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Rufina [12.5K]
2 years ago
13

Help! I'm not sure my answer is correct.Problem is attached!​

Physics
2 answers:
kaheart [24]2 years ago
3 0
  • Let G be xN

\\ \sf\longmapsto x-20=30

\\ \sf\longmapsto x=30+20

\\ \sf\longmapsto x=50N

Option B

NNADVOKAT [17]2 years ago
3 0

Answer:

2) 50 Newtons

Explanation:

Since the net force is 30N to the right and there is a force 20N points to the left already, the force G points to the left should be 50N.

So 50N - 20 N = 30 N

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If an object moving at 12 m/s has a kinetic energy of 72 j what is the mass
EleoNora [17]
Simply, apply the formula E = \frac{1}{2}mv^{2} and insert the values of m = mass, v = velocity and E = Energy.
The result will be 72J =  \frac{1}{2}m(12m/s)^{2}, m = 1 kg
5 0
3 years ago
Why silicon have large forward current as compared to germanium?​
Iteru [2.4K]

Answer:

The structure of Germanium crystals will be destroyed at higher temperature. However, Silicon crystals are not easily damaged by excess heat. Peak Inverse Voltage ratings of Silicon diodes are greater than Germanium diodes. Si is less expensive due to the greater abundance of element.

4 0
3 years ago
In an electric shaver, the blade moves back and forth over a distance of 2.0 mm in simple harmonic motion, with frequency 124 Hz
bagirrra123 [75]

Answer:

(a) A = 1 mm

(b) V_{max}=0.77872 m/s

(c) a_{max}=606.4 m/s^{2}/tex]Explanation:Distance moved back and forth = 2 mm Frequency, f = 124 HzSo, amplitude is the half of the distance traveled back and forth. (a) So, amplitude, A = 1 mm(b) Angular frequency, ω = 2 π f = 2 x 3.14 x 124 = 778.72 rad/s The formula for the maximum speed is given by [tex]V_{max}=\omega \times A

V_{max}=778.72 \times 0.001

V_{max}=0.77872 m/s

(c) The formula for the maximum acceleration is given by

a_{max}=\omega ^{2}A

a_{max}=778.72 ^{2}\times 0.001

[tex]a_{max}=606.4 m/s^{2}/tex]

8 0
3 years ago
1. Which is an advantage of coal energy?​
pentagon [3]
Coals energy is affordable and it is easy to burn.
4 0
3 years ago
Read 2 more answers
The potential energy of a 35 kg cannon ball is 14000 J. How high was the cannon ball to have this much potential energy?
Anettt [7]

From the information given, cannon ball weighs 40 kg and has a potential energy of 14000 J.

We need to find its height.

We will use the formula P.E = mgh

Therefore h = P.E / mg

where P.E is the potential energy,

m is mass in kg,  

g is acceleration due to gravity (9.8 m/s²)

h is the height of the object's displacement in meters.

h = P.E. / mg

h = 14000 / 40 × 9.8

h = 14000 / 392

h = 35.7

Therefore the canon ball was 35.7 meters  high.

6 0
3 years ago
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