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4 years ago

13

If an object has more protons than it does electrons, then it has _________ charge. A. no B. a variable C. a net positive D. a n

et negative

2 answers:

snow_tiger [21]4 years ago

4 0

A) NO
it would have no charge, i think

rewona [7]4 years ago

4 0

The answer would be C

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Can anyone give me the description of the element mercury is positioned on the periodic table of elements and a list of other el

garri49 [273]

I hope this helps you good luck

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3 years ago

Consider the following four objects: a hoop, a solid sphere, a flat disk, a hollow sphere. Each of the objects has mass M and ra

Svetlanka [38]

Answer:

Hoop.

Explanation:

The angular acceleration performed at a given torque:

$\alpha = \frac{\tau}{I}$

The moments of inertia of each element are described below:

Hoop

$I = M\cdot R^{2}$

Solid sphere

$I = \frac{2}{5}\cdot M \cdot R^{2}$

Flat disk

$I = \frac{1}{2}\cdot M \cdot R^{2}$

Hollow sphere

$I = \frac{2}{3}\cdot M \cdot R^{2}$

The greater the moment of inertia, the greater the torque to obtain the same angular acceleration. Therefore, the hoop requires the largest torque to receive the same angular acceleration.

8 0

4 years ago

In a game of pool, a cue ball rolls without slipping toward the stationary eight ball with a momentum of 0.23 kg. After the two

IrinaVladis [17]

This question involves the concepts of the law of conservation of momentum.

The magnitude of the final momentum of the eight ball is "0.22 N.s".

According to the law of conservation of momentum:

$P_{i1}+P_{i2}=P_{f1}+P_{f2}$

where,

$P_{i1}$ = initial momentum of the cue ball = 0.23 N.s

$P_{i2}$ = initial momentum of the eight ball = 0 N.s (since ball is initially at rest)

$P_{f1}$ = final momentum of the cue ball = 0.01 N.s

$P_{f2}$ = final momentum of the eight ball = ?

Therefore,

$0.23\ N.s + 0\N.s = 0.01\ N.s+P_{f2}\\\\P_{f2} = 0.22\ N.s$

Learn more about the law of conservation of momentum here:

brainly.com/question/1113396?referrer=searchResults

3 0

3 years ago

Bru this was a challenge

vredina [299]

Answer:

what was a challenge,,

Explanation:

?

5 0

3 years ago

Read 2 more answers

A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ =

SashulF [63]

Answer:

The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

Explanation:

Given that,

Length of the current element $dl=(0.5\times10^{-3})j$

Current in y direction = 5.40 A

Point P located at $\vec{r}=(-0.730)i+(0.390)k$

The distance is

$|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}$

$|\vec{r}|=0.827\ m$

We need to calculate the magnetic field

Using Biot-savart law

$B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}$

Put the value into the formula

$B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}$

We need to calculate the value of $\vec{dl}\times\vec{r}$

$\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k$

$\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})$

$\vec{dl}\times\vec{r}=0.000175i+0.000365k$

Put the value into the formula of magnetic field

$B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}$

$B=1.670\times10^{-10}i+3.484\times10^{-10}k$

Hence, The magnetic field along x axis is

$B_{x}=1.670\times10^{-10}\ T$

The magnetic field along y axis is zero.

The magnetic field along z axis is

$B_{z}=3.484\times10^{-10}\ T$

7 0

3 years ago