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2 years ago

10

Which of Newton's laws accounts for the following statement? "Negative acceleration is proportional to applied braking force." f

irst law second law third law

2 answers:

Vinil7 [7]2 years ago

7 0

I believe it is Newtons Second Law! :)

Dahasolnce [82]2 years ago

3 0

The Correct Answer Is <em>Newton's Second Law</em>

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A 1,000 kg car is travelling at 6.5 m/s to the North. A 3,500 kg truck is travelling South at the same velocity. What is the tot

Molodets [167]

Answer:

16250 kgm/s due south

Explanation:

Applying,

M = mv................. Equation 1

Where M = momentum, m = mass, v = velocity.

From the car,

Given: m = 1000 kg, v = 6.5 m/s

Substitute these values into equation 1

M = 1000(6.5)

M = 6500 kgm/s

For the truck,

Given: m = 3500 kg, v = 6.5 m/s

Substitute these values into equation 1

M' = 3500(6.5)

M' = 22750 kgm/s.

Assuming South to be negative direction,

From the question,

Total momentum of the two vehicles = (6500-22750)

Total momentum of the two vehicles = -16250 kgm/s

Hence the total momentum of the two vehicles is 16250 kgm/s due south

3 0

2 years ago

dedylja [7]

Pretty sure it’s A. Hope this helps.

4 0

2 years ago

Total these measurements. Your answer should indicate the proper accuracy. Be sure to include the units in your answer. (Remembe

vivado [14]

Answer:

10

Explanation:

This is tough. The last number 0.2 has only one significant figure. So while the sum of all the numbers is 12.3, you must only leave one sig figure. Rounding to the tenths gives 10.

3 0

3 years ago

A parallel-plate capacitor with plates of area 360 cm2 is charged to a potential difference V and is then disconnected from the

Softa [21]

Answer:

$Q=3.9825\times 10^{-9} C$

Explanation:

We are given that a parallel- plate capacitor is charged to a potential difference V and then disconnected from the voltage source.

1 m =100 cm

Surface area =S= $\frac{360}{10000}=0.036 m^2$

$\Delta d=0.8 cm=0.008 m$

$\Delta V=100 V$

We have to find the charge Q on the positive plates of the capacitor.

V=Initial voltage between plates

d=Initial distance between plates

Initial Capacitance of capacitor

$C=\frac{\epsilon_0 S}{d}$

Capacitance of capacitor after moving plates

$C_1=\frac{\epsilon_0 S}{(d+\Delta d)}$

$V=\frac{Q}{C}$

Potential difference between plates after moving

$V=\frac{Q}{C_1}$

$V+\Delta V=\frac{Q}{C_1}$

$\frac{Qd}{\epsilon_0S}+100=\frac{Q(d+\Delta d)}{\epsilon_0S}$

$\frac{Q(d+\Delta d)}{\epsilon_0 S}-\frac{Qd}{\epsilon_0S}=100$

$\frac{Q\Delta d}{\epsilon_0 S}=100$

$\epsilon_0=8.85\times 10^{-12}$

$Q=\frac{100\times 8.85\times 10^{-12}\times 0.036}{0.008}$

$Q=3.9825\times 10^{-9} C$

Hence, the charge on positive plate of capacitor= $Q=3.9825\times 10^{-9} C$

6 0

3 years ago

Acceleration is the magnitude of average velocity.

lina2011 [118]

Answer:

false

Explanation:

just did the question on apex, true was wrong

4 0

3 years ago