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VashaNatasha [74]

3 years ago

11

The ________________ is the measure of how far the pendulum is offset (or pulled back) from a vertical position when it is relea

sed.
A. Period
B.Amplitude
C. Pivot
D. Bob

1 answer:

Dahasolnce [82]3 years ago

6 0

B. Amplitude

It is the maximum distance from the equilibrium point of the pendulum.

You might be interested in

1) What would the average acceleration be for a car at a stoplight that speeds up to 20 m/s in 10 seconds (in m/s^2)

Alexxx [7]

1.)
Velocity is in m/s, and acceleration is in m/s^2 like you said. Because of this, we can calculate this by dividing the speed by the time it took to get to that speed.
(20 meters/second) / 10 seconds = 2 meters/ second^2

2.)
Same thing with the first one.
(100 meters/second) / 4 seconds = 25 meters / seconds^2

7 0

2 years ago

Read 2 more answers

Una persona de 80 kg de masa corre a una velocidad cuya magnitud es de 9m/s. Cual es la magnitud de la cantidad de movimiento? Q

natka813 [3]

Responder:

13,01 m / s

Explicación:

Paso uno:

datos dados

masa de la persona 1 m = 80 kg

velocidad de la persona 1 v = 9 m / s

masa de la persona 2 M = 55kg

velocidad de la persona 2 v =?

Segundo paso:

la expresión del impulso se da como

P = mv

para la primera persona, el impulso es

P = 80 * 9

P = 720N

Paso tres:

queremos que la segunda persona tenga el mismo impulso que la primera, por lo que la velocidad debe ser

720 = 55v

v = 720/55

v = 13,09

v = 13,01 m / s

Por lo tanto, la magnitud de la velocidad debe ser 13.01 m / s.

4 0

3 years ago

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

4 0

3 years ago

The next four questions refer to the situation below.

Anna11 [10]

Answer:

t_{out} = $\frac{v_s - v_r}{v_s+v_r}$ t_{in}, t_{out} = $\frac{D}{v_s +v_r}$

Explanation:

This in a relative velocity exercise in one dimension,

let's start with the swimmer going downstream

its speed is

$v_{sg 1} = v_{sr} + v_{rg}$

The subscripts are s for the swimmer, r for the river and g for the Earth

with the velocity constant we can use the relations of uniform motion

$v_{sg1}$ = D / $t_{out}$

D = v_{sg1} t_{out}

now let's analyze when the swimmer turns around and returns to the starting point

$v_{sg 2} = v_{sr} - v_{rg}$

$v_{sg 2}$ = D / $t_{in}$

D = v_{sg 2} t_{in}

with the distance is the same we can equalize

$v_{sg1} t_{out} = v_{sg2} t_{in}$

t_{out} = t_{in}

t_{out} = $\frac{v_s - v_r}{v_s+v_r}$ t_{in}

This must be the answer since the return time is known. If you want to delete this time

t_{in}= D / $v_{sg2}$

we substitute

t_{out} = \frac{v_s - v_r}{v_s+v_r} ()

t_{out} = $\frac{D}{v_s +v_r}$

7 0

2 years ago

To fill the medication prescription, what information must the pharmacy technician need to obtain? A. The name of the medication

shepuryov [24]

C. Patient info, name of med, dosage & route, special instructions, prescriber’s DEA#, and number of refills

3 0

3 years ago