Answer:
Explanation:
Range of projectile R = 20 m
formula of range
R = u² sin2θ / g
u is initial velocity , θ is angle of projectile
putting the values
20 = u² sin2x 40 / 9.8
u² = 199
u = 14.10 m /s
At the initial point
vertical component of u
= u sin40 = 14.1 x sin 40
= 9.06 m/s
Horizontal component
= u cos 30
At the final point where the ball strikes the ground after falling , its speed remains the same as that in the beginning .
Horizontal component of velocity
u cos 30
Vertical component
= - u sin 30
= - 9.06 m /s
So its horizontal component remains unchanged .
change in vertical component = 9.06 - ( - 9.06 )
= 18.12 m /s
change in momentum
mass x change in velocity
= .050 x 18.12
= .906 N.s
Impulse = change in momentum
= .906 N.s .
The length by which the spring got stretched will be 0.08 m. The force is directly propotional to the distance by which the spring stretched.
<h3>What is spring force?</h3>
The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is
F = kx
The given data in the problem is;
F is the spring force =200
K is the spring constant= 2500 N/m
x is the length by which spring got stretched =?
The stretch of the spring is found as;
Hence the length by which the spring got stretched will be 0.08 m.
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Answer:
6.86 * 10^8 m
Explanation:
Parameters given:
Mass of hot gas, m = 2 kg
Gravitational Force, F = 618.2 N
Mass of Alpha Centauri, M = 2.178 * 10^30 kg
The gravitational force between two masses (the hot gas and Alpha Centauri) , m and M, at a distance, r, given as:
F = (G*M*m) / r²
Where G = gravitational constant
Therefore,
618.2 = (6.67 * 10^(-11) * 2.178 * 10^30 * 2) / r²
=> r² = (6.67 * 10^(-11) * 2.178 * 10^30 * 2) / 618.2
r² = 4.699 * 10^17 m²
=> r = 6.86 * 10^8 m
We are told that the hot gas is on the surface of Alpha Centauri, hence, the distance between both their centers is the radius of Alpha Centauri.
The mean radius of Alpha Centauri is 6.86 * 10^8 m.
We have that for the Question "A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal" it can be said that the minimum force that must be applied on the <em>book is</em>
From the question we are told
A 2kg book is held against a vertical wall. The <em>coefficient </em>of friction is 0.45. What is the minimum force that must be applied on the <em>book</em>, perpendicular to the wall, to prevent the book from slipping down the wal
Generally the equation for the Force is mathematically given as
F=44N
Therefore
the minimum force that must be applied on the <em>book is</em>
F=44N
For more information on this visit
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From the answers provided, I believe the possible answer would be the last option, silicon, oxygen, and one or more metals. Here's my reasoning: the most abundant mineral group found in the Earth's crust is the silicate group. The silicate materials contain both oxygen and silicon. Silicates are the most common minerals in the rock-formation process, and it has, in fact, been estimated that they make up 75 to 90 percent of the Earth's crust. From this piece of evidence, I can guess that the answer will possibly be D, silicon, oxygen, and one or more metals.
It should also be noted that the additional elements that combine with the silicon-oxygen tetrahedron are involved with the other elements commonly found in the Earth's crust and mantle. They are aluminum, calcium, iron, magnesium, potassium and sodium.
