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3 years ago

A speedboat initially at rest accelerates uniformly at 4.0 m/s (squared) for 7.0 s. How fast is the boat moving after 7.0 s?

1 answer:

8 0

Well if the boat initially at rest accelerates at uniformly at 4.0 m/s (squared) then it would be best to muitlply it so 4.0 squared equals 2 by multiplying that by 7.0 your answer would be 14 s

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An electric dipole is formed from ±1.00nC charges spaced 3.00 mm apart. The dipole is at the origin, oriented along the x-axis.

Ronch [10]

Answer:

Value of electric field along the axis and equitorial axis $E=31.25\ N/c$ and $E = 15.625\ N/c$ respectively.

Explanation:

Given :

Distance between charges , $d = 3 \ mm =\dfrac{3}{1000}\ m=3\times 10^{-3}\ m.$

Magnitude of charges , $q=1\ nC = 10^{-9}\ C.$

Dipole moment , $p=qL=10^{-9}\times 3\times 10^{-3}=3\times 10^{-12} \ C\ m.$

Case A) (x,y) = (12.0 cm, 0 cm) :

Electric field of dipole in its axis ,

$E=\dfrac{2kp}{r^3}$

Putting all values and $r=12\times 10^{-2}\ m.$

We get , $E=31.25\ N/c.$

Case B) (x,y) = (0 cm, 12.0 cm) :

Electric field of dipole on equitorial axis ,

$E = \dfrac{kp}{r^3}$

Putting all values and $r=12\times 10^{-2}\ m.$

We get , $E = 15.625\ N/c.$

Hence , this is the required solution.

7 0

3 years ago

A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

3 years ago

What is the length of the color attribute?

astraxan [27]

The length of the colour attribute is 6 characters.

<h3>What is a colour attribute?</h3>

Each colour has a distinct look based on three essential characteristics: hue, chroma (saturation), and value (lightness). It's critical to use all three of these properties when describing colour to appropriately identify it and distinguish it from others.

<h3>What is the use of colour attributes?</h3>

The HTML <font> color Attribute specifies the text color within the font element. Values of Attributes: colour name: It uses the colour name to set the text colour.

It should be noticed that colours are represented by the hex triplet.

Learn more about colour attributes here:

brainly.com/question/18071208

#SPJ4

6 0

1 year ago

I live for love, but i cant have it

luda_lava [24]

Answer:

Dont worry ,

One day you will find the love of your life

Explanation:

8 0

2 years ago

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A 50.0-g Super Ball traveling at 25.0 m/s bounces off a brick wall and rebounds at 22.0 m/s. A

Setler [38]

Explanation:

Average acceleration is change in velocity over time.

a = Δv / Δt

a = (22.0 m/s − (-25.0 m/s)) / 0.00350 s

a = 13,400 m/s²

7 0

3 years ago