In which position are the earth, moon, and sun during a full moon?

What Could be the control in an experiment to measure the effects of gas additives on fuel

alina1380 [7]

Well,

A control in an experiment would basically be the "normal" version of your test subjects.

In a drug testing experiment with people, the control group would be the people who don't take the drug.

In an experiment on the effects of salt on potatoes, the control group would be a potato without salt on it.

So in an experiment to measure the effects of gas additives on fuel, the control would be fuel without additives.

8 0

3 years ago

A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.

Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

k x = m g

$k = \dfrac{mg}{x}$

$k = \dfrac{6.4 \times 9.8}{0.28}$

k = 224 N/m

b) $f = \dfrac{\omega}{2\pi}$

$\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}$

$f =0.94\ Hz$

c) $v_b = -v cos \omega t$

$v_b = -5.1 \times cos (5.92 \times 0.42)$

$v_b = 4.04\ m/s$

d) $a_{max} = v \omega$

$a_{max} = 4.04 \times 5.92$

$a_{max} =23.94\ m/s^2$

e) $Y =- A sin (\omega t)$

$A = \dfrac{v}{\omega}$

$A = \dfrac{4.04}{5.92}$

A = 0.682 m

$Y =- 0.682 \times sin (5.92 \times 0.42)$

$Y =- 0.42$

Force = $m \omega^2 |Y|$

= $6.4 \times 5.92^2\times 0.42$

F = 94.20 N

4 0

3 years ago

1. A step-up transformer increases 15.7V to 110V. What is the current in the secondary as compared to the primary? Assume 100 pe

Serjik [45]

1. $I_2 = 0.14 I_1$

Explanation:

We have:

$V_1 = 15.7 V$ voltage in the primary coil

$V_2 = 110 V$ voltage in the secondary coil

The efficiency of the transformer is 100%: this means that the power in the primary coil and in the secondary coil are equal

$P_1 = P_2\\V_1 I_1 = V_2 I_2$

where I1 and I2 are the currents in the two coils. Re-arranging the equation, we find

$\frac{I_2}{I_1}=\frac{V_1}{V_2}=\frac{15.7 V}{110 V}=0.14$

which means that the current in the secondary coil is 14% of the value of the current in the primary coil.

2. 5.7 V

We can solve the problem by using the transformer equation:

$\frac{N_p}{N_s}=\frac{V_p}{V_s}$

where:

Np = 400 is the number of turns in the primary coil

Ns = 19 is the number of turns in the secondary coil

Vp = 120 V is the voltage in the primary coil

Vs = ? is the voltage in the secondary coil

Re-arranging the formula and substituting the numbers, we find:

$V_s = V_p \frac{N_s}{N_p}=(120 V)\frac{19}{400}=5.7 V$

5 0

3 years ago

A car passes point “A” and then 120 meters later. It’s velocity was measured 21 m/s. If it’s acceleration was constant at 0.853

Norma-Jean [14]

Recall that

${v_f}^2-{v_i}^2=2a\Delta x$

where $v_i$ and $v_f$ are the initial and final velocities, respecitvely; $a$ is the acceleration; and $\Delta x$ is the change in position.

So we have

$\left(21\dfrac{\rm m}{\rm s}\right)^2-{v_i}^2=2\left(0.853\dfrac{\rm m}{\mathrm s^2}\right)(120\,\mathrm m)$

$\implies v_i\approx\boxed{15.4\dfrac{\rm m}{\rm s}}$

(Normally, this equation has two solutions, but we omit the negative one because the car is moving in one direction.)

7 0

3 years ago

If the action force is a player kicking a soccer ball, then what is the reaction force?

LiRa [457]

The force acting on his feet.

4 0

3 years ago