Jack tries to place magnets on his refrigerator at home, but they won’t stick. What could be the reason?

1 answer:

5 0

The most probable reason why the magnets won't stick on the refrigerator is that the body of the refrigerator and the magnets have like poles. If both have negative or both have positive poles facing each other, they will repel. In principle, magnets are attracted to opposite poles and like poles repel.

You might be interested in

An automobile moves at a constant speed over the crest of a hill traveling at a speed of 88.5 km/h. At the top of the hill, a pa

Sergio039 [100]

Answer:

$r=61.65m$

Explanation:

Since the package remains in contact with the car's seat, the package's speed is equal to the car's speed. At the top on the mountain the package's centripetal force must be equal to its weight:

$mg=F_c$

The centripetal force is defined as:

$F_c=ma_c=\frac{mv^2}{r}$

Here v is the linear speed of the object and r is the radius of curvature. We need to convert the linear speed to $\frac{m}{s}$ :

$88.5\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=24.58\frac{m}{s}$

Now, we calculate r:

$mg=\frac{mv^2}{r}\\r=\frac{v^2}{g}\\r=\frac{(24.58\frac{m}{s})^2}{9.8\frac{m}{s^2}}\\\\r=61.65m$

3 0

3 years ago

A satellite that goes around the earth once every 24 hours iscalled a geosynchronous satellite. If a geosynchronoussatellite is

alexgriva [62]

Answer:

R=4.22*10⁴km

Explanation:

The tangential speed $v$ of the geosynchronous satellite is given by:

$v=\frac{2\pi R}{T}$

Because $2\pi R$ is the circumference length (the distance traveled) and T is the period (the interval of time).

Now, we know that the centripetal force of an object undergoing uniform circular motion is given by:

$F_c=\frac{mv^{2} }{R}$

If we substitute the expression for $v$ in this formula, we get:

$F_c=\frac{m(\frac{2\pi R}{T})^{2}}{R}=\frac{4m\pi ^{2}R}{T^{2}}$

Since the centripetal force is the gravitational force $F_g$ between the satellite and the Earth, we know that:

$F_g=\frac{GMm}{R^{2}}\\\\\implies \frac{GMm}{R^{2}}=\frac{4m\pi ^{2}R}{T^{2}}\\\\R^{3}=\frac{GMT^{2}}{4\pi^{2}} \\\\R=\sqrt[3]{\frac{GMT^{2}}{4\pi^{2}} }$

Where G is the gravitational constant ( $G=6.67*10^{-11} Nm^{2}/kg^{2}$ ) and M is the mass of the Earth ( $M=5.97*10^{24}kg$ ). Since the period of the geosynchronous satellite is 24 hours (equivalent to 86400 seconds), we finally can compute the radius of the satellite: