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aksik [14]
3 years ago
12

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.
You might be interested in
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:

<u>a)Kinematics equation for the Stone, dropped:</u>

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

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

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

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

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

P_{1}V_{1} = P_{2}V_{2}

Substituting into the formula, we have;

P_{1} * 0.3 = 80 * 0.9

0.3P_{1} = 72

P_{1} = \frac {72}{0.3}

P_{1} = 240

Therefore, the initial pressure of the gas is 240 kPa

3 0
3 years ago
Consider the uniform electric field \vec{E} =(4.00~\hat{j}+3.00~\hat{k})\times 10^3~\text{N/C} ​E ​⃗ ​​ =(4.00 ​j ​^ ​​ +3.00 ​k
Tema [17]

Answer:

Electric flux, \phi=6.668\times 10^4\ Nm^2/C

Explanation:

It is given that,

Electric field, E=(4j+3k)\times 10^3\ N/C

We need to find the electric flux through a circular area of radius 2.66 m that lies in the xy-plane. A=Ak

The electric flux is given by :

\phi=E{\cdot}A

\phi=(4j+3k)\times 10^3{\cdot}Ak

Since, k.k=i.i=j.j = 1

So,

\phi=3\times 10^3\times \pi\times (2.66)^2\ k

\phi=6.668\times 10^4\ Nm^2/C

So, the electric flux through a circular area is \phi=6.668\times 10^4\ Nm^2/C.Hence, this is the required solution.

6 0
3 years ago
1 Calculate the size of the quantum involved in the excitation of (a) an electronic motion of frequency 1.0 × 1015 Hz, (b) a mol
torisob [31]

Answer: a) E= 6.63x10^-19J

E= 3.97×10^2KJ/mol

b) E = 3.31×10^-19J

E= 18.8×10^4 KJ/mol

C) E = 1.32×10^-33J

E= 8.01×10^-10KJ/mol

Explanation:

a) E = h ×f

h= planks constant= 6.626×10^-34

E=(6.626×10^-34)×(1.0×10^15)

E=6.63×10^-19J

1mole =6.02×10^23

E=( 6.63×10^-19)×(6.02×10^23)

E=3.97×10^2KJ/mol

b) E =(6.626×10^-34)/(1.0×10^15)

E=3.13×10^-19J

E= 3.13×10^-19) ×(6.02×10^23)

E= 18.8×10^3KJ/MOL

c) E= (6.626×10^-34) /0.5

E= 1.33×10^-33J

E= (1.33×10^-33) ×(6.02×10^23)

E= 8.01×10^-10KJ/mol

8 0
3 years ago
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