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3 years ago

Which statement is true regarding coppers ability to conduct electricity

1 answer:

6 0

Copper is the second best known electrical conducting substance. The first best one is silver. We almost always use copper because silver costs too much.

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Assume that g is 10 N/kg and that air resistance and other

nirvana33 [79]

Answer:

1) Given acceleration due to gravity as 10N/kg

N/kg is the same as m/s². From W=mg

making g subject ... you have g = W/m

where m is in kg and W is in N

hence we have N/kg

So g=10ms-²

Gravitational Potential Energy=mgh

At height 4m

PE=6 x 10 x 4

=240J.

At height 6m

PE=6 x 10 x 6

=360J.

2) KE=1/2mv²

=1/2 x 6 x 5² =75J

When the speed is doubled... it becomes 2x

so V = 2x5=10ms-¹

KE=1/2 x 6 x 10² =300J.

3) KE=1/2mv²

KE is given as 100J

mass=0.5kg

Making V the subject

you have v² =2KE/m

v²=2 x 100/0.5

v²=400

taking square root

v=20ms-¹.

Have a great day✌

3 0

3 years ago

The table below shows the difference between the average temperatures in April and October in four locations.

nevsk [136]

i dont know if im right but

b. 1.7

6 0

3 years ago

If a scientist gets unexpected results during the first trial of an experiment,

My name is Ann [436]

Answer:

A. Repeat the experiment to be sure the results are valid.

5 0

1 year ago

Read 2 more answers

How would you design an experiment to condition a rabbit to salivate to the ringing of a cell phone?

Klio2033 [76]

Food causes the response (salivation). Pair the food each time with a cell phone ringing. Continue this process with repetition.
• Eventually, the rabbit will learn to salivate at the ringing

5 0

3 years ago

Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in

charle [14.2K]

Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

$t$ = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s

$s$ = distance to be moved by the rock long the horizontal = 98 yards

$y$ = displacement to be moved by the rock during the time of flight along the vertical = 0 yard

<u>Assume:</u>

$u$ = magnitude of initial velocity of the rock

$\theta$ = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

$\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\$

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

$\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\$

On dividing equation (1) by (2), we have

$\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ$

Now, putting this value in equation (2), we have

$u\cos 51.34^\circ\times 5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s$

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

3 0

3 years ago