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3 years ago

Match the particles with their characteristics,

Physics

2 answers:

In-s [12.5K]3 years ago

6 0

Positive charge=proton
Negative charge=electron
No charge/neutral=neutron

Pepsi [2]3 years ago

4 0

Answer:

Neuton has no charge =proton has positive charge >electron has negative charge.

Explanation:

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An apple falls out of a tree from a height of 2.3m. what is the impact speed of the apple

Anuta_ua [19.1K]

Answer:6.71 m/s

Explanation:

Given

Apple fall from a height of $h=2.3 m$

We need to find the impact speed of apple which can be given by using

$v^2-u^2=2gh$

where v=final velocity

u=initial velocity

h=Displacement

Assuming initial velocity to be zero

substituting the value we get

$v^2-0=2\times 9.8\times 2.3$

$v=\sqrt{2\times 9.8\times 2.3}$

$v=6.71\ m/s$

5 0

3 years ago

What is the displacement of a spring if it has a spring constant of 10 N/m, and a force of 2.5 N is applied?

pochemuha

O.25 m is the displacement

5 0

3 years ago

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Two rocks are at the top of a building. Rock 1 is dropped from rest while Rock 2 is thrown horizontaly at a velocity of 5 ms.

LuckyWell [14K]

Answer:

I'm pretty sure the answer is 0 m/s²

Explanation:

The horizontal velocity of the second rock is 5 m/s, so if we pretend air resistance doesn't exist, it will maintain that horizontal velocity, meaning that there is no horizontal acceleration.

7 0

3 years ago

You launch a cannonball at an angle of 35° and an initial velocity of 36 m/s (assume y = y₁=

velikii [3]

Answer:

Approximately $4.2\; {\rm s}$ (assuming that the projectile was launched at angle of $35^{\circ}$ above the horizon.)

Explanation:

Initial vertical component of velocity:

$\begin{aligned}v_{y} &= v\, \sin(35^{\circ}) \\ &= (36\; {\rm m\cdot s^{-1}})\, (\sin(35^{\circ})) \\ &\approx 20.6\; {\rm m\cdot s^{-1}}\end{aligned}$ .

The question assumed that there is no drag on this projectile. Additionally, the altitude of this projectile just before landing $y_{1}$ is the same as the altitude $y_{0}$ at which this projectile was launched: $y_{0} = y_{1}$ .

Hence, the initial vertical velocity of this projectile would be the exact opposite of the vertical velocity of this projectile right before landing. Since the initial vertical velocity is $20.6\; {\rm m\cdot s^{-1}}$ (upwards,) the vertical velocity right before landing would be $(-20.6\; {\rm m\cdot s^{-1}})$ (downwards.) The change in vertical velocity is:

$\begin{aligned}\Delta v_{y} &= (-20.6\; {\rm m\cdot s^{-1}}) - (20.6\; {\rm m\cdot s^{-1}}) \\ &= -41.2\; {\rm m\cdot s^{-1}}\end{aligned}$ .

Since there is no drag on this projectile, the vertical acceleration of this projectile would be $g$ . In other words, $a = g = -9.81\; {\rm m\cdot s^{-2}}$ .

Hence, the time it takes to achieve a (vertical) velocity change of $\Delta v_{y}$ would be:

$\begin{aligned} t &= \frac{\Delta v_{y}}{a_{y}} \\ &= \frac{-41.2\; {\rm m\cdot s^{-1}}}{-9.81\; {\rm m\cdot s^{-2}}} \\ &\approx 4.2\; {\rm s} \end{aligned}$ .

Hence, this projectile would be in the air for approximately $4.2\; {\rm s}$ .

8 0

1 year ago

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According to Newton's Second Law, the force of the club hitting the golf ball will cause it to accelerate. At the moment of impa

vazorg [7]

Answer:

Option B

Explanation:

<h3>According to Newton's third law, for every reaction there will be equal and opposite reaction</h3>

Here in this case the force of the club hitting the golf ball will be in one direction and the force acting on club due to golf ball will be in opposite direction and magnitude of this force will be same as the magnitude of the force of the club hitting the golf ball

In this case the action will be the force of the club hitting the golf ball and reaction will be the force acting on club due to golf ball

∴ The club pushes against to golf ball with a force equal and opposite to the force of the golf ball on the club

8 0

3 years ago

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