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qwelly [4]
3 years ago
6

Skylar goes to a pumpkin patch and pick out a pumpkin that has a mass of 6000 grams how many kilograms is the pumpkin

Physics
2 answers:
Alex3 years ago
6 0
1000g = 1kg
6000g =6000 × 1/1000
=6kg
myrzilka [38]3 years ago
3 0
The answer is 2
721.55422 kilograms
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This graph indicates that the kinetic energy of an object is increasing. What is the most reasonable explanation for this observ
soldi70 [24.7K]

B) The object's velocity doubled.

Explanation:

The graph is missing: find it in attachment.

The kinetic energy of an object is the energy possessed by the object due to its motion. It is calculated as

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is the velocity of the object

We notice that:

  • The  kinetic energy is directly proportional to the mass
  • The kinetic energy is proportional to the square of the velocity

In the graph, one of the two quantities (either mass or speed) is represented on the x-axis, while the quantity on the y-axis is the kinetic energy.

First of all, we notice that the relationship is not linear: this means that the quantity on the x-axis cannot be the mass, so it must be the velocity.

Moreover, we notice that when the quantity on the x-axis increases from 1 to 2 (so, it doubles), the kinetic energy increases by a factor of 4. This means that the object's velocity has doubled, therefore

B) The object's velocity doubled.

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

5 0
3 years ago
Linh builds a circuit from the diagram shown. Which bulb could Linh remove from the circuit to make all of the other bulbs stop
LenKa [72]

Answer:

4

Explanation:

In order for the current to continue flowing through the circuit (and for the bulbs to continue shining), there must be a closed path containing the battery where current can flow. Let's see the effect of removing each bulb on the circuit:

- 1: when removing bulb 1 only, the current can still flow through the path battery-bulb 3- bulb 4

- 2: when removing bulb 2 only, the current can still flow through the path battery-bulb 3- bulb 4

- 3: when removing bulb 3 only, the current can still flow through the path battery-bulb 1-bulb 2- bulb 4

- 4: when removing bulb 4 only, the current can no longer flow. In fact, there is no closed path that contains the battery now, so the current will not flow and all the bulbs will stop shining.

4 0
3 years ago
Read 2 more answers
What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 47.0 nC placed between q1
lesya [120]

Answer:

Incomplete question, check attachment for completed question

Explanation:

The force of attraction between two forces are given as

F=kQq/r²

4 0
3 years ago
Starting from your campsite you walk 3.0 km east, 6.0 km north, 1.0 km east, and then 4.0 km west. How far are you from your cam
Hatshy [7]
Think of it like a graph. You start at the origin which is (0,0).  go three to the east which now you are (3,0). Then, six to the north. Now, you are at (3,6).  1 to the east, ((4,6).  Then you go 4 to the west which is back tracking. So, you end at (0,6) which is saying you are now 6 km north from your campsite. 

Hope this helps!
6 0
3 years ago
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates
34kurt

The given question is incomplete. The complete question is as follows.

A parallel-plate capacitor has capacitance C_{0} = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.

What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed 3.00 \times 10^{4} V/m?

Explanation:

It is known that relation between electric field and the voltage is as follows.

             V = Ed

Now,  

              Q = CV

or,           Q = C \times Ed

Therefore, substitute the values into the above formula as follows.

              Q = C \times Ed

                  = 8.50 pF \times (\frac{10^{-12} F}{1 pF})(3 \times 10^{4} m/s)(1 mm)(\frac{10^{-3} m}{1 mm})

                  = 2.55 \times 10^{-10} C

Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is 2.55 \times 10^{-10} C.

3 0
4 years ago
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