All categories

2 years ago

Can something move horizontally and vertically at the same time? Yes! Why? No! Why not?

Physics

2 answers:

artcher [175]2 years ago

6 0

Answer:

yes

Explanation:

THE VERTICAL & HORIZONTAL MOTIONS ARE INDEPENDENT. THE HORIZONTAL VELOCITY DOES NOT AFFECT THE VERTICAL MOTION.

Elina [12.6K]2 years ago

4 0

Answer:

yes they can

hrfifjhrfcu hfbrhfuchfdc

You might be interested in

Identify the prefix that would be used to express<br> 2,000,000,000 bytes of computer memory?

Reika [66]

Answer:

It would be 2 GB

4 0

2 years ago

Read 2 more answers

Compare and contrast the CONFLICT (choose one) in the short story you read with the elements appearing in The Watsons Go to Birm

forsale [732]

Answer:

they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta

Explanation:

use in your own words teachers know when your not trust me.

4 0

2 years ago

Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini

amid [387]

Answer:

Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

$E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}$

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

$E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}$

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

$V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r$

= $\frac{q}{4\pi e_{0} } [\frac{1}{R} -\frac{r^{2}-R^{2} }{2R^{3} } ]$

∴NOTE: Graph is attached

8 0

3 years ago

Which is the absolute location of St. Louis, Missouri?

devlian [24]

Answer:

The first is the closest in absolute values.

The second two are vague and the last is approximate.

6 0

2 years ago

g A 2.3 kg block is attached to the spring, and it is released from rest 0.7 m from the spring's equilibrium position. Neglectin

DedPeter [7]

Answer:

3.71 m/s

Explanation:

From the law of conservation of linear momentum, since we are neglecting minor energy losses due to friction then we can express it as $mgh=0.5mv^{2}$ since all the potential energy is transformed to kinetic energy

Making v the subject of the formula then $v=\sqrt {2gh}$ and here m is the mass of the block, g is acceleration due to gravity, h is the height. Substituting 0.7 m for h and 9.81 for g then we obtain that $v=\sqrt {2\times 9.81\times 0.7}=3.705941176 m/s\approx 3.71 m/s$

6 0

3 years ago