For the examples below, explain what a physical change or a chemical change has occurred. Justify your answerA. When you mix baking soda and vinegar, carbon dioxide is released.

B. You build a tall sand castle at the beach. After wave washes it over, the sand castle turns into a big pile of sand

C. Boiling water turns a raw egg into a hard boiled egg

D. Max divided cookie dough into small pieces on a cookie sheet.

E. A loaf of freshly baked bread tastes better and looks much different than a lump of bread dough

F. A glass of water is left in the Sun. In the time, the water evaporates leaving the glass empty.

Answers

Answer 1

Answer: Letters A, E, and F undergo chemical change because new substances are formed as to carbon dioxide, bread, and water vapor respectively, whereby letters B, C, and D undergo physical change because the tall sand castle remains as sand, the raw egg remains as an egg as it hardens (boiled) and the cookies remain as cookies no matter how many times it was divided into smaller pieces. The later substances' chemical composition were never changed nor formed new substances as they undergo physical processes.

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a truck is moving around a circular curve @ a uniform velocity of 13m/s. if the centifrical force on truck is 3,300N with truck mass of 1600kg. What is curve radius?

Answers

Answer

81.94 m

Explanation

The centripetal force of an object moving in a circular path is given by:

F = mv²/r Where m is the mass of the object, v is the constant velocity and r is the radius of the curve.

F = mv²/r

3,300 = (1600×13²)/r

3,300 = 270,400/r

r = 270,400/3,300

= 81.94 m

Which of the following answers describe how to calculate the acceleration of an object? A. Divide the change in the objects velocity by the time it takes to make that change.
B. Divide the time the object travels by the distance it travels.
C. Calculate the distance the object travels and divide that by the time it takes to get there.
D. Calculate the velocity of an object and divide that by the time the object is in motion.

Answers

A. Divide the change in the objects velocity by the time it takes to make that change.

Explanation:

Acceleration is a vector quantity whose magnitude is given by the following equation:

$a=(\Delta v)/(\Delta t)$

where

$\Delta v$ is the change in velocity of the object

$\Delta t$ is the time it took to make the change in velocity

Therefore, we see that this equation corresponds to choice A.

I think it's A I hope this is right but I just studied this so I think this is the right answer

Whats a snoogley please help

Answers

Somthing that you cuddle

A force is always required to move an object from rest

Answers

True a resultant force is needed

Newton's universal law of gravitation says that every object exerts a force on another object. The shuttle has a gravitational force with Earth, the moon, and whatever planet it is near. This force prevents the shuttle from drifting.Question 1 options:
True
False

Answers

the correct answer is true

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.

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