What is the acceleration of a baseball when it is moving downward after being tossed directly upward?

Ryan runs 50 m north and then 50 m east. What is his displacement?

san4es73 [151]

Answer:

Explanation:

Distance and direction of an object's change in position from a starting point. Displacement. 3. Jermaine runs exactly 2 laps around a 400 meter track. What is the displacement? 0 ... David walks 3 km north, and then turns east and walks 4 km. ... A person walks 50 meters directly north, stops, and then travels 32 meters

3 0

3 years ago

If two asteroids moved closer together, what would be the result on the gravitational force each asteroid exerts on the other?

spayn [35]

Gravitational force depends on inverse square law. That is, gravitational force is inversely proportional to square of distance between asteroids.
As distance between them decreases, gravitational force increases. Hence A is correct.

7 0

2 years ago

Read 2 more answers

A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 31.0 nC , respecti

allsm [11]

Answer:

The charge on the third object is − 21.7nC

Explanation:

From Gauss's Law

Φ = Q/ε₀

where;

Φ is the total electric flux through the shell = − 533 N⋅m²/C

Q is the total charge Q in the shell = ?

ε₀ is the permittivity of free space = 8.85 x 10⁻¹²

From this equation; Φ = Q/ε₀

Q = Φ * ε₀ = − 533 * 8.85 x 10⁻¹²

Q = −4.7 X 10⁻⁹ C = -4.7nC

Q = q₁ + q₂ + q₃

− 4.7nC = − 14.0 nC + 31.0 nC + q₃

− 4.7nC − 17nC = q₃

− 21.7nC = q₃

Therefore, the charge on the third object is − 21.7nC

8 0

3 years ago

) each plate of a parallel-plate air-filled capacitor has an area of 0.0020 , and the separation of the plates is an electric fi

Dvinal [7]

I attached the full question.
We know that for a parallel-plate capacitor the surface charge density is given by the following formula:
$\sigma=\varepsilon_0 \frac{V}{d}$
Where V is the voltage between the plates and d is separation.
Voltage is by definition:
$V=Ed$
Voltage is analog to the mechanical work done by the force.
Above formula is correct only If the field is constant, and we can assume that it is since no function has been given.
The charge density would then be:
$\sigma=\varepsilon_0 \frac{Ed}{d}=\varepsilon_0E\\
\sigma= 8.85\cdot10^{-12}\cdot 2.1\cdot 10^6= 0.0000185\frac{c}{m^2}$
Please note that elecric permittivity of air is very close to elecric permittivity of vacum, it is common to use them <span>interchangeably</span>.

6 0

3 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

$T^2=\dfrac{4\pi ^2}{GM}a^3$

G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

3 0

3 years ago