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3 years ago

A car is initially traveling at 17 m/s when the driver steps down hard on the gas pedal

Physics

1 answer:

ddd [48]3 years ago

6 0

Answer:

29 m/s

Explanation:

Acceleration is the rate of change of velocity, i.e, the change in velocity per unit time.

The driver made the car to accelerate at 3 m/s^2. That means than in every second the velocity of the car increased by 3 m/s. Therefore in four seconds its velocity would increase by 3m/s x 4 = 12m/s. Starting with an initial velocity of 17 m/s the final velocity would be 17 m/s + 12 m/s = 29 m/s.

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For no apparent reason, a poodle is running at a constant speed of 5.00 m/s in a circle with radius 2.9 m . Let v⃗ 1 be the velo

Nostrana [21]

At a constant speed of 5.00 m/s, the speed at which the poodle completes a full revolution is

$\left(5.00\,\dfrac{\mathrm m}{\mathrm s}\right)\left(\dfrac{1\,\mathrm{rev}}{2\pi(2.9\,\mathrm m)}\right)\approx0.2744\,\dfrac{\mathrm{rev}}{\mathrm s}$

so that its period is $T=3.644\,\frac{\mathrm s}{\mathrm{rev}}$ (where 1 revolution corresponds exactly to 360 degrees). We use this to determine how much of the circular path the poodle traverses in each given time interval with duration $\Delta t$ . Denote by $\theta$ the angle between the velocity vectors (same as the angle subtended by the arc the poodle traverses), then

$\Delta t=0.4\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.4\,\mathrm s}\theta\implies\theta\approx39.56^\circ$

$\Delta t=0.2\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{0.2\,\mathrm s}\theta\implies\theta\approx19.78^\circ$

$\Delta t=7\times10^{-2}\,\mathrm s\implies\dfrac{3.644\,\mathrm s}{360^\circ}=\dfrac{7\times10^{-2}\,\mathrm s}\theta\implies\theta\approx6.923^\circ$

We can then compute the magnitude of the velocity vector differences $\Delta\vec v$ for each time interval by using the law of cosines:

$|\Delta\vec v|^2=|\vec v_1|^2+|\vec v_2|^2-2|\vec v_1||\vec v_2|\cos\theta$

$\implies|\Delta\vec v|=\begin{cases}3.384\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.4\,\mathrm s\\1.718\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=0.2\,\mathrm s\\0.6038\,\frac{\mathrm m}{\mathrm s}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

and in turn we find the magnitude of the average acceleration vectors to be

$\implies|\vec a|=\begin{cases}8.460\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.4\,\mathrm s\\8.588\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=0.2\,\mathrm s\\8.625\,\frac{\mathrm m}{\mathrm s^2}&\text{for }\Delta t=7\times10^{-2}\,\mathrm s\end{cases}$

So that takes care of parts A, C, and E. Unfortunately, without knowing the poodle's starting position, it's impossible to tell precisely in what directions each average acceleration vector points.

5 0

3 years ago

Moving a magnet quickly in and out of a coil of wire produces _____.

Oksanka [162]

<span>An electric current.</span>

8 0

4 years ago

A small plastic sphere of mass 2.0 g is charged to −5.0 nC and held 2.5 cm above a small glass bead of mass 0.015 g resting on a

LenKa [72]

Answer:uhhh

Explanation:

Uhhhh

5 0

4 years ago

In a certain experiment, cylindrical samples of diameter 4 cm and length 7 cm are used. The two thermocouples in each sample are

sergeinik [125]

Answer:

K = .3941 × 10³ W/m.K

Explanation:

Qcond = K A ΔT÷ L

∴K = Qcond ×L ÷ A ΔT

J ÷ S = P

P = I × V =Qcond

∴Qcond = I × V

= 0.6 A × 110 V

=66 W

L = 0.12 m

ΔT = 8 °C

Qcond =33 V

Area = (πD²) ÷ 4

= [π (4 × 10⁻² )²] ÷ 4

= 1.256 × 10⁻³ m²

∴A = 1.256 × 10 ⁻³³ m²

So K = ( Qcond × L ) ÷ A ΔT

= (33) (0.12 ) ÷ (1.256 ×10⁻³ ) × 8

= 0.3941 × 10³ W/m .K

7 0

3 years ago

Suppose you do work on an object from rest to a certain speed v. Why does it take four times the amount of work to increase its

Otrada [13]

Answer: it is due to friction.

Explanation: if you were to push an object that is at rest, there is static friction that you would first need to overcome. An example would be pushing a car as it is in neutral. Pushing it takes more force and work to overcome that friction.
But once it overcomes static, it will then turn to kinetic friction. Once the car starts to move, you notice that it is easier to push. This is due to the fact that kinetic friction would equal normal force times the constant while the static friction is greater than or equal to the normal force times the constant.

Hope this helps!!!

8 0

4 years ago