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2 years ago

10

What is the magnitude of velocity for a 3,100 kg car possessing 4,100 kg•m/s of momentum?

2 answers:

Margaret [11]2 years ago

8 0

Answer:

3.75

Explanati

azamat2 years ago

3 0

Answer:

my c-ock and balls be rotating im gunna be hxrny all day every day im gunna f-u-ck u and r-ape ur family girls should go back to the kitchen and hope not to be beat by their husband uh yea mhm girls desurve to be r-aped f-u-ck buck suck c-u-ck

Explanation:

You might be interested in

What radiation do remote controls use?

dalvyx [7]

Answer:

infra red radiation is used

3 0

3 years ago

The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc

Mnenie [13.5K]

Answer:

The image distance is 20.0 cm.

Explanation:

Given that,

Power = 1.55 dp

Distance between book to eye = 26.0+3.00=29.0 cm

We need to calculate the focal length

Using formula of focal length

$f = \dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{1}{1.55}$

$f=0.645\ m$

$f=64.5\ cm$

We need to calculate the image distance

Using lens formula

$\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}$

$\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{-u}$

Put the value into the formula

$\dfrac{1}{v}=\dfrac{1}{64.5}-\dfrac{1}{-29}$

$\dfrac{1}{v}=\dfrac{187}{3741}$

$v=20.0\ cm$

Hence, The image distance is 20.0 cm.

5 0

3 years ago

Choose the answer choice that BEST completes the following sentence. About four million tons of __________________ are converted

yaroslaw [1]

Answer:

<em>About four million tons of </em><em><u>matter</u></em><em> are converted to energy in the Sun’s core every second</em>

Explanation:

<u>The Sun </u>

It's estimated that The Sun fuses near 600 million tons $(6.10^9\ kg)$ of hydrogen into helium every second. As a result, 4 million tons of matter are converted to energy every second according to Einstein's equation $E=mc^2$

4 0

3 years ago

The radius of the thinner wire is 0.22 mm and the radius of the thick wire is 0.55 mm. There are 4.0 x 1028 mobile electrons per

Natasha_Volkova [10]

Answer:

0.2631 N/C

Explanation:

Given that:

The radius of the wire r = 0.22 mm = 0.22 × 10⁻³ m

The radius of the thick wire r' = 0.55 mm = 0.55 × 10⁻³ m

The numbers of electrons passing through B, N = 6.0 × 10¹⁸ electrons

Electron mobility μ = 6.0 x 10-4 (m/s)/(N/C)

= 0.0006

The number of electron flow per second is calculated as follows:

$I = \frac{q}{t}$

$I = \frac{Ne}{t}$

$I = \frac{6*10^{18}(1.6*10^{-18})C}{1 \ s}$

$I = 0.96 \ A$

The magnitude of the electric field is:

E = $\frac{I}{ \mu n eA}$

E = $\frac{I}{ \mu n e(\pi r^2)}$

E = $\frac{0.96}{(0.0006 m/s N/C ) (4*10^{28})(1.6*10^{-19}C)(0.55*10^{-3}m)^2}$

E = 0.2631 N/C

8 0

3 years ago

A dog leaps horizontally off a 70 m cliff with a speed of 6 m/s, how far from the base will the dog land?

iris [78.8K]

Answer:

$\Delta x=22.67786838m$

Explanation:

Let's use projectile motion equations. First of all we need to find the travel time. So we are going to use the next equation:

$y-y_0=v_o*sin(\theta)*t-\frac{1}{2}*t^2$ (1)

Where:

$y=Final\hspace{3}position\hspace{3}at\hspace{3}y-axis$

$y_o=Initial\hspace{3}position\hspace{3}at\hspace{3}y-axis$

$v_o=initial\hspace{3}velocity$

$t=travel\hspace{3}time$

$g=gravity\hspace{3}constant$

$\theta=Initial\hspace{3}launch\hspace{3}angle$

In this case:

$\theta=0$

Because the dog jumps horizontally

Let's asume the gravity constant as:

$g=9.8$

$y=0$

Because when the dog reach the base the height is 0

$y_o=70$

$v_o=6$

Now let's replace the data in (1)

$y_o-\frac{1}{2} *(9.8)*t^2+70$

Isolating t:

$t=\pm\sqrt{\frac{2*70}{9.8} } =3.77964473$

Finally let's find the horizontal displacement using this equation:

$\Delta x=v_o*cos(\theta)*t$

Replacing the data:

$\Delta x=6*1*3.77964473=22.67786838m$

8 0

3 years ago