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2 years ago

Pls help me, tomorrow is the deadline of my assignment

Physics

1 answer:

elena-14-01-66 [18.8K]2 years ago

3 0

Answer:

6- B

7- I think A

8-B

9-D

10-A

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Which of the following is a unit of acceleration?

olasank [31]

${\tt{\red{\underline{\underline{\huge{Answer:}}}}}}$

$\longrightarrow$ The rate of change of velocity per unit time is called acceleration.

$\longrightarrow$ Its SI unit is m/s².

$\huge\boxed{\fcolorbox{blue}{red}{Thank you}}$

7 0

2 years ago

A 0.450 kg soccer ball has a kinetic energy of 119 J.

Anastaziya [24]

Answer:

V is approximately = 23m/s

Explanation:

Kinetic energy = ½ mv²

Where m= mass = 0.450kg

V= velocity =?

K. E = 119J

Therefore

K. E = ½ mv²

Input values given

119= ½ × 0.450 × v²

Multiply both sides by 2

119 ×2 = 2 × 1/2 × 0.450 × v²

238= 0.450v²

Divide both sides by 0.450

238/0.450 = 0.450v²/0.450

v² = 528.89

Square root both sides

Sq rt v² = sq rt 528.89

V = 22.998m/s

V is approximately = 23m/s

I hope this was helpful, please rate as brainliest

8 0

3 years ago

Pls help ASAP. This is astronomy but there isn't an option :)

marin [14]

From largest to smallest they are: Universe, galaxy, solar system, star, planet, moon and asteroid.

Explanation:Let's describe them from smallest to largest. In fact the size order is not exact as there are exceptions.An asteroid is a rocky body which lies in the asteroid belt between Mars and Jupiter. They are typically quite small object. The largest asteroid Ceres has been reclassified as a dwarf planet.A moon is typically a rocky body which is in orbit around a planet. Some moons such as our Moon are quite large and are typically bigger than asteroid. Some moons can actually be smaller than some asteroids.A planet is a nearly spherical body which is in orbit around the Sun. Planets are larger than moons.A star is what planets orbit around. It is the source of light and heat. Our Sun is a star which is many times bigger than all of the planets.A solar system is a star and all of its planets, asteroids, comets and other bodies. It is significantly bigger than a star.A galaxy, such as our Milky Way Galaxy, is a collection of solar systems orbiting around a central core. Most galaxies have a supermassive black hole at their centres.Galaxies also form clusters which are large scale structures.The universe is everything. It contains billions of galaxies. Lots of information RIGHT!!!!

YOUR VERY WelCoMe!!!! :) :) :) :) :0 :)

7 0

3 years ago

What is the best inference for the speed of the car after 3 seconds

SVEN [57.7K]

Answer:

B. 17m/s

Explanation:

This question contains a graph that illustrates the relationship between the speed of a car over time. The graph shows that one can make an inference of the amount of time it takes for the car to cover a particular speed and vice versa.

In this case, after 3 seconds, the speed of the car will be 17 m/s. This inference was got by tracing the position of 3s in the x-axis to the value on the y-axis. Doing this, the best inference for the speed of the car after 3 seconds is 17m/s.

8 0

2 years ago

A car traveling on a flat (unbanked), circular track accelerates uniformly from rest with a tangential acceleration of 1.90 m/s2

Ahat [919]

Answer:

Approximately $0.608$ (assuming that $g = 9.81\; \rm N\cdot kg^{-1}$ .)

Explanation:

The question provided very little information about this motion. Therefore, replace these quantities with letters. These unknown quantities should not appear in the conclusion if this question is actually solvable.

Let $m$ represent the mass of this car.
Let $r$ represent the radius of the circular track.

This answer will approach this question in two steps:

Step one: determine the centripetal force when the car is about to skid.
Step two: calculate the coefficient of static friction.

For simplicity, let $a_{T}$ represent the tangential acceleration ( $1.90\; \rm m \cdot s^{-2}$ ) of this car.

<h3>Centripetal Force when the car is about to skid</h3>

The question gave no information about the distance that the car has travelled before it skidded. However, information about the angular displacement is indeed available: the car travelled (without skidding) one-quarter of a circle, which corresponds to $90^\circ$ or $\displaystyle \frac{\pi}{2}$ radians.

The angular acceleration of this car can be found as $\displaystyle \alpha = \frac{a_{T}}{r}$ . ( $a_T$ is the tangential acceleration of the car, and $r$ is the radius of this circular track.)

Consider the SUVAT equation that relates initial and final (tangential) velocity ( $u$ and $v$ ) to (tangential) acceleration $a_{T}$ and displacement $x$ :

$v^2 - u^2 = 2\, a_{T}\cdot x$ .

The idea is to solve for the final angular velocity using the angular analogy of that equation:

$\left(\omega(\text{final})\right)^2 - \left(\omega(\text{initial})\right)^2 = 2\, \alpha\, \theta$ .

In this equation, $\theta$ represents angular displacement. For this motion in particular:

$\omega(\text{initial}) = 0$ since the car was initially not moving.
$\theta = \displaystyle \frac{\pi}{2}$ since the car travelled one-quarter of the circle.

Solve this equation for $\omega(\text{final})$ in terms of $a_T$ and $r$ :

$\begin{aligned}\omega(\text{final}) &= \sqrt{2\cdot \frac{a_T}{r} \cdot \frac{\pi}{2}} = \sqrt{\frac{\pi\, a_T}{r}}\end{aligned}$ .

Let $m$ represent the mass of this car. The centripetal force at this moment would be:

$\begin{aligned}F_C &= m\, \omega^2\, r \\ &=m\cdot \left(\frac{\pi\, a_T}{r}\right)\cdot r = \pi\, m\, a_T\end{aligned}$ .

<h3>Coefficient of static friction between the car and the track</h3>

Since the track is flat (not banked,) the only force on the car in the horizontal direction would be the static friction between the tires and the track. Also, the size of the normal force on the car should be equal to its weight, $m\, g$ .

Note that even if the size of the normal force does not change, the size of the static friction between the surfaces can vary. However, when the car is just about to skid, the centripetal force at that very moment should be equal to the maximum static friction between these surfaces. It is the largest-possible static friction that depends on the coefficient of static friction.

Let $\mu_s$ denote the coefficient of static friction. The size of the largest-possible static friction between the car and the track would be:

$F(\text{static, max}) = \mu_s\, N = \mu_s\, m\, g$ .

The size of this force should be equal to that of the centripetal force when the car is about to skid:

$\mu_s\, m\, g = \pi\, m\, a_{T}$ .

Solve this equation for $\mu_s$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g}$ .

Indeed, the expression for $\mu_s$ does not include any unknown letter. Let $g = 9.81\; \rm N\cdot kg^{-1}$ . Evaluate this expression for $a_T = 1.90\;\rm m \cdot s^{-2}$ :

$\mu_s = \displaystyle \frac{\pi\, a_T}{g} \approx 0.608$ .

(Three significant figures.)

7 0

3 years ago

Pls help me, tomorrow is the deadline of my assignment ​

Pls help me, tomorrow is the deadline of my assignment