Known :
l = 7 cm
w = 4 cm
Asked :
h = ...?
Answer :
V = B triangle × h (long)
35 = ½ × 4 × h × 7
35 = ½ × 28 × h
35 = 14 h
h = 35 ÷ 14
h = 2,5 cm
Sorry if I am wrong, I only study
Answer:
10 kg
Explanation:
Assuming a frictionless surface, then force F=ma where F is the applied force, m is the mass and a is acceleration. Making m the subject of the formula then 
Substituting 100 N for the applied force F and 10 m/s^2 for acceleration a then the value of m will be 
Therefore, in terms of kilograms, the bookshelf weighs 10 Kg
To solve this problem we will use the basic concept given by the Volume of a sphere with which the atom approaches. The fraction in percentage terms would be given by the division of the total volume of the nucleus by that of the volume of the atom, that is,






Therefore the percent of the atom's volume is occupied by mass is 
Explanation:
Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.
→ The tangential component of acceleration is rate of increase in the speed of plane so,

→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).
dy/dx = d(0.4x²)/dx
= 0.8x
Take the derivative again,
d²y/dx² = d(0.8x)/dx
= 0.8
at x= 5 Km
dy/dx = 0.8(5)
= 4
![p = \frac{[1+ (\frac{dy}{dx})^{2}]^{\frac{3}{2} } }{\frac{d^{2y} }{dx^{2} } }](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B%5B1%2B%20%28%5Cfrac%7Bdy%7D%7Bdx%7D%29%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%20%7D%7B%5Cfrac%7Bd%5E%7B2y%7D%20%7D%7Bdx%5E%7B2%7D%20%7D%20%7D)
now insert the values,
![p = \frac{[1+(4)^{2}]^{\frac{3}{2} } }{0.8} = 87.62 km](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B%5B1%2B%284%29%5E%7B2%7D%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D%20%20%7D%7B0.8%7D%20%20%3D%2087.62%20km)
→ Now the normal component of acceleration is given by

= (200)²/(87.6×10³)
aₙ = 0.457 m/s²
→ Now the total acceleration is,
![a = [(a_{t})^{2} +(a_{n} )^{2} ]^{0.5}](https://tex.z-dn.net/?f=a%20%3D%20%5B%28a_%7Bt%7D%29%5E%7B2%7D%20%2B%28a_%7Bn%7D%20%29%5E%7B2%7D%20%5D%5E%7B0.5%7D)
![a = [(0.8)^{2} + (0.457)^{2}]^{0.5}](https://tex.z-dn.net/?f=a%20%3D%20%5B%280.8%29%5E%7B2%7D%20%2B%20%280.457%29%5E%7B2%7D%5D%5E%7B0.5%7D)
a = 0.921 m/s²
Answer: 1.28 sec
Explanation:
Assuming that the glow following the collision was produced instantaneously, as the light propagates in a straight line from Moon to the Earth at a constant speed, we can get the time traveled by the light applying velocity definition as follows:
V = ∆x / ∆t
Solving for ∆t, we have:
∆t = ∆x/v = ∆x/c = 3.84 108 m / 3.8 108 m/s = 1.28 sec