction is 33.334 m/s

4 0

3 years ago

An object is dropped from a platform 100 feet high. Ignoring wind resistance, what will its speed be when it reaches the ground?

ICE Princess25 [194]

<span>if we assume the origin is at the dropping point and the object is merely dropped and not thrown up or down then y0 = 0 and v0 = 0. The equation reduces to </span>

<span>y = 0 + 0t + ½gt² </span>
<span>y = ½gt² </span>

<span>t = √(2y/g) </span>

<span>in the ft - lb - s system </span>

<span>y = -100 ft </span>
<span>g = -32.2 ft / s² </span>

<span>t = √(2y/g) </span>
<span>t = √(2(-100) / (-32.2)) </span>
<span>t = 2.5 s</span>

7 0

4 years ago

Read 2 more answers

A metal wire has a resistance of 13.00 at a temperature of 25.0 degree celsius

Nady [450]

Explanation:

what exactly are you asking for?

4 0

3 years ago

A particle moves along the x axis from the origin. The magnitude of the position vector at time t is

Maurinko [17]

1) The average velocity is $-2.1\cdot 10^5 m/s$

2) The instantaneous velocity is $64t-260t^3$

Explanation:

1)

The average velocity of an object is given by

$v=\frac{d}{t}$

where

d is the displacement

t is the time elapsed

In this problem, the position of the particle is given by the function

$x(t) = 32t^2 - 65t^4$

where t is the time.

The position of the particle at time t = 6 sec is

$x(6) = 32(6)^2 - 65(6)^4=-83,088 m$

While the position at time t = 12 sec is

$x(12)=32(12)^2-65(12)^4=-1,343,232 m$

So, the displacement is

$d=x(12)-x(6)=-1,343,232-(-83,088)=-1,260,144 m$

And therefore the average velocity is

$v=\frac{-1,260,144 m}{12 s- 6 s}=-2.1\cdot 10^5 m/s$

2)

The instantaneous velocity of a particle is given by the derivative of the position vector.

The position vector is

$x(t) = 32t^2 - 65t^4$

By differentiating with respect to t, we find the velocity vector:

$v(t) = x'(t) = 2\cdot 32 t - 4\cdot 65 t^3 = 64t - 260 t^3$

Therefore, the instantaaneous velocity at any time t can be found by substituting the value of t in this expression.

Learn more about velocity:

brainly.com/question/5248528

#LearnwithBrainly

6 0

4 years ago

In a displacement/time graph the slope of the line is equal to the ________.

mamaluj [8]

In a displacement/time graph, the slope of the line is equal to the velocity

3 0

3 years ago