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3 years ago

Which of these actions would increase heat transfer between two objects?

Physics

2 answers:

sergejj [24]3 years ago

7 0

Answer:

b- increases the area of their contact

Explanation:

i did it in edge! ; )

Kaylis [27]3 years ago

3 0

Answer:

Increasing the area of their contact.

Explanation:

Just took the test and it was right.

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A golfer takes two putts to get his ball into the hole he is on green. The first putt displaces the ball 4.5m east, and the seco

LenaWriter [7]

I think its the one going it at 4.5m

8 0

3 years ago

A gry is an old English measure for length, defined as 1/10 of a line, where line is another old English measure for length, def

Valentin [98]

Answer:

0.2448 point²

Explanation:

1 gry = 1/10 line

1 line = 1/12 inch

=> 1 gry in inches = 1/10 * 1/12 = 1/120 inch

=> 1 inch = 120 gry

1 point = 1/72 inch

=> 1 inch = 72 points

Therefore,

120 gry = 72 points

=> 1 gry = 3/5 point

Therefore,

1 gry² = (3/5)² point²

1 gry² = 9/25 point²

This means that 0.68 gry² will be:

0.68 gry² = 0.68 * 9/25 point²

=> 0.68 gry² = 0.2448 point²

6 0

3 years ago

What does the area under a speed-time graph represent

Tom [10]

Answer: It represents the whole distance traveled. Hope this helps!

Explanation:

4 0

3 years ago

You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an

a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, $\mu=6\times 10^{-2}$

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

$KE=PE+W$

$\dfrac{1}{2}mv^2=mgh+W$

W is the work done by the friction.

$W=f\times d$

$f=\mu mg\ cos\theta$

$W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}$

$W=\dfrac{\mu mgh}{tan\theta}$

$\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}$

$\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}$

$\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}$

v = 8.19 m/s

So, the speed of the box is 8.19 m/s. Hence, this is the required solution.

8 0

3 years ago

An astronaut on the moon drops a feather from rest (v = 0). The acceleration due to gravity is 1.67 m/s 2 . If the feather beg

olga_2 [115]

Given :

Initial velocity , u = 0 m/s .

Acceleration due to gravity on moon , $g_m=1.67\ m/s^2$ .

Height , h = 2 m .

To Find :

Final position after falling for 1.5 seconds .

Solution :

We know , by equation of motion :

$s=ut+\dfrac{at^2}{2}$

Here , $a = g_m$ .

So , equation will transform by :

$s=ut+\dfrac{g_mt^2}{2}\\\\s=0+\dfrac{1.67\times 1.5^2}{2}\ m\\\\s=1.88\ m$

Therefore , the height form moon's surface is 1.88 m .

Hence , this is the required solution .

6 0

3 years ago