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3 years ago

With what speed must a ball be thrown directly upward so that it remains in the air for 10 seconds?

Physics

1 answer:

MA_775_DIABLO [31]3 years ago

7 0

Answer:

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Two long, parallel wires are separated by 2.2 mm. Each wire has a 32-AA current, but the currents are in opposite directions. Pa

Alex

Answer:

$B=1.1636*10^{-3}T$

Explanation:

Given data

$d_{wires}=2.2mm=0.022m\\ I_{current}=32A\\$

To find

Magnitude of the net magnetic field B

Solution

The magnitude of the net magnetic field can be find as:

$B=2*u\frac{I}{2\pi r}\\ B=2*(4\pi*10^{-7} )\frac{32}{2\pi (0.022/2)} \\ B=1.1636*10^{-3}T$

3 0

3 years ago

An object moves in a circle at a constant speed of 1.0 m/s. The radius of the circle is 1.0 m. If a force of 1.0 N acts toward t

Vlad1618 [11]

Answer:5

Explanation:

Given

speed of object $v=1\ m/s$

radius of circle $r=1\ m$

Force towards the center $F=1\ N$

Work done is given by the dot product of Force and displacement

and we know know displacement of the object is along the circle which is perpendicular to the force acting therefore Work done will be zero

$W=F\cdot s\cos 90$

$W=0$

4 0

3 years ago

Atom in the ground state is said to be

koban [17]

Answer:

Ground-state atom

Explanation:

When an atom is not excited, it is in its ground-state, which we refer as "standard" or "normal" state.

(Hopefully that helped you!)

GOOD LUCK

Astrophysicist Dr. D

3 0

3 years ago

If 5.7 grams of gold has 3 cubic centimeters than whats the density of gold in g/cm3

Aleonysh [2.5K]

Density: 1.9 g/cm3

density = mass/volume
d = 5.7/3
d = 1.9

3 0

3 years ago

A meteoroid passes through a position in space where its speed is very small relative to Earth's and it is at a perpendicular di

Oliga [24]

Answer:

$v=7506.4m/s$

Explanation:

If we call 1 the position in space where the meteoroid of mass m speed is very small relative to Earth's (whose mass is $M_E=5.98\times10^{24}kg$ and radius is $R_E=6371000m$ ) and it is at a perpendicular distance of $h_1=19R_E$ above Earth's surface, and 2 the position when it is $h_2=R_E$ above the ground, then, since in this case mechanical energy is conserved, we can write:

$E_1=E_2$

which means:

$K_1+U_1=K_2+U_2$

where K is the kinetic energy and U the gravitational potential energy. Since $K_1=0J$ we can write:

$\frac{-GM_Em}{r_1}=\frac{mv_2^2}{2}+\frac{-GM_Em}{r_2}$

which means:

$v_2=\sqrt{2GM_E(\frac{1}{r_2}-\frac{1}{r_1})}$

And since $r=R_E+h$ (the distance of an object to the center of the Earth is Earth's radius plus the height of the object), we have:

$v_2=\sqrt{2GM_E(\frac{1}{R_E+h_2}-\frac{1}{R_E+h_1})}=\sqrt{2GM_E(\frac{1}{2R_E}-\frac{1}{20R_E})}=\sqrt{\frac{GM_E}{R_E}(1-\frac{1}{10})}=\sqrt{\frac{0.9GM_E}{R_E}}$

This for our values is:

$v_2=\sqrt{\frac{0.9GM_E}{R_E}}=\sqrt{\frac{0.9(6.67\times10^{-11}Nm^2/kg^2)(5.98\times10^{24}kg)}{6371000m}}=7506.4m/s$

3 0

3 years ago