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3 years ago

What two things must you know to find the momentum of an object? :( anyone?

2 answers:

7 0

<em>Linear Momentum is a vector quantity.</em>
<em> p = mass * Velocity</em>
<em>We need to know the mass, and velocity (magnitude and direction)</em>

miss Akunina [59]3 years ago

4 0

You will need too know the mass and velocity

You might be interested in

How many electrons pass a given point in the circuit in 3 min? The fundamental charge is 1.602 × 10−19 C. amps: 0.415 , volts: 1

morpeh [17]

The no of electron passing through the circuit is current by time. The no of electron passing are 46.63 x 10¹⁹ electrons.

<h3>What is charge?</h3>

The charge is the physical quantity which attracts or repels another object when comes into its field.

The time 3 min has 3 x 60 =180 seconds.

Total Charge Q = current I x time t

Q = 0.415 x 180

Q =74.7 coulombs

The no of electrons is the ratio of total charge divided by the fundamental charge.

n = 74.7/ 1.602 × 10⁻¹⁹

n= 46.63 x 10¹⁹ electrons

Thus, electrons pass a given point in the circuit in 3 min are 46.63 x 10¹⁹ .

Learn more about charge.

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4 0

2 years ago

Name 7 characteristics that are unique to mammals.

SSSSS [86.1K]

Mammals are endothermic vertebrates
Have hair and fur on the body
Have mammary glands
Four chambered hearts
Have sebaceous (fat secreting glands), sudoriferus (sweat), and scent glands.
Have heterodont dentation (different types of teeth)
Posses diaphragm
Posses one single jaw bone
Have three small bones in the middle of the ear

3 0

3 years ago

Read 2 more answers

Pls help me 10 points

lapo4ka [179]

Answer:

d is the right answer

it's was helpful to you

3 0

3 years ago

Read 2 more answers

A test rocket is launched vertically from ground level (y=0), at time t=0.0s. The rocket engine provides constant upward acceler

Luden [163]

From the definition of average velocity,

$\bar v=\dfrac{\Delta y}{\Delta t}=\dfrac{49\,\mathrm m}t$ ,

and the fact that constant acceleration means

$\bar v=\dfrac{v_f+v_0}2=\dfrac{30\,\frac{\mathrm m}{\mathrm s}}2=15\,\dfrac{\mathrm m}{\mathrm s}$

we can solve for the time $t$ :

$15\,\dfrac{\mathrm m}{\mathrm s}=\dfrac{49\,\mathrm m}t\implies t=3.3\,\mathrm s$

7 0

3 years ago

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect

Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a) 0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b) r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 $\pi$ $r^{2}$ = $\frac{Q1}{E_{0} }$

So,

Rearranging the above equation to get Electric field, we will get:

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$

Multiply and divide by $r1^{2}$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ x $\frac{r1^{2} }{r1^{2} }$

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

c) r > r2 :

Electric Field = ?

E x 4 $\pi$ $r^{2}$ = $\frac{Q1 + Q2}{E_{0} }$

Rearranging the above equation for E:

E = $\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

As we know from above, that:

$\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Then, Similarly,

$\frac{Q2}{E_{0} . 4 \pi. r^{2} }$ = (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

So,

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

Replacing the above equations to get E:

E = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ ) + (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Now, for

d) Under what conditions, E = 0, for r > r2?

For r > r2, E =0 if

σ1 x $r1^{2}$ = - σ2 x $r2^{2}$

4 0

3 years ago