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dangina [55]
2 years ago
12

If copper (E⁰ = ⁺0.34) and lithium (E⁰ = ⁻3.05) are used to make a battery, will electrons flow from lithium to copper or vice v

ersa? Please include an explanation because I seriously do not understand this topic T-T
Physics
1 answer:
Lemur [1.5K]2 years ago
4 0

Answer:

give me brainliest first and ill give u the correct anwers

Explanation:

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12. A satellite is put into an orbit at a distance from the center of the Earth equal to twice the distance from the center of t
Kitty [74]

Answer:

Explanation:

F = GmM/d²

As gravity force is proportional to the inverse of the square of the distance,

doubling the distance will reduce the weight to a forth.

F' = GmM/(2d)²

F' = ¼GmM/d²

F' = ¼F = ¼(4000)

F' = 1000 N

6 0
3 years ago
A ball of mass 2.0 kg falls vertically and hits the ground with speed 7.0 ms-1 as shown below.
Sauron [17]

Answer:

8.0 Ns

Explanation:

Change in momentum is given as:

Final momentum - Initial momentum

= m*v - m*u

Where m = mass of ball

v = final velocity

u = initial velocity

Change in momentum = (2.0 * 3.0) - (2.0 * 7.0)

= 6.0 - 14.0 = -8.0 Ns

The magnitude will be |-8.0| = 8.0 Ns

7 0
4 years ago
Read 2 more answers
What is the force of a 13kg ball that has been dropped and has fallen for 1 second?
ankoles [38]

Answer:

127.4 newtons

Explanation:

Assuming g = 9.8:

F = ma = 13(9.8) = 127.4 N

6 0
3 years ago
2. For a rotating rigid body, which of the following statements is NOT correct?
AfilCa [17]

Answer:

                                                dasgfwe

Explanation:

6 0
4 years ago
A 10-kg projectile is fired straight up with an initial velocity of 500 m/s. (a) What is the projectile’s potential energy at th
Naily [24]
(a) the initial kinetic energy of the projectile is equal to:
K_i= \frac{1}{2}mv^2= \frac{1}{2}(10 kg)(500 m/s)^2=1.25 \cdot 10^6 J
The projectile is fired straight up, so at the top of its trajectory, its velocity is zero; this means that it has no kinetic energy left, so for the law of conservation of energy, all its energy has converted into potential energy, which is equal to
U_f=K_i= 1.25 \cdot 10^6 J

b) If the projectile is fired with an angle of 45^{\circ}, its velocity has 2 components, one in the x-direction and one in the y-direction:
v_x = v_0 \cos 45^{\circ} =(500 m/s) \cos 45^{\circ} =353.6 m/s
v_y = v_0 \sin 45^{\circ} = (500 m/s)(\sin 45^{\circ} )=353.6 m/s

This means that at the top of its trajectory, only the vertical velocity will be zero (because the horizontal velocity is constant, since the motion on the x-axis is a uniform motion). Therefore, at the top of the trajectory, the projectile will have some kinetic energy left:
K_f =  \frac{1}{2}m v_x^2 = \frac{1}{2} (10kg)(353.6 m/s)^2=6.25 \cdot 10^5 J
For the conservation of energy, the initial energy mechanical energy must be equal to the mechanical energy at the highest point:
K_i = K_f + U_f
the initial kinetic energy is the same as point a), so we can re-arrange this equation to find the new potential energy at the top of the trajectory:
U_f = K_i - K_f = 1.25 \cdot 10^6 J - 6.25 \cdot 10^5 J = 6.25 \cdot 10^5 J
7 0
3 years ago
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