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3 years ago

14

The equation for distance is d = st. If a car has a speed of 85 m/s and travels

1 answer:

Ludmilka [50]3 years ago

4 0

Answer:

D) 10 200 m

Explanation:

Speed is known as the rate of change of distance.

$speed=\frac{distance}{time}$

distance = speed * time

= 85 ms⁻¹ * 120 s

= 10200 m

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For a certain spring, k = 15N/m. A weight is hung from the spring, stretching it from 0.3m to 0.4m. What force did the weight pr

Mashcka [7]

The stretch of the spring is
$\Delta x = 0.4 m-0.3 m=0.1 m$
The constant of the spring is k=15 N/m, so we can find the force produced by the weight by using Hook's law:
$F=k\Delta x=(15 N/m)(0.1 m)=1.5 N$

8 0

3 years ago

A diver shines a flashlight upward from beneath the water (n = 1.33) at a 32.7° angle to the vertical. At what angle does the li

suter [353]

Answer:

45.93°

Explanation:

The angle of incidence is given as 32.7°

The refractive index of the water that is $n_1=1.33$

Refractive index of the air that is $n_2=1$ (because the refractive index of air is 1 )

We have to find the angle at which the light leave the water means angle of refraction

So according to snell's law $n_1sini=n_2sinr$

$1.33sin32.7=1\times sinr$

$sinr=0.7185$

r =45.93°

So the light leave the water at an angle of 45.93°

3 0

3 years ago

A little help with this one

Aloiza [94]

The answer would be A because as you get further away from sound the intensity decreases.

3 0

3 years ago

A space expedition discovers a planetary system consisting of a massive star and several spherical planets. The planets all have

Juliette [100K]

Answer:

T/√8

Explanation:

From Kepler's law, T² ∝ R³ where T = period of planet and R = radius of planet.

For planet A, period = T and radius = 2R.

For planet B, period = T' and radius = R.

So, T²/R³ = k

So, T²/(2R)³ = T'²/R³

T'² = T²R³/(2R)³

T'² = T²/8

T' = T/√8

So, the number of hours it takes Planet B to complete one revolution around the star is T/√8

7 0

3 years ago

A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a

stira [4]

Answer:

Explanation:

Given

acceleration is given by

$a=-g-c_Dv^2$

where $\ddot{y}=a$

$\dot{y}=v$

Also acceleration is given by

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ds=\frac{v}{a}dv$

$\int ds=\int \frac{v}{-g-0.001v^2}dv$

$\Rightarrow Let -g-0.001v^2=t$

$-0.001\times 2vdv=dt$

$vdv=-\frac{dt}{0.002}$

$at\ v_0=50\ m/s,\ t=-g-0.001(50)^2$

$t=-g-2.5$

at $v=0,\ t=-g$

$\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}$

$\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}$

$s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}$

$s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})$

$s=113.608\ m$

when air drag is neglected maximum height reached is

$h=\frac{v_0^2}{2g}$

$h=\frac{50^2}{2\times 9.8}$

$h=127.55\ m$

3 0

3 years ago