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3 years ago

7

Brainliest..... plz help!!!!!!!!!

1 answer:

alisha [4.7K]3 years ago

4 0

Tamam hocam nasılsınız hocam proje konularını atabilirmisiniz ne kadar çok seviyorum çok öpüyorum iyi akşamlar yarın görüşürüz iyi geceler Allah rahatlık var bir şey var ya bir insan değilim ki bir şey olmaz sen kaç gibi müsait olur musun lütfen bana yardımcı olabilir misin bir insan bir

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hey pls help with this HDJSJDH im rlly bad at this kinda stuff :,) i'll mark the correct answer brainliest

timama [110]

I am pretty sure it’s A
The cue exerts force onto the white ball which pushes the blue ball into the direction of the hole.

7 0

3 years ago

Read 2 more answers

A hockey puck slides on the ice and eventually stops. How would Newton interpret this behavior?

azamat

That would be a the first law of newton's laws of motion because it stops from an external force

6 0

3 years ago

Consider the expression below. Assume m is an integer. 6m(2m + 18) Enter an expression in the box that uses the variable m and m

Anika [276]

6x2=12m
6x18=108
12m+108
Simplified: m+9 bc 12/12 and 108/12

8 0

4 years ago

Q 19.23: A proton is initially moving at 3.0 x 105 m/s. It moves 3.5 m in the direction of a uniform electric field of magnitude

DENIUS [597]

Answer:

The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

Explanation:

Given;

initial velocity of proton, $v_p_i$ = 3 x 10⁵ m/s

distance moved by the proton, d = 3.5 m

electric field strength, E = 120 N/C

The kinetic energy of the proton at the end of the motion is calculated as follows.

Consider work-energy theorem;

W = ΔK.E

$W =K.E_f - K.E_i$

where;

K.Ef is the final kinetic energy

W is work done in moving the proton = F x d = (EQ) x d = EQd

$K.E_f =EQd + \frac{1}{2}m_pv_p_i^2$

$m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C$

$K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5 \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\$

$K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J$

Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.

3 0

3 years ago

A uniform solid sphere has a moment of inertia I about an axis tangent to its surface. What is the moment of inertia of this sph

arsen [322]

Answer:

option E

Explanation:

given,

I is moment of inertia about an axis tangent to its surface.

moment of inertia about the center of mass

$I_{CM} = \dfrac{2}{5}mR^2$ .....(1)

now, moment of inertia about tangent

$I= \dfrac{2}{5}mR^2 + mR^2$

$I= \dfrac{7}{5}mR^2$ ...........(2)

dividing equation (1)/(2)

$\dfrac{I_{CM}}{I}= \dfrac{\dfrac{2}{5}mR^2}{\dfrac{7}{5}mR^2}$

$\dfrac{I_{CM}}{I}=\dfrac{2}{7}$

$I_{CM}=\dfrac{2}{7}I$

the correct answer is option E

4 0

3 years ago