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2 years ago

How do you think the ball in pinball will travel? draw a line to trace the path the ball will take after it is launched

Physics

1 answer:

Montano1993 [528]2 years ago

5 0

The magnetic properties of the ball help to travel in the pin ball.

<h3>What is pinball?</h3>

A Pinball is a steel sphere of diameter 3 cm. It weighs 2.8 ounces. The ball can reach speed upto 90 miles per hour on the traditional steel pinball.

The magnetic properties of the ball help to travel in the pinball , as some machines uses the magnets to trap the ball in certain places on the playfield.

In some of the machines the ceramic pinball is called Powerball.

Pinball move around the table, hitting bumpers and some targets to score the points.

You'll get only three balls to play, when your third ball falls down then your game will be over.

So, the ball in the pinball will travel because of then magnetic properties of ball.

For more pinball content visit here

brainly.com/question/27458951

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Iron has a mass of 7.87 g per cubic centimeter of volume, and the mass of an iron atom is 9.27 × 10-26 kg. If you simplify and t

Alecsey [184]

Known values; mass per / volume and mass of each atom

average volume= volume per atom. 
density*mass per atom = volume per atom 

1cm^3 / 7.87 gram * 9.27 *10^-26 kg /atom* 1000 gram / kg=
=1.178 * 10^-22 cm^3 per atom 

Convert that to mass 
1 cm^3 * (1 m / 100 cm)^3 
average volume= 1.178 *10^-29 m^3 per atom 

distance: 

Volume of a cube is L^3 

so L = (1.178*10^-29)^(1/3) 

Half way through the cube is distance 1/2 L. 
Second cube is 1/2 L 
So distance between the center of 2 cubes = L 

=2.28 *10^-10 m

4 0

3 years ago

When the rate of motion is the same direction the speed will decrease. True False

slega [8]

A)true is the answer

6 0

3 years ago

The tools to distinguish the thickness of two objects of thickness of 1.92mm and the other with thickness of 1.93mm is ?

melomori [17]

Answer:

the instrument that gives this precision is the micrometer screw

Explanation:

The high precision measurements of small parts are the general vernier and the micrometer screw.

In these two instruments the same principle is used: there is a fixed rule and a mobile one that increases precision.

Let's analyze the absolute error or precision of each instrument

* For the vernier, the precision of the fixed rule is 1 mm and there are 20 divisions (the most common); therefore the precision of the instrument is

Δx = 1 mm / 20

Δx = 0.05 mm

* For the micrometer screw, the precision of the fida rule is 0.5 mm and the number of divisions is 50, therefore the precision of the screw is

Δx = 0.5mm / 50

Δx = 0.01 mm

consequently the instrument that gives this precision is the micrometer screw

8 0

3 years ago

Darna rolls the 7.05-kg ball down the lane and it hits the 1.52-kg pin head on. The ball was moving at 8.24 m/s before the colli

Amiraneli [1.4K]

Answer:

The velocity of the ball after the collision is 5.39 m/s

Explanation:

Hi there!

To solve this problem, we will use the conservation of momentum: the momentum of the system ball-pin remains the same before and after the collision. The momentum of the system is calculated as follows:

momentum before the collision (initial momentum) = mb · vb1 + mp · vp1

momentum after the collision (final momentum) = mb · vb2 + mp · vp2

Where:

mb = mass of the ball = 7.05 kg

vb1 = velocity of the ball before the collision = 8.24 m/s

mp = mass of the pin = 1.52 kg

vp1 = velocity of the pin before the collision = 0 m/s.

vb2 = velocity of the ball after the collsion = unknown.

vp2 = velocity of the pin after the collision = 13.2 m/s

Since momentum is conserved, then:

initial momentum = final momentum

mb · vb1 + mp · vp1 = mb · vb2 + mp · vp2

Solving for vb2:

mb · vb1 + mp · vp1 - mp · vp2 = mb · vb2

(mb · vb1 + mp · vp1 - mp · vp2) / mb = vb2

Since the pin is initially at rest, vp1 = 0:

(mb · vb1 - mp · vp2) / mb = vb2

(7.05 kg · 8.24 m/s - 1.52 kg · 13.2 m/s) / 7.05 kg = vb2

vb2 = 5.39 m/s

The velocity of the ball after the collision is 5.39 m/s

4 0

3 years ago

A box of mass 13kg sits on a table with a coefficient of static friction of 0.85. What is the maximum force of static friction?

Alexus [3.1K]

Answer:

108.3 N

Explanation:

The maximum force of static friction acting on the box is given by:

$F=\mu N$

where

$\mu = 0.85$ is the coefficient of static friction

N is the normal reaction of the table on the box

Since the box is in equilibrium along the vertical direction, the normal reaction N is equal to the weight of the box, so:

$N=mg=(13 kg)(9.8 m/s^2)=127.4 N$

And so, the maximum force of static friction is

$F=(0.85)(127.4 N)=108.3 N$

7 0

3 years ago