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3 years ago

Complete this sentence. Average speed is worked out from dividing _ by

Physics

2 answers:

mestny [16]3 years ago

4 0

Average speed is worked out from dividing the total distance by the total time.

Dmitry [639]3 years ago

3 0

Average speed is worked out from dividing distance by time.

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A suspension bridge is 60.0 m above the level base of a gorge. A stone is thrown or dropped from the bridge. Ignore air resistan

Alexxx [7]

Answer:

a) $t_{1}=3.49s$

b) $t_{2}=2.00s$

c)Xmax=80.71m

Explanation:

a)Kinematics equation for the Stone, dropped:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h=60m$ initial position is bridge height

$v_{o}=0m/s$ the stone is dropped

The ball reaches the ground, y=0, at t=t1:

$0=h-1/2*g*t_{1}^{2}$

$t_{1}=\sqrt{2h/g}=\sqrt{2*60/9.83}=3.49s$

b)Kinematics equation for the Stone, with a initial speed of 20m/s:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h=60m$ initial position is bridge height

$v_{o}=-20m/s$ the stone is thrown straight down

The ball reaches the ground, y=0, at t=t1:

$0=h+v_{o}t_{2}-1/2*g*t_{2}^{2}$

$0=60-20t_{2}-1/2*9.83*t_{2}^{2}$

t2=-6.01 this solution does not have physical sense

t2=2.00

c)Kinematics equation for the Stone, with a initial speed of 20m/s with an angle of 30° above the horizontal:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h=60m$ initial position is bridge height

$v_{o}=20sin(30)=10m/s$ the stone is thrown with an angle of 30° above the horizontal

The ball reaches the ground, y=0, at t=t3:

$0=h+v_{o}t_{3}-1/2*g*t_{3}^{2}$

$0=60+10t_{3}-1/2*9.83*t_{3}^{2}$

t3=-2.62 this solution does not have physical sense

t3=4.66

the movement in x:

v=constant=20cos(30)m/s

x(t)=v*t

Xmax=v*t3=20cos(30)*4.66=80.71m

3 0

3 years ago

PLEASE HELP ME!!!

denis-greek [22]

The reason for that is that P-waves (primary waves) travel faster than S-waves (secondary waves).

If we call $v_p$ the speed of the primary waves and $v_s$ the speed of the secondary waves, and we call $S$ the distance of the seismogram from the epicenter, we can write the time the two waves take to reach the seismogram as
$t_P = \frac{S}{v_P}$
$t_S= \frac{S}{v_S}$

So the lag time between the arrival of the P-waves and of the S-waves is
$\Delta t = t_S-t_P= \frac{S}{v_S}- \frac{S}{v_P}= S(\frac{1}{v_S}- \frac{1}{v_P})$
We see that this lag time is proportional to the distance S, therefore the larger the distance, the greater the lag time.

6 0

3 years ago

With the temperature held constant, the piston of a cylinder containing a gas is pulled out so that the volume increases from 0.

klio [65]

Answer:

P1 = 240 kPa.

Explanation:

Given the following data;

Initial volume = 0.3 m³

Final volume, V2 = 0.9 m³

Final pressure, P2 = 80 kPa

To find the initial pressure, we would use Boyle's law;

Boyles states that when the temperature of an ideal gas is kept constant, the pressure of the gas is inversely proportional to the volume occupied by the gas.

Mathematically, Boyles law is given by the formula;

$PV = K$