The answer is B. The atmosphere provides the earth a moderate and livable climate. The atmosphere is a storage of gases. It contains air which has oxygen that enables us to breath and live in this planet. Also, the atmosphere protects the people by absorbing harmful electromagnetic radiation from the sun.
Answer:
262 kN/C
Explanation:
If the electrons is moving parallel, thus it has a retiline movement, and because the velocity is varing, it's a retiline variated movement. Thus, the acceleration can be calculated by:
v² = v0² + 2aΔS
Where v0 is the initial velocity (2.0x10⁷ m/s), v is the final velocity (4.0x10⁷ m/s), and ΔS is the distance (1.3 cm = 0.013 m), so:
(4.0x10⁷)² = (2.0x10⁷)² + 2*a*0.013
16x10¹⁴ = 4x10¹⁴ + 0.026a
0.026a = 12x10¹⁴
a = 4.61x10¹⁶ m/s²
The electric force due to the electric field (E) is:
F = Eq
Where q is the charge of the electron (-1.602x10⁻¹⁹C). By Newton's second law:
F = m*a
Where m is the mass, so:
E*q = m*a
The mass of one electrons is 9.1x10⁻³¹ kg, thus, the module of electric field strenght (without the minus signal of the electron charge) is:
E*(1.602x10⁻¹⁹) = 9.1x10⁻³¹ * 4.61x10¹⁶
E = 261,866.42 N/C
E = 262 kN/C
Answer:
24 m/s
Explanation:
Applying Newton's second law of motion,
F = ma................ Equation 1
Where F = force acting on the object, m = mass of the object a = acceleration of the object.
First we calculate for a.
make a the subject of the formula in equation 1 above
a = F/m............... Equation 2
Given: F = 3 N, m = 0.5 kg.
Substitute into equation 2
a = 3/0.5
a = 6 m/s²
Secondly we calculate the velocity by using,
v = u+at................... Equation 3
Where v = Final velocity of the object, u = initial velocity of the object, t = time interval
Given: u = 0 m/s (at rest), a = 6 m/s², t = 4 s
Substitute into equation 3
v = 0+6(4)
v = 24 m/s
Answer:
3.64×10⁸ m
3.34×10⁻³ m/s²
Explanation:
Let's define some variables:
M₁ = mass of the Earth
r₁ = r = distance from the Earth's center
M₂ = mass of the moon
r₂ = d − r = distance from the moon's center
d = distance between the Earth and the moon
When the gravitational fields become equal:
GM₁m / r₁² = GM₂m / r₂²
M₁ / r₁² = M₂ / r₂²
M₁ / r² = M₂ / (d − r)²
M₁ / r² = M₂ / (d² − 2dr + r²)
M₁ (d² − 2dr + r²) = M₂ r²
M₁d² − 2dM₁ r + M₁ r² = M₂ r²
M₁d² − 2dM₁ r + (M₁ − M₂) r² = 0
d² − 2d r + (1 − M₂/M₁) r² = 0
Solving with quadratic formula:
r = [ 2d ± √(4d² − 4 (1 − M₂/M₁) d²) ] / 2 (1 − M₂/M₁)
r = [ 2d ± 2d√(1 − (1 − M₂/M₁)) ] / 2 (1 − M₂/M₁)
r = [ 2d ± 2d√(1 − 1 + M₂/M₁) ] / 2 (1 − M₂/M₁)
r = [ 2d ± 2d√(M₂/M₁) ] / 2 (1 − M₂/M₁)
When we plug in the values, we get:
r = 3.64×10⁸ m
If the moon wasn't there, the acceleration due to Earth's gravity would be:
g = GM / r²
g = (6.672×10⁻¹¹ N m²/kg²) (5.98×10²⁴ kg) / (3.64×10⁸ m)²
g = 3.34×10⁻³ m/s²
