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3 years ago

8

A cyclist on a training ride records the distance she travels away from home. The data only shows the first150 minutes of the ri

de before her cycling computer ran out of battery.

1 answer:

Varvara68 [4.7K]3 years ago

3 0

Answer:

A) 58 km

B) 30 mins

Explanation:

In pic details

graph in pic

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Find electric field at point p which is a distance l away from the both +q and -q

denis-greek [22]

Answer:

$\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

$=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }$

Where

q is the charge
r is the distance
$E$ is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }$

F2 $=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

⇒

F1+F2= $\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }$

8 0

4 years ago

Which is an example of a mixture?

Doss [256]

Answer:

An example of a mixture would be salt water

3 0

3 years ago

Read 2 more answers

True or False? A model train traveling at a constant speed around a circular track has a constant velocity.

RSB [31]

For both, remember that velocity is a vector force, meaning that it has both a magnitude and direction. Since direction is part of the vector, if the direction changes so does the velocity in general. The changing of velocity is acceleration, so for 1 the answer is False.
For 2 I'm not sure what you meant, but the answer is the acceleration changes but the velocity does not.

8 0

3 years ago

Read 2 more answers

Rana writes a summary about a mass on a spring in simple harmonic motion as it moves upward from the equilibrium position toward

IrinaVladis [17]

Answer:

The magnitude of acceleration should be increasing.

Velocity is positive as the mass is moving towards the maximum positive displacement. Velocity would be decreasing. Acceleration is negative as velocity is decreasing. Additionally, the magnitude of acceleration would be increasing (becomes more negative.)

Explanation:

As the mass in this question moves upwards, the displacement of this mass is becoming more positive. Hence, the velocity of this mass would be positive.

In a simple harmonic motion, velocity is:

maximized at the equilibrium position (where displacement is $0$ ,) and
$0$ when displacement is maximized.

The mass in this question is moving from $0$ displacement (where velocity is maximized) towards maximum displacement (where velocity is $0$ .) Thus, the velocity of the mass would be decreasing.

Since the velocity of this mass is decreasing, the acceleration of this mass would be negative. In a simple harmonic motion, acceleration is:

$0$ at the equilibrium position, and
maximum when displacement is maximized, but opposite to the direction of displacement.

The mass in this question is moving from the equilibrium position (where acceleration is $0$ ) towards maximum displacement (where acceleration is most negative.) Thus, the magnitude of the acceleration of this mass would be increasing, and the acceleration of the mass would become more negative.

8 0

2 years ago

Mars’s moon Phobos orbits the planet at a distance of 9380 km from its center, and it takes 7 hours and 39 minutes to complete o

sukhopar [10]

Answer: 0.107

Explanation:

We can solve this problem with Kepler's Third Law of Planetary motion:

$T^{2}=4 \pi^{2} \frac{r^{3}}{G M_{MARS}}$ (1)

Where:

$T=7 h 39 min$ is the orbital period of Phobos around Mars

$G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$ is the Gravitational Constant

$M_{MARS}$ is the mass of Mars

$r=9380 km \frac{1000 m}{1 km}=9,380,000 m$ is the semimajor axis of the orbit Phobos describes around Mars (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)

Well, firstly we have to convert the orbital period to seconds:

$T=7 h 39 min=(7 h \frac{3600 s}{1 h}) + (39 min \frac{60 s}{1 min})=25200 s + 2340 s=27540 s$

Now, we have to find $M_{MARS}$ from (1):

$M_{MARS}=\frac{4 \pi^{2} r^{3}}{G T^{2}}$ (2)

$M_{MARS}=\frac{4 \pi^{2} (9,380,000 m)^{3}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}) (27540 s)^{2}}$ (3)

$M_{MARS}=6.436(10)^{23} kg$ (4) This is the mass of Mars

On the other hand, it is known the mass of the Earth is:

$M_{EARTH}=5.972(10)^{24} kg$ (5)

Then, if we want to know the ratio of Mars’s mass to the mass of the earth, we have to divide $M_{MARS}$ by $M_{EARTH}$ :

$\frac{M_{MARS}}{M_{EARTH}}=\frac{6.436(10)^{23} kg}{5.972(10)^{24} kg}$

Finally:

$\frac{M_{MARS}}{M_{EARTH}}=0.107$

7 0

4 years ago