1.A poorly tuned Yugo can accelerate from rest to a speed of 28 m/s in
20 s.
a) What is the average acceleration of the car?
b) What distance does it travel in this time?

Answers

Answer 1

Answer:

The average acceleration of the car is 1.4 m / s^2.
The distance travel at this time is 560 m.

Explanation:

Average acceleration is defined as the rate of change of velocity to the rate of change of period. Change of velocity refers to the difference between final velocity and initial velocity. The unit of acceleration is meter/sec/sec (or) m s^2.

Avg Acceleration=(Final velocity - Initial velocity) /(Final time - Initial time)

a. To find the average acceleration of the car, (initial velocity and initial time is 0, the final velocity is 28 m/s, final time = 20 s)

Avg acceleration of the car = (28 - 0) / (20 - 0)

= 28 / 20 = 1.4 m / s^2.

b. The distance traveled is given by the product of speed and time

Distance traveled = Speed * Time

The Distance traveled by the car = 28 * 20 = 560 m.

Answer 2

Answer:

Final answer:

By applying basic physics formulas, the car's average acceleration can be determined as 1.4 m/s², and the distance it travels in 20 seconds is 280 meters.

Explanation:

To answer this question, you'll need to apply basic physics formulas. Let's go step by step:

a) What is the average acceleration of the car?

Acceleration is defined as the rate of change of velocity per unit of time. The formula is Acceleration = (Final speed - Initial speed) / Time. Here, the initial speed is 0 (because the car is starting from rest), final speed is 28 m/s and time is 20 s.

So, Acceleration = (28 m/s - 0) / 20 s = 1.4 m/s². That's your answer for part a.

b) What distance does it travel in this time?

The distance traveled by the car can be found using the formula of linear motion: Distance = Initial speed x Time + 0.5 x Acceleration x Time². Since the car starts from rest, the initial speed is zero, and this component of the formula will be zero.

So, Distance = 0 + 0.5 x 1.4 m/s² x (20 s)² = 280 m. So, the car travels 280 meters in that time.

Learn more about Physics Acceleration and Distance here:

brainly.com/question/32817533

#SPJ3

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What do you think that astronomers mean when they use the term observable universe? (Hint: Think of the time it takes for light from very distant objects to reach Earth.)

Answers

What does "observable universe" mean?

Light does not travel instantaneously between points in space. It has a finite speed "c", measured experimentally to be about 3 x 108 meters/second (or about 1.1 x 109 kilometers/hour. Flying at this rate you could get from NYC to Tokyo in about 1/30th of a second.)

Since light takes time to travel, we never actually see the current moment. Looking down at your hand, you do not see it as it is right now, but rather as it was a miniscule fraction of a moment earlier. Now, this interval is so small, given the short distance between your retina and your hand, that the difference is utterly negligible. In fact, bound by Earth's meager scope, the phenomenon isn't really worth mentioning.

The discrepancy becomes significant, however, when exploring much larger distances. Light years, for example. A light year is the distance light travels in one year. If you look at a star that's 50 light years away, you are seeing it as it was 50 years ago. Thus the deeper you peer into space, the farther you are seeing back in time. If this star had exploded 49 years ago, in a spectacular event called a Supernova, we would not know it until 1 more year from now.

Likewise, any event that happened beyond a certain point in the past is unknowable to us if the signal from it hasn't had time to reach us. It is not that our telescopes are too weak, or our instrumentation insensitive. We simply do not yet have access to the information. (No matter how prolific a reader you may be, you'd be hard pressed to read a friend's email if it has yet to arrive in your inbox.)

As a consequence of this limitation, astronomers often refer to the observable universe, a term referring to the volume of space that we are physically able to detect. The question of what lies outside this observable region is a tempting one to ponder. Yet inspiring though it may be, there is a certain futility in such a pursuit.

What happened to the sound made by the tuning fork hit the prongs harder

Answers

the sound waves and vibration will increase causing the sound to be louder. just like if you strum a guitar lightly it is a soft sound, if you strum it hard it will be loud and play not so smootly

Why is a conductor important in science terms

Answers

A conductor is important to science terms because conductors conduct heat which means they keep something hot. Insulators are the opposite.

