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3 years ago

In the impulse approximation, __________.

Physics

1 answer:

Gwar [14]3 years ago

7 0

Answer:

(B) the external forces acting during a collision can be neglected

Explanation:

Impulse is the product of force and time for which it acts. From Newton's second law, impulse is the change in momentum. In a collision of bodies, there is a change in velocity, either in direction or magnitude. Hence, there is a change in momentum and an impulse.

The forces between the object contribute to the impulse and cannot be neglected.

Since the forces act over a time interval, they may be time-dependent. In such cases, the impulse is not a rectangular pulse (which is the graph of a time-independent force).

Impulse calculations are performed in isolated systems. Hence, external forces are neglected.

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PLEASE HELP! I think it's F but I'm not sure... Am I right???

mario62 [17]

c.E ps... that's the answerrr

Explanation:

that's the answer E

7 0

3 years ago

Read 2 more answers

The resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is:a) 180 d

r-ruslan [8.4K]

Answer:

0 degrees

Explanation:

Let $F_1\ and\ F_2$ are two forces. The resultant of two forces acting on the same point is given by :

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos\theta}$

Where $\theta$ is the angle between two forces

When $\theta=0$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2\ cos(0)}$

$F_R=\sqrt{F_1^2+F_2^2+2F_1F_2}$

When $\theta=90^{\circ}$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(90)}$

$F_R=\sqrt{F_1^2+F_2^2}$

When $\theta=180^{\circ}$ i.e. when two forces are parallel to each other,

$F_R=\sqrt{F_1^2+F_2^2+F_1F_2\ cos(180)}$

$F_R=\sqrt{F_1^2+F_2^2-2F_1F_2}$

It is clear that the resultant of two forces acting on the same point simultaneously will be the greatest when the angle between them is 0 degrees. Hence, this is the required solution.

8 0

3 years ago

A bowling ball is launched off the top of a 240-foot tall building. The height of the bowling ball above the ground t seconds af

natima [27]

Answer:

When the ball hits the ground, its velocity is -128 ft/s.

Explanation:

Hi there!

First, let's find the time it takes the ball to reach the ground (the value of t for which s(t) = 0):

s(t) = -16t² + 32t + 240

0 = -16t² + 32t + 240

Solving the quadratic equation with the quadratic formula:

t = 5.0 s (the other solution of the equation is rejected because it is negative).

Now, we have to find the velocity of the ball at t = 5.0 s.

The velocity of the ball is the change of height over time (the derivative of s(t)):

v = ds/dt = s'(t) = -32t + 32

at t = 5.0 s:

s'(5.0) = -32(5.0) + 32 = -128 ft/s

When the ball hits the ground, its velocity is -128 ft/s.

4 0

3 years ago

You throw a 5.5 g coin straight down at 4.0 m/s from a 25-m-high bridge.

yuradex [85]

Answer

A) 1,347.5 Joules

B) 22.49 m/s

Explanation

Part A

The work done by the gravity is known as potential energy.

It is given by;

P.E = mgh

Where m is mass, g is acceleration due to gravity and h is the vertical height.

P.E = 5.5 × 9.8 ×25

= 1,347.5 Joules.

Part B

Using the Newton's third Law of motion,

V² = U² +2as

Where v is final velocity, u is the initial velocity, and s is the displacement of the stone.

V² = 4² + (2×9.8×25)

= 16 + 490

= 506

V = √506

= 22.49 m/s

6 0

4 years ago

Read 2 more answers

A grating has 380 lines/mm. How many orders of the visible wavelength 457 nm can it produce in addition to the m = 0 order? Plea

egoroff_w [7]

Answer:

5 orders.

Explanation:

The condition for diffraction maxima,

$sin \theta = mN\lambda\\m=\frac{sin \theta}{N\lambda}$

$m=\frac{sin 90^{\circ}}{(380\frac{lines}{mm}(\frac{1000mm}{1m} )) 457\times 10^{-9} m }\\m=5.75\simeq5$

So maximum m is 5 which means it produce 5 orders.

Therefore, the number of possible order will produce by visible light of wavelength 457 nm is 5 orders.

6 0

4 years ago