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3 years ago

How do you think a graph of deceleration would differ from a graph showing acceleration

2 answers:

6 0

A graph of acceleration would have to be a graph with an increasing slope, since you're going faster and faster

a graph of deceleration would have to be a graph with a negative slope since you are slowing down over time

Levart [38]3 years ago

6 0

Simply, a deceleration starts from top and goes down, and acceleration goes from bottom to top.

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Write the following number in scientific notation: 658000

Mice21 [21]

6.58 x 105 is the answer

8 0

3 years ago

A compact car, mass 660 kg, is moving at 1.00 ✕ 102 km/h toward the east. The driver of the compact car suddenly applies the bra

Maslowich

Answer: $59.43\times 10^3 kg-m/s$

Explanation:

Given

mass of car $m=660 kg$

Initial velocity of car $=102 km/h\approx 28.33 m/s$ towards east

Time taken to stop $t=2.1 s$

Force exerted $F_{avg}=4.8\times 10^3 N$

change in momentum is given by impulse imparted to the car

$Impulse(J)=-F\cdot t$

$J=-4.8\times 10^3\times 2.1$

$J=-59.49\times 10^3 kg-m/s$

negative Sign indicates that impulse is imparted opposite to the direction of motion

magnitude of momentum $J=59.49\times 10^3$

6 0

3 years ago

Read 2 more answers

Two protons are maintained at a separation of nm. Calculate the electric potential due to the two particles at the midpoint betw

Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero

Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

$\displaystyle{U=\frac{q}{4\pi \epsilon_0r}}$ (1)

where $q$ is the charge of the particle, $\epsilon_0$ the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and $r$ is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

$\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}$

Substituting the values $q=1.602 \cdot10^{-19}\ C$ , $\displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}}$ and $r=0.5 \cdot 10^{-9} m$ we obtain: