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2 years ago

Non examples of convert

Physics

1 answer:

german2 years ago

8 0

Answer:

another. persist. continue. last. remain.

Explanation:

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Pls help me fast and correct

bekas [8.4K]

Answer:

Explanation:

due to position at height is the answer

4 0

2 years ago

Help with 7,8,9 please!!

Marysya12 [62]

# 7 the answer is 4) because you need opposites in order to attract

# 8 I'm going to guess it is 4) because electricity is needed in order a light bolt to work.(don't take my word for this one I'm not sure but, the other answer are correct)

# 9 the answer is 4) because the comb dose collect negative charges from the hair in order to make the hair have positive charge.

Hope this helped : )

4 0

3 years ago

The unit light-year is a measure of

stiv31 [10]

A light-year is a unit of distance.

The definition of a light-year is the distance light travels in one year.

5 0

3 years ago

Read 2 more answers

A 0.473 kg ice puck, moving east with a speed of 2.76 m/s, has a head-on collision with a 0.819 kg puck initially at rest. Assum

Gekata [30.6K]

Answer:

The final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

Explanation:

Let us consider east as positive direction and west as negative direction .

Given

mass of puck 1 , $m_1= 0.473 kg$

mass of puck 2 , $m_2= 0.819 kg$

initial speed of puck 1 , $u_1=2.76\frac{m}{s}$

initial speed of puck 2 , $u_2=0.00\frac{m}{s}$

Final speed of puck 1 and puck 2 be $v_1\, and\, v_2$ respectively

Apply conservation of linear momentum

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

=> $0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2$

=> $1.594=0.5775\times v_1+ v_2$ -----(A)

Since collision is perfectly elastic , coefficient restitution e=1

$u_2-u_1=v_1-v_2$

=> $0-2.76=v_1-v_2$ ------(B)

From equation (A) and (B)

$v_1=-0.739\frac{m}{s}$

and $v_2=2.02\frac{m}{s}$

Thus the final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

3 0

3 years ago

A particle moves along the curve below. y = sqrt(1 + x^3) As it reaches the point (2, 3), the y-coordinate is increasing at a ra

blagie [28]

Answer:7 cm/s

Explanation:

Given

Particle move along curve

$y=\sqrt{1+x^3}$

As it reaches the (2,3) its y coordinate is increasing at 14 cm/s

Differentiating y w.r.t time

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3x^2}{2\sqrt{1+x^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

Now at (2,3)

$\frac{\mathrm{d} y}{\mathrm{d} t}=\frac{3\cdot 2^2}{2\sqrt{1+2^3}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$14=\frac{3\times 4}{2\times \sqrt{9}}\times \frac{\mathrm{d} x}{\mathrm{d} t}$

$\frac{\mathrm{d} x}{\mathrm{d} t}=7 cm/s$

7 0

3 years ago