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2 years ago

6

A roller coaster car rapidly picks up speed as it rolls down a slope. As it starts down the slope, its speed is 4m/s. But 3 seco

nds later at the bottom of the slope its speed is 22m/s. What is its average acceleration?

1 answer:

slamgirl [31]2 years ago

7 0

Answer:

6 m/s^2

Explanation:

The average acceleration of the car is given by:

$a=\frac{v-u}{t}$

where

v is the final speed

u is the initial speed

t is the time elapsed

For the car in this problem,

v = 22 m/s

u = 4 m/s

t = 3 s

Therefore, the acceleration of the car is

$a=\frac{22 m/s-4 m/s}{3 s}=6 m/s^2$

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A parallel-plate air capacitor is made from two plates 0.210 m square, spaced 0.815 cm apart. it is connected to a 120 v battery

GuDViN [60]

Answer:

at the beginning: $2.3\cdot 10^{-10} F$

when the plates are pulled apart: $1.1\cdot 10^{-10} F$

Explanation:

The capacitance of a parallel-plate capacitor is given by

$C=k \epsilon_0 \frac{A}{d}$

where

k is the relative permittivity of the medium (for air, k=1, so we can omit it)

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the permittivity of free space

A is the area of the plates of the capacitor

d is the separation between the plates

In this problem, we have:

$A=0.210 m^2$ is the area of the plates

$d=0.815 cm=8.15\cdot 10^{-3} m$ is the separation between the plates at the beginning

Substituting into the formula, we find

$C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{8.15\cdot 10^{-3} m}=2.3\cdot 10^{-10} F$

Later, the plates are pulled apart to $d=1.63 cm=0.0163 m$ , so the capacitance becomes

$C=(1)(8.85\cdot 10^{-12}F/m)\frac{0.210 m^2}{0.0163 m}=1.1\cdot 10^{-10} F$

4 0

2 years ago

Reading or writing on a phone is especially dangerous while driving because:

algol [13]

Answer:

Explanation:

When we are driving we need a lot of attention and concentration. Also one involved in driving should be consious and courteous

Thus, whenever a person is drives, and when he is disactracted by Mobile phones it will destroy his presence of mind.

It will good if use mobile after stopping the vehicle

Thanks

5 0

2 years ago

Read 2 more answers

When this pendulum swings, at which point is its kinetic energy the<br> highest? *

Eddi Din [679]

Answer:

The bottom/center of the pendulum

Explanation:

As it swings, the pendulum will have maximum potential energy at the top of its arc.

As it comes back towards the center that potential energy will convert into kinetic energy until it reaches the middle of its swing (when the pendulum is fully vertical) where all potential energy has been converted into kinetic energy.

This is when the kinetic energy is the highest

As it begins to move away from the center of its arc, that kinetic energy will convert into potential energy again, and the process repeats

5 0

2 years ago

What is the monthly cost of using a 750 W refrigerator that runs for 8 hours every day, if the cost per kWh is $0.23?

disa [49]

Answer:

Assuming there are 28 days in each month,

750W = 0.75kW

Cost of electric bill = 0.75 × 8 × 28 × $0.23

= $38.64

8 0

1 year ago

Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

2 years ago