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kirza4 [7]
3 years ago
10

How many protons does the largest transition metal have in its nucleus?

Physics
1 answer:
xxMikexx [17]3 years ago
3 0
The largest transition metal is copernicium with 112 protons.
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Which two options are forms of potential energy
blsea [12.9K]

Answer:

e and b

Explanation:

6 0
3 years ago
Read 2 more answers
Someone help please.....
Westkost [7]

Answer:

0.0928km/min (4dp)

Explanation:

To find the jogger's speed in km per minute, we just need to divide the number of km jogged by the time in minutes it took to jog that distance. This will give us the distance they jogged every minute which is their speed.

4km in 32 minutes:

4/32 = 0.125km/min

2km in 22 minutes:

2/22 = 0.091 (3dp)km/min

1km in 16 minutes:

0.0625km/min

Now to find the average speed of these 3 speeds, we just add them all together and divide by how many values there are (3 values).

Average (mean)  = \frac{0.125+0.091+0.0625}{3}

Average = 0.2785/3

Average speed of jogger = 0.0928 (4dp) km/min

Hope this helped!

8 0
4 years ago
A force F= (6.70 N)i + (5.60 N)j + (6.60 N)k act on a 9.80 kg mobile object that moves from an initial position of di = (9.10 m)
Mandarinka [93]

Answer:

W = 0.11J

P = 0.0186W

Explanation:

Please see attachment below.

4 0
3 years ago
The lens in the eyepiece of a reflecting telescope breaks. How will this most affect the function of the telescope?
Serggg [28]

Answer:

B

Magnified images will not be created.

Explanation:

I did it and this was the correct answer

5 0
3 years ago
. A projectile is fired with an initial velocity of 113 m słatan angle of 60.0
bonufazy [111]

Answer:

a) 9.99 s

b) 538 m

c) 20.5 s

d) 1160 m

Explanation:

Given:

x₀ = 0 m

y₀ = 49.0 m

v₀ = 113 m/s

θ = 60.0°

aₓ = 0 m/s²

aᵧ = -9.8 m/s²

a) At the maximum height, the vertical velocity vᵧ = 0 m/s.  Find t.

vᵧ = aᵧ t + v₀ᵧ

(0 m/s) = (-9.8 m/s²) t + (113 sin 60.0° m/s)

t ≈ 9.99 s

b) At the maximum height, the vertical velocity vᵧ = 0 m/s.  Find y.

vᵧ² = v₀ᵧ² + 2aᵧ (y − y₀)

(0 m/s)² = (113 sin 60° m/s)² + 2 (-9.8 m/s²) (y − 49.0 m)

y ≈ 538 m

c) When the projectile lands, y = 0 m.  Find t.

y = y₀ + v₀ᵧ t + ½ aᵧ t²

(0 m) = (49.0 m) + (113 sin 60° m/s) t + ½ (-9.8 m/s²) t²

You'll need to solve using quadratic formula:

t ≈ -0.489, 20.5

Since negative time doesn't apply here, t ≈ 20.5 s.

d) When the projectile lands, y = 0 m.  Find x. (Use answer from part c).

x = x₀ + v₀ₓ t + ½ aₓ t²

x = (0 m) + (113 cos 60° m/s) (20.5 s) + ½ (0 m/s²) (20.5 s)²

x ≈ 1160 m

4 0
3 years ago
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