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2 years ago

Acceleration Practice

Physics

2 answers:

Trava [24]2 years ago

8 0

Answer:

0.85 m/s²

Explanation:

Acceleration is change in velocity over change in time. In mathematically, it can be expressed as:

$\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{\vec v_2 - \vec v_1}{t_2-t_1}}$

Our final velocity is given to be 17 m/s in 20 seconds. Initial velocity is at starting point which is 0 m/s in 0 second. Therefore:

$\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{17-0}{20-0}}\\\\\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = \dfrac{17}{20}}\\\\\displaystyle{\vec{a} = \dfrac{\Delta \vec{v}}{\Delta t} = 0.85 \ \, \sf{m/s^2}}$

Therefore, the acceleration of a horse from starting point to 17 m/s in 20 seconds is 0.85 m/s²

just olya [345]2 years ago

4 0

Answer:

0.85 m/s^2

Explanation:

Acceleration = (final velocity - starting velocity) / time

Final velocity = 17

Starting velocity = 0

Time = 20

1. Substitute:

Acceleration = (starting velocity - final velocity) / time --> Acceleration = ( 17 - 0) / 20

2. Solve:

Acceleration = ( 17 - 0) / 20 -->

Acceleration = 0.85

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You are driving your 1700 kg car at 21 m/s down a hill with a 5.0∘ slope when a deer suddenly jumps out onto the roadway. You sl

podryga [215]

Answer:

d = 27.7 m

Explanation:

Here the car is driving on the inclined plane

So here we can say that work done by the gravity and work done by friction is equal to change in kinetic energy of the system

So here we can write it as

$mgsin\theta \times d - F_f \times d = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$

now we have

m = 1700 kg

$v_f = 0$

$v_i = 21 m/s$

$F_f = 1.5 \times 10^4 N$

$(1700)(9.81)sin5 (d) - (1.5\times 10^4)d = 0 - \frac{1}{2}(1700)(21^2)$

$1453.5 d - (1.5\times 10^4)d = -374850$

$d = 27.7 m$

6 0

3 years ago

Students in an introductory physics lab are performing an experiment with a parallel-plate capacitor made of two circular alumin

Zarrin [17]

Answer:

$\rm 9.186\times 10^{-7}\ C.$

Explanation:

<u>Given:</u>

Diameter of the plates of the capacitor, D = 21 cm = 0.21 m.
Distance of separation between the plates, d = 1.0 cm = 0.01 m.
Minimum value of electric field that produces spark, $\rm E=3\times 10^6\ N/C.$

When the dimensions of the plate of the capacitor is comparatively much larger than the distance of separation between the plates, then, according to the Gauss' law of electrostatics, the value of the electric field strength in the region between the plates of the capacitor is given by

$\rm E=\dfrac{\sigma}{\epsilon_o}.$

where,

$\rm \sigma$ = surface charge density of the plate of the capacitor = $\dfrac qA$ .

$\rm q$ = magnitude of the charge on each of the plate.
$\rm A$ = surface area of each of the plate = $\rm \pi \times (Radius)^2=\pi \times\left ( \dfrac{D}{2}\right )^2= \pi \times \left ( \dfrac{0.21}{2}\right )^2=3.46\times 10^{-2}\ m^2.$
$\epsilon_o$ = electrical permittivity of free space, having value = $8.85\times 10^{-12}\rm \ C^2N^{-1}m^{-2}.$

For the minimum value of electric field that produces spark,

$\rm E = \dfrac{q}{A\epsilon_o}\\\Rightarrow q = E\ A\epsilon_o\\=3\times 10^6\times 3.46\times 10^{-2}\times 8.85\times 10^{-12}\\=9.186\times 10^{-7}\ C.$

It is the maximum value of the magnitude of charge which can be added up to each of the plates of the capacitor.

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3 years ago

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makkiz [27]

made from pure metals . . . no;
they've been made from all kinds of weird compounds and alloys.

conduct electricity with zero resistance . . . yes;
that's why they're called "superconductors".

produce a strong magnetic field . . . possible, but not because it's a superconductor;
just like any other conductor, the magnetic field depends on the current that's flowing in the conductor.

no loss of energy in the transfer of electricity . . .
there's no loss of energy in the current flowing in the superconductor;
but if you tried to transfer the current out of the superconductor into
something else, then there would be some loss.

7 0

3 years ago

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Liono4ka [1.6K]

Answer:

Safety

Explanation:

Expressways are banked to resist centifugal action

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3 years ago

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Answer:

Explanation:

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3 years ago