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3 years ago

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On a distance-time graph, time is shown on the y-axis. A) True B) False

1 answer:

Artist 52 [7]3 years ago

7 0

Answer:

false : In distance time graph,time is shown on the x -axis

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What an object is made of and the color of light that strikes it determine the. A. apparent color of the object.. B. transparenc

Crank

What an object is made of and the color of light that strikes it determine the apparent color of the object. The correct option among all the options that are given in the question is the first option or option "A". I hope that this is the answer that you were looking for and it has come to your help.

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3 years ago

Read 2 more answers

Two boats leave a dock together . Each travels in a straight line . The angle between their courses measures 54° 10¢ . One boat

nata0808 [166]

Answer:

218.93 km

Explanation:

$\theta$ = Angle between the paths of the two boats = 54.10°

Distance = Speed × Time

Distance traveled by one boat = $36.2\times 3\ km$

Distance traveled by other boat = $45.6\times 3\ km$

From the triangle law of vectors we have

$d=\sqrt{(36.2\times 3)^2+(45.6\times 3)^2+2\times 36.2\times 3\times 45.6\times 3\times cos54.10}\\\Rightarrow d=218.93\ km$

The distance they will be apart after 3 hours is 218.93 km

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4 years ago

Determine the approximate force (N) used to pull a sled up a 400 m hill using 1900 J of work.

Sergeu [11.5K]

The work done to pull the sled up to the hill is given by
$W=Fd$
where
F is the intensity of the force
d is the distance where the force is applied.

In our problem, the work done is $W=1900 J$ and the distance through which the force is applied is $d=400 m$ , so we can calculate the average force by re-arranging the previous equation and by using these data:
$F= \frac{W}{d}= \frac{1900 J}{400 m} = 4.75 N \sim 5 N$

4 0

4 years ago

The average distance from Earth to the Moon is 384,000 km.

Fofino [41]

Answers:

1) Time it takes to travel to the moon

Velocity $V$ is defined as:

$V=\frac{d}{t}$

Where:

$d=384,000 km$ is the average distance from Earth to the Moon

$V=800 km/h$ is the velocity of the spacecraft

$t$ the time

Isolating $t$ :

$t=\frac{d}{V}$

$t=\frac{384,000 km}{800 km/h}$

$t=480 h$ This is the time in hours

2) Time in days

A day is equivalent to 24 hours:

$1 day= 24 h$

Hence:

$t=480 h (\frac{1 day}{24 h})$

$t=480 h (\frac{1 day}{24 h})=20 days$ This is the time in days

3) Time in months

A month is equivalent to 30.4 days:

$1 month= 30.4 days$

Hence:

$t=480 h (\frac{1 day}{24 h})$

$t=20 days (\frac{1 month}{30.4 days})=0.65 months$ This is the time in months

6 0

3 years ago

Solving a series circuit, did I do this correctly?

nirvana33 [79]

The total resistance in the circuit is 16 Ohms.
The total current in the circuit is 0.5 Ampere.
The current at $R_1$ is 0.5 Ampere.
The current at $R_3$ is 0.5 Ampere.
The voltage drop at $R_1$ is 4 volts.
The voltage drop at $R_2$ is 2.5 volts.
The voltage drop at $R_3$ is 1.5 volts.
The total power consumed by the circuit is 4watts
The power consumed at $R_1$ is 2 watts
The power consumed at $R_2$ is 1.25 watts

Given:

The voltage across the circuit = V = 8 V

The resistors connected are in series:

$R_1=8 \Omega, R_2=5\Omega ,R_3=3 \Omega$

To find:

The values of from 1 to 10.

Solution

The voltage across the circuit = V = 8 V

The total resistance in the circuit = $R_{eq}$

$R_{eq}=R_1+R_2+R_3\\=8 \Omega +5 \Omega + 3\Omega =16\omega$

The total current in the circuit = I

$V=IR_{eq}\\I=\frac{V}{R_{eq}}=\frac{8 V}{16 \Omega}=0.5 A$ (Ohm's law)

For series combinations, the current in each resistor remains the same.

So, the current in $R_1, R_2 \&R_3$ :

$I_1= I_2= I_3=I=0.5 A\\$

The voltage drop across at $R_1$ = $V_1$

The current across $R_1$ = I = 0.5 A

$V_1=I\times R_1\\\\=0.5A\times 8\Omega = 4 V$

The voltage drop across at $R_2$ = $V_2$

The current across $R_2$ = I = 0.5 A

$V_2=I\times R_2\\\\=0.5A\times 5\Omega = 2.5 V$

The voltage drop across at $R_3$ = $V_3$

The current across $R_3$ = I = 0.5 A

$V_3=I\times R_3\\\\=0.5A\times 3\Omega = 1.5 V$

The total power consumed by circuit:

$P= V\times I \\\\= 0.5 A\times 8 V = 4 watt$

Power consumed at $R_1$ :

$P_1=V_1\times I\\\\= 4V\times 0.5A = 2 watt$

Power consumed at $R_2$ :

$P_2=V_2\times I\\\\= 2.5 V\times 0.5A = 1.25 watt$

Power consumed at $R_3$ :

$P_3=V_3\times I= \\\\1.5 V\times 0.5A = 0.75 watt$

Learn more about, current, voltage, resistance, and power of the circuit here:

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2 years ago