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alex41 [277]
3 years ago
13

g f, as a pioneer, you wished to warm your room by taking an object heated on top of a stove into it, which of the following 15

lb objects, each heated to 100 degrees celcius, would be the best choice and why. The specific heat capacity (in J/(g.K)) is given in parenthesis. Iron (0.450), copper (0.387), granite (0.79), gold (0.129), or water (4.18)
Physics
1 answer:
Bingel [31]3 years ago
3 0

Answer:

Granite

Explanation:

The best choice would be granite.

This is because granite has the highest specific heat capacity of 0.79 J/(g.K) and would thus store more heat in it for the same mass and same temperature change as the other materials.

Since it contains more heat, it would be able to heat up the room to a higher temperature than the other materials.

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It's 12.1 m/s, assuming that's the launch velocity that's given.
For projectile motion, velocity's y-component is parabolic/quadratic. It's x-component is constant, so you don't need to know it. 
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A certain 48 V electric forklift can lift up to 7000 lb at a
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How many electrons flow through a point in a wire in 7.00 s if there is a constant current of I = 4.35 A?
OLEGan [10]

Answer:

1.90×10²⁰ Electrons

Explanation:

From the question,

Q = It.................... Equation 1

Where Q = charge flowing through the wire, I = current, t = time

Given: I = 4.35 A, t = 7.00 s

Substitute these values into equation 1

Q = 4.35(7.00)

Q = 30.45 C.

But,

1 electron contains 1.6×10⁻¹⁹ C

therefore,

30.45 C = 30.45/1.6×10⁻¹⁹  electrons

= 1.90×10²⁰ Electrons

8 0
3 years ago
A trick shot archer shoots an arrow with a velocity of 30.0 m/s at an angle of 20.0 degrees with respect to the horizontal. An a
svlad2 [7]

Answer:

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

the apple should be tossed after \Delta t=0.0173\ s

Explanation:

Given:

  • velocity of arrow in projectile, v=30\ m.s^{-1}
  • angle of projectile from the horizontal, \theta=20^{\circ}
  • distance of the point of tossing up of an apple, d=30\ m

<u>Now the horizontal component of velocity:</u>

v_x=v\ cos\ \theta

v_x=30\times cos\ 20^{\circ}

v_x=28.191\ m.s^{-1}

<u>The vertical component of the velocity:</u>

v_y=v.sin\ \theta

v_y=30\times sin\ 20^{\circ}

v_y=10.261\ m.s^{-1}

<u>Time taken by the projectile to travel the distance of 30 m:</u>

t=\frac{d}{v_x}

t=\frac{30}{28.191}

t=1.0642\ s

<u>Vertical position of the projectile at this time:</u>

h=v_y.t-\frac{1}{2}g.t^2

h=10.261\times 1.0642-\frac{1}{2} \times 9.8\times 1.0642^2

h=5.3701\ m

<u>Now this height should be the maximum height of the tossed apple where its velocity becomes zero.</u>

v'^2=u'^2-2g.h

0^2=u'^2-2\times 9.8\times 5.3701

u'=10.259\ m.s^{-1} is the initial velocity of tossing the apple.

<u>Time taken to reach this height:</u>

v'=u'-g.t'

0=10.259-9.8\times t'

t'=1.0469\ s

<u>We observe that </u>t>t'<u> hence the time after the launch of the projectile after which the apple should be tossed is:</u>

\Delta t=t-t'

\Delta t=1.0642-1.0469

\Delta t=0.0173\ s

8 0
4 years ago
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