Answer:
20 [N], in the opposite direction of the first force.
Explanation:
We know that newton's second law stipulates that the sum of forces on a body must be equal to the product of mass by acceleration.
![SumF = m*a\\30 + F = 2*5\\F = 30 - (2*5)\\F = - 20 [N]](https://tex.z-dn.net/?f=SumF%20%3D%20m%2Aa%5C%5C30%20%2B%20F%20%3D%202%2A5%5C%5CF%20%3D%2030%20-%20%282%2A5%29%5C%5CF%20%3D%20-%2020%20%5BN%5D)
The negative sign means that the other force acting on the body must be in the opposite direction to the force of 30 [N]
We have,
- Jane mass is 55 kg
- His body covered with 700 nails all of them having a surface area of 1.00 mm² each = 700 × 1 = 700 mm² = 700/1000000 = 7/10000
We know that,
Let's calculate force as we already have area;
- F = ma
- F = 55 × 9.8 { Acceleration due to gravity }
- F = 539 N
Now, if should she would be on 700 nails then pressure will be;
- P = F/A
- P = 539/7 × 10000
- P = 5390000/7
- P = 770,000 Pascal
And if should would be on a 1 nail only,
- P = F/A
- P = 539/1 × 1000000
- P = 539000000 Pascal
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The first one is C. Hope this helps!! If you need anymore help just message me and I will try to get back to you quickly and help in any way i can!
No, because in oxygen depraved rooms, if you drop a feather and a bowling ball at the same height and time, they will fall at the same speed and have the same amount of impact.
Answer:
a) the distance between her and the wall is 13 m
b) the period of her up-and-down motion is 6.5 s
Explanation:
Given the data in the question;
wavelength λ = 26 m
velocity v = 4.0 m/s
a) How far from the wall is she?
Now, The first antinode is formed at a distance λ/2 from the wall, since the separation distance between the person and wall is;
x = λ/2
we substitute
x = 26 m / 2
x = 13 m
Therefore, the distance between her and the wall is 13 m
b) What is the period of her up-and-down motion?
we know that the relationship between frequency, wavelength and wave speed is;
v = fλ
hence, f = v/λ
we also know that frequency is expressed as the reciprocal of the time period;
f = 1/T
Hence
1/T = v/λ
solve for T
Tv = λ
T = λ/v
we substitute
T = 26 m / 4 m/s
T = 6.5 s
Therefore, the period of her up-and-down motion is 6.5 s