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FromTheMoon [43]

2 years ago

13

Write a collision scenario here. If you choose your own collision, you can have neither, one, or both of the objects break. Be s

ure to describe which object (if any) breaks.

1 answer:

OleMash [197]2 years ago

5 0

Answer:

My scenario would be A Car vs. a guard rail on a road. You have a car that is coming down a Highway at a speed of 43 Mph Miles per hour (69.2018 Kmh)

And it hits a steel guardrail and the car smashes in at the front and the guardrail is only bent while the car has the bumper and the hood along with the headlights and windshield along with the passenger side window break.

Explanation:

This is caused by so much force reacting from one object to another but also depends on molecular density.

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Which of the following statements are true about magnets?

timama [110]

The correct answer is 3

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3 years ago

What is matter, atom, molecules?

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matter is a solid liquid or gas , a atom is a basic unit of a chemical element and a molecule is a group of atoms together

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3 years ago

One terminal of a car battery is said to be connected to “ground.” since it is not really connected to the ground, what is meant

GalinKa [24]

This expression means that the negative terminal (-) is connected to the metal chassis or engine, which means that all voltages used for the electrical devices in the car are measured with respect to the car's chassis or engine.
Today's vehicles have a negative ground system, which means that the vehicle's steel frame or chassis is directly connected to the negative side of the battery via the negative battery cable.

3 0

3 years ago

A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

a

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

b

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$

$F_{31 +32} = 2.16 *10^{-5} (4i + 3j) - 12*10^{-5} j$

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of $F_{31 +32}$ is mathematically evaluated as

$|F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}$

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction is obtained as

$tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}$

$\theta = tan ^{-1} [-0.63889]$

$\theta = - 32.57 ^o$

$\theta = 360 - 32.57$

$\theta =327.43 ^o$

5 0

3 years ago

a spring is stretched 20 cm by a 30.0 n force. determine the work done in stretching the spring from 0 to 40 cm?

Alekssandra [29.7K]

The work done when a spring is stretched from 0 to 40cm is 4J.

What is work done?

Work done is the magnitude of force multiplied by displacement of an object. It is also the amount of energy transferred to an object when work is done on that.

The work done on the spring to stretch to 40cm is,

F = kx

where F is force, k is force constant.

k = F / x = 10 N / 20 * 10^-2 m = 50 N/m

W = 0.5 * k * (x)^2

where W = work done, k = force constant.

W = 0.5 x 50 x (40 x 10^-2)^2 = 4 J.

Therefore, the work done on the spring when it is stretched to 40cm is 4J.

To learn more about work done click on the given link brainly.com/question/25573309

#SPJ4

3 0

1 year ago