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Veseljchak [2.6K]
3 years ago
8

(02.09 MC) An object starts at rest. Its acceleration over 30 seconds is shown in the graph below: An acceleration versus time g

raph is shown with acceleration on the y axis from zero to seven meters per second per second and time on the x axis from zero to 30 second. The graph starts at the origin and goes to five meters per second in the first five seconds, then goes back to zero at time 10 seconds, then stays at zero until 20 seconds, then goes up to 6 meters per second until 30 seconds. Use the graph above to determine the change in speed of the object between 20 and 30 seconds? (2 points) Select one: a. 6 m/s b. 30 m/s c. 60 m/s d. 180 m/s
Physics
2 answers:
Irina-Kira [14]3 years ago
3 0

Answer:

c. 60 m/s

Explanation:

As you can see in the graph, or the description of the graph, the change in speed since the acceleration at 20 seconds was 0 will be all the acceleration that we added up after that second, since we accelerated at 6 m/s until 30 seconds, that means that the object remained 10 seconds accelerating at 6m/s, so 10seconds times 6 m/s that is 60 m/s in the change in speed between the 20th second and the 30th second.

Verizon [17]3 years ago
3 0

Answer:

c. 60 m/s

Explanation: It makes sense if you think about it!

It took me a while but than it made sense!! I hope this helps you out!

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Microwave ovens have a plate that rotates at a rate of about 7.0 rev/min. What is this in revolutions per second?
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Answer:

The value is  w =  0.1167 \ rev/second

Explanation:

From the question we are told that

    The rate at which the plate rotates is  w =7.0 \ rev/min

Generally the revolution per second is mathematically represented as

       w =  \frac{7.0}{60}

=>    w =  0.1167 \ rev/second

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3 years ago
The basis for defining the length of a day is the fact that question 21 options:
abruzzese [7]
B.earth rotates on its axis in 24 hours; i.e., it rotates 15° of longitude per hour.
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The projectile launcher shown below will give the object on the right an initial horizontal speed of 5.9 m/s. While the other ob
Crazy boy [7]

Answer:

104.3 cm  or 179.7

Explanation:

First find time that it takes for the object to hit the ground

\sqrt{(2H)/g}  ->   \sqrt{(2 x 179)/ 9.8} = 6.04s\\*

Then find xf of projectile xf= 5.9(6.04) = 37.7\\\\

not 100% sure if the projectile is going away from the object or towards it but you either do 142- 37.7   or    142+37.7  

hope that helps

4 0
2 years ago
An object with charge q = −6.00×10−9 C is placed in a region of uniform electric field and is released from rest at point A. Aft
sergeinik [125]

Answer:

a) 80 V

b) The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

Explanation:

<u>Given :</u>

We are given an object with charge q = -6.00 x I0^-9 C starts moving from the rest at point A, which means its kinetic energy at point A is zero ( K_{A}= 0) to the point B at distance l = 0.500m where its kinetic energy is (  K_{B}= 5.00 x 10^-7J) . Also, the electric potential of q at point A is VA = + 30.0 v.

<u>Required :</u>

<em>(a) We are asked to find the electric potential VB </em>

<em>(b) We want to determine the magnitude and the direction of the electric field E. </em>

<u> Solution </u>

(a) We are given the values for VA,K_{B} and q so we want to find a relationship between these three parameters and VB to get the value of VB.

As we have two states, at points A and B , where the charge moved from A to B due to the applied electric field. The mechanical energy of the object is conservative during this travel, and we can apply eq(1) in this situation:

                                   K_{A} +U_{A} =K_{B} +U_{B} .........................................(1)                                          

Where K_{A}= 0 and the potential energy U of the charge is given by U = q V

where V is the electric potential.  So, equation (1) will be in the form :

                                  0+qVA=K_{B} +qVB                      (Divide by q)

                                         VA=K_{B} /q + VB                  (solve for VB)

                                         VB=VA- K_{B}/q .......................................(2)

We get the relation between VB, VA and K_{B}, now we can plug our values for VA, K_{B} and q into equation (2) to get VB

                                         VB=VA- K_{B}/q

                                              =30V-(5.00 x 10^-7J)/(-6.00 x I0^-9)

                                              =80 V

(b) After we calculated VB we can use equation a to get the electric field E that applied to the charge q, where the potential difference between the two points equals the integration of the electric field multiplied by the distance l between the two points

                                   VA-VB =\int\limits^1_0 {E} \, dl...................................(a)

                                               =E\int\limits^1_0 {} \, dl

                                   VA-VB=El                      (solve for E)

                                            E= VA-VB/l..................................(3)

Now let us plug our values for VA, Vs and l into equation (3) to get the electric field E

                                            E= VA-VB/l

                                              =-100 N/C

The magnitude of the electric field is 100 N/C and the direction of the electric field is from point B to point A where the electric field is toward the negative charge

5 0
3 years ago
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