Help please!
Which of the following are valid names for the equinoxes?

Answers

Vernal Equinox and Autumn Equinox and Winter and Summer Solstice. :D Hope that helps.

There is a Spring equinox (March in the northern hemisphere), and a Fall equinox (September).

3. A block of mass m1=1.5 kg on an inclined plane of an angle of 12° is connected by a cord over a mass-less, frictionless pulley to a second block of a mass m2= 1.2 kg hanging vertically. If the hanging block falls 0.92min 1.23 s, what is the coefficient of friction between m1 and the inclined plane?

Please show work

Answers

Answer:

$\mu=0.377$

Explanation:

we need to start by drawing the free body diagram for each of the masses in the system. Please see attached image for reference.

We have identified in green the forces on the blocks due to acceleration of gravity ( $w_1$ and $w_2$ ) which equal the product of the block's mass times "g".

On the second block ( $m_2$ ), there are just two forces acting: the block's weight ( $m_2\,*\,g$ ) and the tension (T) of the string. We know that this block is being accelerated since it has fallen 0.92 m in 1.23 seconds. We can find its acceleration with this information, and then use it to find the value of the string's tension (T). We would need both these values to set the systems of equations for block 1 in order to find the requested coefficient of friction.

To find the acceleration of block 2 (which by the way is the same acceleration that block 1 has since the string doesn't stretch) we use kinematics of an accelerated object, making use of the info on distance it fell (0.92 m) in the given time (1.23 s):

$x_f-x_i=v_i\,t-(1)/(2) a\,t^2$ and assume there was no initial velocity imparted to the block:

$x_f-x_i=v_i\,t-(1)/(2) a\,t^2\n-0.92\,m=0\,-(1)/(2) a\,(1.23)^2\na=(0.92\,*\,2)/(1.23^2) \na=1.216 \,(m)/(s^2)$

Now we use Newton's second law in block 2, stating that the net force in the block equals the block's mass times its acceleration:

$F_(net)=m_2\,a\nw_2-T=m_2\,a\nm_2\,g-T=m_2\,a\nm_2\,g-m_2\,a=T\nm_2\,(g-a)=T\n1.2\,(9.8-1.216)\,N=T\nT=10.3008\,N$

We can round this tension (T) value to 10.3 N to make our calculations easier.

Now, with the info obtained with block 2 (a - 1.216 $(m)/(s^2)$ , and T = 10.3 N), we can set Newton's second law equations for block 1.

To make our study easier, we study forces in a coordinate system with the x-axis parallel to the inclined plane, and the y-axis perpendicular to it. This way, the motion in the y axis is driven by the y-component of mass' 1 weight (weight1 times cos(12) -represented with a thin grey trace in the image) and the normal force (n picture in blue in the image) exerted by the plane on the block. We know there is no acceleration or movement of the block in this direction (the normal and the x-component of the weight cancel each other out), so we can determine the value of the normal force (n):

$n-m_1\,g\,cos(12^o)=0\nn=m_1\,g\,cos(12^o)\nn=1.5\,*\,9.8\,cos(12^o)\nn=14.38\,N$

Now we can set the more complex Newton's second law for the net force acting on the x-axis for this block. Pointing towards the pulley (direction of the resultant acceleration $a$ ), we have the string's tension (T). Pointing in the opposite direction we have two forces: the force of friction (f ) with the plane, and the x-axis component of the block's weight (weight1 times sin(12)):

$F_(net)=m_1\,a\nT-f-w_1\,sin(12)=m_1\,a\nT-w_1\,sin(12)-m_1\,a=f\nf=[10.3-1.5\,*\,9.8\,sin(12)-1.5\,*1.216]\,N\nf=5.42\,N$

And now, we recall that the force of friction equals the product of the coefficient of friction (our unknown $\mu$ ) times the magnitude of the normal force (14.38 N):

$f=\mu\,n\n5.42\,N=\mu\,*\,14.38\,N\n\mu=(5.42)/(14.38)\n\mu=0.377$

with no units.

what happens if heat flows into matter when no change of state is involved? a. both thermal energy and temperature increase b. both thermal energy and temperature decrease c. thermal energy decreases, temperature increases d. thermal energy increases, temperature decreases.