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2 years ago

250 W 12 km/h4. Convert yoor answer from 3 to meters per second.

Physics

1 answer:

taurus [48]2 years ago

5 0

Answer: 3.33 m/s

Explanation:

Assuming the questions is to convert 12 km/h to meter per second (m/s), let's begin:

In order to make the conversion, we have to know the following:

$1 km=1000 m$

And:

$1 h=3600 s$

Keeping this in mind, we can make the conversion:

$12 \frac{km}{h} \frac{1000 m}{1 km} \frac{1h}{3600 s}$

Then:

$12 \frac{km}{h}= 3.33 \frac{m}{s}$

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this is a 3 part questionOn vacation, your 1400-kg car pulls a 560-kg trailer away from a stoplight with an acceleration of 1.85

mamaluj [8]

ANSWER:

(a) 1036 N

(b) -1036 N

STEP-BY-STEP EXPLANATION:

Given:

Mc = 1400 kg

Mt = 560 kg

a = 1.85 m/s^2

(a)

Force by car on trailer:

$\begin{gathered} F_c=m\cdot a \\ F_c=560\cdot1.85 \\ F_c=1036\text{ N} \end{gathered}$

(b)

$\begin{gathered} F_t=-F_c \\ F_t=-1036\text{ N} \end{gathered}$

(c)

$\begin{gathered} F_n=1400\cdot1.85 \\ F_n=2590\text{ N} \end{gathered}$

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3 months ago

A resistor with an unknown resistance is connected in parallel to a 13 Ω resistor. When both resistors are connected in parallel

larisa86 [58]

Answer:

R2 = 10.31Ω

Explanation:

For two resistors in parallel you have that the equivalent resistance is:

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\\$ (1)

R1 = 13 Ω

R2 = ?

The equivalent resistance of the circuit can also be calculated by using the Ohm's law:

$I=\frac{V}{R_{eq}}\\\\R_{eq}=\frac{V}{I}$ (2)

V: emf source voltage = 23 V

I: current = 4 A

You calculate the Req by using the equation (2):

$R_{eq}=\frac{23V}{4A}=5.75\Omega$

Now, you can calculate the unknown resistor R2 by using the equation (1):

$\frac{1}{R_2}=\frac{1}{R_{eq}}-\frac{1}{R_1}\\\\R_2=\frac{R_{eq}R_1}{R_1-R_{eq}}\\\\R_2=\frac{(5.75\Omega)(13\Omega)}{13\Omega-5.75\Omega}=10.31\Omega$

hence, the resistance of the unknown resistor is 10.31Ω

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250 W 12 km/h4. Convert yoor answer from 3 to meters per second.​

250 W 12 km/h4. Convert yoor answer from 3 to meters per second.