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2 years ago

15

What are the opposite ends of a magnet called? A. north and south terminals B. north and south poles C. magnetic fields D. magne

tic forces

2 answers:

Bezzdna [24]2 years ago

6 0

I think the answer is B.

Hope this helps.

Soloha48 [4]2 years ago

6 0

Pretty sure b I hope this helps

~~~~~~MK

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Which of the following is an example of rotation

SOVA2 [1]

I believe it is D. Earth spinning on it's axis.

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2 years ago

Why is the battery run flat with reference to the energy transformations that take place

andreev551 [17]

Answer:

Rechargeable batteries can still go flat after repeated use because the materials involved in the reaction lose their ability to charge and regenerate

Explanation:

they go flat because they a rechargeable and sometimes don't always work

thank you hope u have a good day

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1 year ago

Using this information...

Pepsi [2]

$19.2\:\text{m/s}$

Explanation:

At the top of the tree, the velocity of the pebble is purely horizontal so we can calculate it as

$v_{y} = v_{0y} = v_0\cos 40° = (25\:\text{m/s})(0.766)$

$\:\:\:\:\:= 19.2\:\text{m/s}$

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2 years ago

Remember KE=M*V2/2…  A model airplane moves twice as fast as another identical model airplane. Compared with the kinetic energy

pishuonlain [190]

Answer: 4 times as much

Explanation:

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1 year ago

Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fil

laila [671]

Answer:

$W_n_e_t=7.648512 \approx 7.6J$

$K.E=0.8J$

$v=0.7844645406 \approx 0.78m/s$

Explanation:

From the question we are told that

Mass of pitcher $M= 2.6kg$

Force on pitcher $f=8.8N$

Distance traveled $48cm=>0.48m$

Coefficient of friction $\mu=0.28$

a)Generally frictional force is mathematically given by

$F=\mu N$

$F=0.28*2.6*9.8$

$F=7.1344N$

Generally work done on the pitcher is mathematically given as

$W_n_e_t=W_f+W_F$

$W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N$

$W_n_e_t=4.224-3.424512$

$W_n_e_t=0.799488\approx 0.8J$

b)Generally K.E can be given mathematically as

$K.E= W_n_e_t$

Therefore

$K.E=0.8J$

c)Generally the equation for kinetic energy is mathematically represented by

$K.E=1/2mv^2$

$0.8=1/2mv^2$

Velocity as subject

$v=\sqrt{\frac{K.E*2}{m} }$

$v=\sqrt{\frac{0.8*2}{2.6} }$

$v=0.7844645406 \approx 0.78m/s$

6 0

2 years ago