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Rina8888 [55]
2 years ago
14

WHAT ARE THREE TECHNICAL ADVANCEMENTS

Physics
1 answer:
Snezhnost [94]2 years ago
6 0

the iphone

the airplane

5G

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A 2.00 kg block hangs from a spring. A 300 g body hung below the block stretches the spring 2.00 cm farther. (a) What is the spr
Leokris [45]

The spring constant is 147 N/m

Given the mass of the block is 2.00 kg , the mass of the body is 300 g and the length of the spring is 2.00 cm

We need to find the spring constant

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.

The force required to stretch an elastic object such as a metal spring is directly proportional to the extension of the spring

We know that F = kx

300(9.8)= k (0.02)

k = 147.15 N/m

Rounding off to the nearest is 147N/m

The spring constant is 147N/m

Learn more about Hooke's law here

brainly.com/question/15365772

#SPJ4

7 0
2 years ago
A generator converts mechanical energy into _____________________ energy.
erica [24]
Electrical energy is your answer.
6 0
3 years ago
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Which is the best description of inertia? Inertia is
valentinak56 [21]

Answer:

The best description of inertia is an objects velocity

8 0
3 years ago
When the distance between two charges is halved, the electrical force between them?
Llana [10]
If the distance between two charges is halved, the electrical force between them increases by a factor 4.

In fact, the magnitude of the electric force between two charges is given by:
F= k \frac{q_1 q_2}{r^2}
where
k is the Coulomb's constant
q1 and q2 are the two charges
r is the separation between the two charges

We see that the magnitude of the force F is inversely proportional to the square of the distance r. Therefore, if the radius is halved:
r'= \frac{r}{2}
the magnitude of the force changes as follows:
F'=k \frac{q_1 q_2}{r'^2}=k \frac{q_1 q_2}{( \frac{r}{2})^2 }=k \frac{q_1 q_2}{ \frac{r^2}{4} } =4k  \frac{q_1 q_2}{r^2}=4 F
so, the force increases by a factor 4.
3 0
3 years ago
An electron with a speed of 1.7 × 107 m/s moves horizontally into a region where a constant vertical force of 4.9 × 10-16 N acts
Alex Ar [27]

Answer:

 y = 77.74 10⁻⁵ m

Explanation:

For this exercise we can use Newton's second law

        F = m a

        a = F / m

        a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹

        a = 0.538 10¹⁵ m / s

This is the vertical acceleration of the electron.

Now let's use kinematics to find the time it takes to move the

         x= 29 mm = 29 10⁻³ m

On the x axis

            v = x / t

            t = x / v

            t = 29 10⁻³ / 1.7 10⁷

            t = 17 10⁻¹⁰ s

Now we can look for vertical distance at this time.

            y = v_{oy} t + ½ a t²

            y = 0 + ½ 0.538 10¹⁵ (17 10⁻¹⁰)²

            y = 77.74 10⁻⁵ m

3 0
3 years ago
Read 2 more answers
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