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2 years ago

12

A 5.00 kilogram mass is traveling at 100 meters per second. Determine the speed of the mass after an impulse of 30 Newton * seco

nds is applied.

2 answers:

valentinak56 [21]2 years ago

7 0

Given: Mass m = 5.0 Kg; Initial Velocity Vi = 100 m/s

Impulse Ft = 30 N.s

Required: Final Velocity = ?

Formula: Ft = m(Vf -Vi)

Vf = Vi + Ft/m

Vf = 100 m/s + 30 N.s/5 Kg

Vf = 100 m/s + 6 m/s

Vf = 106 m/s

luda_lava [24]2 years ago

6 0

Impulse is the change in momentum, J. It is measured in the same unit as momentum, kg•m/s or N•s
The formula for impulse is
J = ∆p = pf - pi
30Ns = pf - pi
pi = m*vi
pi = (5kg)(100m/s)
pi = 500kg-m/s
pf - 500kg-m/s = 30Ns
pf = 530kg-m/s
pf = m*vf
v f = pf /m
v f = (530kg-m/s)/(5kg)
v f = 106m/s (<span>speed of the mass after an impulse)</span>

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The dimensions of the rectangle are:

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b = $100/\pi$ m

<h3>What is a perimeter in math?</h3>

The perimeter is the length of the outline of a shape. To find the perimeter of a rectangle or square you have to add the lengths of all the four sides.

<h3>How do we find a perimeter of a rectangle?</h3>

The perimeter of a rectangle,denoted by P is given by the formula, P=2l+2b, where l is the length and b is the breadth of the rectangle.

<h3>Given:</h3>

As per the question:

Perimeter of the room is given as P = 200 m

The region is rectangular having a semicircle at each end.

Now,

Let 'l' be the length of the rectangle, 'b' be its breadth and 'r' be the radius of the semi-circle at each end.

Then, Area of the given rectangle, A = lb

Perimeter of the room, P is = $\pi r+l+\pi r+l=2\pi r+2l=\pi b+2l$

Therefore, $\pi b+2l=200$

$b=(200-2l)/\pi$

Now,

Area, A = $l(200-2l)/\pi=(200l-2l^{2} )/\pi$

Now, differentiate A w.r.t l:

Again differentiating w.r.t 'l', we get:

$d^{2} A/dl^{2} =-4l/\pi$ < 0

Thus we get maximum are when $dA/dl=0$

Therefore,

$(200-4l)/\pi=0$

l = 50 m

Now, from

$\pi b+2l=200$

$\pi b=200-2*50$

$b=100/\pi$

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1 year ago

Could I please get some help on this question I don’t understand .

Oksana_A [137]

Answer:

12.5 m/s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Height (h) = 8 m

Final velocity (v) at 8 m above the lowest point =?

NOTE: Acceleration due to gravity (g) = 9.8 m/s²

The velocity of the roller coaster at 8 m above the lowest point can be obtained as follow:

v² = u² + 2gh

v² = 0² + (2 × 9.8 × 8)

v² = 0 + 156.8

v² = 156.8

Take the square root of both side

v = √156.8

v = 12.5 m/s

Therefore, the velocity of the roller coaster at 8 m above the lowest point is 12.5 m/s.

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2 years ago

Consider the motion of a 4.00-kg particle that moves with potential energy given by U(x) = + a) Suppose the particle is moving w

gtnhenbr [62]

Correct question:

Consider the motion of a 4.00-kg particle that moves with potential energy given by

$U(x) = \frac{(2.0 Jm)}{x}+ \frac{(4.0 Jm^2)}{x^2}$

a) Suppose the particle is moving with a speed of 3.00 m/s when it is located at x = 1.00 m. What is the speed of the object when it is located at x = 5.00 m?

b) What is the magnitude of the force on the 4.00-kg particle when it is located at x = 5.00 m?

Answer:

a) 3.33 m/s

b) 0.016 N

Explanation:

a) given:

V = 3.00 m/s

x1 = 1.00 m

x = 5.00

$u(x) = \frac{-2}{x} + \frac{4}{x^2}$

At x = 1.00 m

$u(1) = \frac{-2}{1} + \frac{4}{1^2}$

= 4J

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$= \frac{1}{2} * 4(3)^2$

= 18J

Total energy will be =

4J + 18J = 22J

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$u(5) = \frac{-2}{5} + \frac{4}{5^2}$

$= \frac{4-10}{25} = \frac{-6}{25} J$

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= 2Vf²

Total energy =

2Vf² - 0.024

Using conservation of energy,

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Vf² = (22+0.24) / 2

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$\frac{-du(x)}{dx} = \frac{-d}{dx} [\frac{-2}{x}+ \frac{4}{x^2}$

$= \frac{2}{x^2} - \frac{8}{x^3}$

At x = 5.0 m

$\frac{2}{5^2} - \frac{8}{5^3}$

$F = \frac{2}{25} - \frac{8}{125}$

= 0.016N

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