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4 years ago

a car advertisement states that a certain car can accelerate from rest to 70km/h in 7 seconds. find the cars acceleration

1 answer:

4 0

<span>a = (v2 - v1)/t= acceleration formula

a = (70 - 0)/7
a = 10 km/hr/sec
-----
It's better to use as few units as possible.
10 km/hr = 10 km*1000 m/km/(1 hr*3600 sec/hr) = 25/9 m/sec
a= 25/9 m/sec/se</span>

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LASIK eye surgery uses pulses of laser light to shave off tissue from the cornea, reshaping it. A typical LASIK laser emits a 1.

Fittoniya [83]

Answer:

143 kW

Explanation:

Given that

Diameter of the beam, d = 1 mm

Wavelength of the beam, λ = 193 nm

Time used by the pulse, t = 14 ns

Energy of the pulse, U = 2 mJ

Recall that Power can be mathematically calculated using the relation,

Power = Work Done / Time,

To solve this, we apply the formula

P = U / Δt

P = 2*10^-3 J / 14*10^-9 s

P = 142857 W

P = 143 kW

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3 years ago

3. When two things are the same temperature:

JulijaS [17]

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3 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant