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3 years ago

Does bouncing or sticking produce more impulse?

Physics

2 answers:

Elden [556K]3 years ago

4 0

Answer:

Rebounding involves a change in the direction of an object; the before- and after-collision direction is different. ... As mentioned above, if cars rebound upon collision, the momentum change will be larger and so will the impulse. A greater impulse will typically be associated with a bigger force.

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Alex Ar [27]3 years ago

4 0

Answer:

yes

Explanation:

Bouncing is a way to increase impulse. Because an object that bounces changes directions the force of impulse must be absorbed then generated by the target object. (Impulse is nearly doubled.)

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A 20.0 cm tall object is placed 50.0 cm in front of a convex mirror with a radius of curvature of 34.0 cm. Where will the image

Neko [114]

Answer:

1.Theimage will be located at -0.13m or -13 cm

2.The height of the image will be 0.052m or 5.2cm

Explanation:

Given that;

Height of object, h=20 cm = 0.2m

Object distance in front of convex mirror, o,= 50 cm =0.5m

Radius of curvature, r, =34 cm =0.34m

Let;

Image distance, i,=?

Image height, h'=?

You know that focal length,f, is half the radius of curvature,hence

f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)

f= -0.17m

Apply the relationship that involves the focal length;

$=\frac{1}{o} +\frac{1}{i} =\frac{1}{f}$

$=\frac{1}{0.5} +\frac{1}{i} =-\frac{1}{0.17}$

Re-arrange to get i

$\frac{1}{i} =-2-5.88\\\\\\\frac{1}{i} =-7.88\\\\i=-0.13m$

This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror

Apply the magnification formula

magnification, m=height of image÷height of object

$m=\frac{h'}{h} =-\frac{i}{o}$

substitute the values to get the height of image h'

$\frac{h'}{0.20} =-\frac{-0.13}{0.5} \\\\\\h'=\frac{0.13*0.20}{0.5} \\\\\\h'=\frac{0.025}{0.5} =0.052m\\\\\\h'=5.2cm$

5 0

4 years ago

A basketball player shoots toward a basket 5.3 m away and 3.0 m above the floor. If the ball is released 1.9 m above the floor a

Snezhnost [94]

Answer:

Vi = 8.28 m/s

Explanation:

This problem is related to the projectile motion.

As we know there are two components of motion associated with this, the horizontal component and vertical component.

The horizontal distance covered by the ball is

Vx*t = x

Vx*t = 5.3

Vx = 5.3/t eq. 1

Also we know that

Vx = Vicos(60)

Vx = Vi*0.5 eq. 2

equate eq. 1 and eq. 2

5.3/t = Vi*0.5

5.3/0.5 = Vi*t

Vi*t = 10.6 eq. 3

The vertical distance is

Vy = y1 + Vyi*t - 0.5gt²

also we know that

Vyi = Visin(60)

Vyi = Vi*0.866

It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance

3 = 1.9 + Vi*0.866*t - 0.5gt²

3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²

3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²

1.1 = 0.866(Vi*t) - 4.9t²

0.866(Vi*t) = 4.9t² + 1.1

substitute Vi*t = 10.6 in above equation

0.866(10.6) = 4.9t² + 1.1

9.18 = 4.9t² + 1.1

4.9t² = 8.08

t² = 8.08/4.9

t² = 1.648

t = 1.28 sec

Finally, initial speed can be found by substituting the value of t into eq. 3

Vi*t = 10.6

Vi = 10.6/t

Vi = 10.6/1.28

Vi = 8.28 m/s

8 0

4 years ago

45 grams in kilo grams

Hunter-Best [27]

Answer:

0.045 kg

Explanation:

1000 grams = 1kg

7 0

3 years ago

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According to Boyle’s Law, increasing the pressure of a gas will have which effect on its volume? increase decrease random effect

umka21 [38]

Volume will decrease if the heat remains constant

3 0

4 years ago

What is the weight of an 1000 gram mass

Reil [10]

That depends on where you weigh it.

-- On Earth, it weighs 9.807 newtons (2.205 pounds).

-- On the moon, it weighs 1.623 newtons (5.84 ounces).

-- On Jupiter, it weighs 24.79 newtons (5.57 pounds).

BTW ... 1,000 grams of mass is called ' one kilogram '.

8 0

4 years ago

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