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2 years ago

Which statements accurately describe electronic tools? Check all that apply.

Physics

2 answers:

Nitella [24]2 years ago

8 0

Answer:

Electronic tools collect large amounts of data.

Electronic tools display data quickly and clearly.

Electronic tools provide accurate data analysis.

Explanation:

In the 21st century, electronic tools are useful in every area of life. From education to banking to statistical data analysis to sophisticated technological designs etc.

These electronic tools could collect, process, analyse and display large data with great precision within a very short time.

levacccp [35]2 years ago

3 0

Answer:

Electronic tools provide accurate data

Explanation:

The sentence "Electronic tools provide accurate data analysis" is the only correct of all the options shown, because it is the only sentence that describes a characteristic present in all the electronic tools. The other options are not all electronic tools, but each option represents a type of electronic tool, depending on the time and technology used to build the tool.

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What is the difference between an elastic and inelastic collision

barxatty [35]

Answer:

In an elastic collision, the total kinetic energy is conserved, while in an inelastic collision, it is not

Explanation:

Let's define the two types of collision:

- Elastic collision: an elastic collision is a collision in which:

1) the total momentum of the system is conserved

2) the total kinetic energy of the system is conserved

Typically, elastic collisions occur when there are no frictional forces acting on the objects in the system, so that no kinetic energy is lost into thermal energy. An example of elastic collision is the collision between biliard balls.

- Inelastic collision: an inelastic collision is a collision in which:

1 ) the total momentum of the system is conserved

2) the total kinetic energy of the system is NOT conserved

In an elastic collision, part of the total kinetic energy is lost (=converted into thermal energy) due to the presence of frictional forces. An example of inelastic collision is the accident between two cars, in which part of the energy is converted into heat.

4 0

3 years ago

A stone is thrown straight up from the edge of a roof, 925 feet above the ground, at a speed of 20 feet per second. Remembering

Black_prince [1.1K]

<h2>Answer: 469 feet</h2>

Explanation:

This problem is a good example of Vertical motion, where the main equation for this situation is:

$y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}$ (1)

Where:

$y$ is the height of the stone at 6s (the value we want to find)

$y_{o}=925ft$ is the initial height of the stone

$V_{o}=20ft/s$ is the initial velocity of the stone

$t=6s$ is the time at which we need to find the height

$g=32ft/s^{2}$ is the acceleration due to gravity

Having this clear, let's find $y$ from (1):

$y=925ft+(20ft/s)(6s)-\frac{1}{2}(32ft/s^{2})(6s)^{2}$ (2)

Finally:

$y=469ft$ This is the height of the stone at t=6s

4 0

2 years ago

The more you heat an object, the

zhannawk [14.2K]

The answer is D. The temperature obviously doesnt rise slower or faster, and if you are heating an object, it would make no sense to say that less heat is being transferred.

8 0

3 years ago

Read 2 more answers

Two stones are launched from the top of a tall building. One stoneis thrown in a direction 30.0^\circ above the horizontal with

Butoxors [25]

Answer:

Part A)

t(1) > t(2), the stone thrown 30 above the horizontal spends more time in the air.

Part B)

x(f1) > x(f2), the first stone will land farther away from the building.

Explanation:

<u>Part A)</u>

Let's use the parabolic motion equation to solve it. Let's define the variables:

y(i) is the initial height, it is a constant.
y(f) is the final height, in our case is 0
v(i) is the initial velocity (v(i)=16 m/s)
θ1 is the first angle, 30°
θ2 is the first angle, -30°

For the first stone

$y_{f1}=y_{i1}+v*sin(\theta_{1})t_{1}-0.5gt_{1}^{2}$

$0=y_{i1}+16*sin(30)t_{1}-0.5*9.81*t_{1}^{2}$

$0=y_{i1}+8t_{1}-4.905*t_{1}^{2}$ (1)

For the second stone

$0=y_{i2}+16*sin(-30)t_{2}-4.905t_{2}^{2}$

$0=y_{i2}-8t_{2}-4.905t_{2}^{2}$ (2)

If we solve the equation (1) we will have:

$t_{1}=\frac{-8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can do the same procedure for the equation (2)

$t_{1}=\frac{8\pm \sqrt{64+19.62*y_{i}}}{-9.81}$

We can analyze each solution to see which one spends more time in the air.

It is easy to see that the value inside the square root of each equation is always greater than 8, assuming that the height of the building is > 0. Now, to get positive values of t(1) and t(2) we need to take the negative option of the square root.

Therefore, t(1) > t(2), it means that the stone thrown 30 above the horizontal spends more time in the air.

We can use the equation of the horizontal position here.

<u>First stone</u>

$x_{f1}=x_{i1}+vcos(30)t_{1}$

$x_{f1}=0+13.86*t_{1}$

$x_{f1}=13.86*t_{1}$

<u>Second stone</u>

$x_{2}=x_{i2}+vcos(-30)t_{2}$

$x_{1}=0+13.86*t_{1}$

$x_{1}=13.86*t_{2}$

Knowing that t(1) > t(2) then x(f1) > x(f2)

Therefore, the first stone will land farther away from the building.

They land at different points at different times.

I hope it helps you!

3 0

3 years ago

The moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is

sleet_krkn [62]

Answer:

I = I₀ + M(L/2)²

Explanation:

Given that the moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is I₀.

The parallel axis theorem for moment of inertia states that the moment of inertia of a body about an axis passing through the centre of mass is equal to the sum of the moment of inertia of the body about an axis passing through the centre of mass and the product of mass and the square of the distance between the two axes.

The moment of inertia of the body about an axis passing through the centre of mass is given to be I₀

The distance between the two axes is L/2 (total length of the rod divided by 2

From the parallel axis theorem we have

I = I₀ + M(L/2)²

5 0

3 years ago