Answer:
the distance traveled by the car is 42.98 m.
Explanation:
Given;
mass of the car, m = 2500 kg
initial velocity of the car, u = 20 m/s
the braking force applied to the car, f = 5620 N
time of motion of the car, t = 2.5 s
The decelaration of the car is calculated as follows;
-F = ma
a = -F/m
a = -5620 / 2500
a = -2.248 m/s²
The distance traveled by the car is calculated as follows;
s = ut + ¹/₂at²
s = (20 x 2.5) + 0.5(-2.248)(2.5²)
s = 50 - 7.025
s = 42.98 m
Therefore, the distance traveled by the car is 42.98 m.
The weight is 45 N, because the three chains hold the sign, and each contributes 15 N.
Notice that the mass would be the weight/acceleration of gravity, m = 45/9.8 kg. But they ask the weight (force, so Newtons)
First law of motion<span>- sometimes referred to as the </span>law<span> of inertia. An object at rest stays at rest and an object in </span>motion<span> stays in </span>motion<span> with the same speed and in the same direction unless acted upon by an unbalanced force.</span>
Answer:
W = 1,307 10⁶ J
Explanation:
Work is the product of force by distance, in this case it is the force of gravitational attraction between the moon (M) and the capsule (m₁)
F = G m₁ M / r²
W = ∫ F. dr
W = G m₁ M ∫ dr / r²
we integrate
W = G m₁ M (-1 / r)
We evaluate between the limits, lower r = R_ Moon and r = ∞
W = -G m₁ M (1 /∞ - 1 / R_moon)
W = G m1 M / r_moon
Body weight is
W = mg
m = W / g
The mass is constant, so we can find it with the initial data
For the capsule
m = 1000/32 = 165 / g_moon
g_moom = 165 32/1000
.g_moon = 5.28 ft / s²
I think it is easier to follow the exercise in SI system
W_capsule = 1000 pound (1 kg / 2.20 pounds)
W_capsule = 454 N
W = m_capsule g
m_capsule = W / g
m = 454 /9.8
m_capsule = 46,327 kg
Let's calculate
W = 6.67 10⁻¹¹ 46,327 7.36 10²² / 1.74 10⁶
W = 1,307 10⁶ J
Answer:
Explanation:
Its definitely an Attractive force since the two charges are Unlike.
From Coulombs Law
F=kq1q2/R²
Given
K=9x10^9
R=1m
q1=2C
q2=-1C
F=(9x10^9 x 2 x -1)/1²
F= - 1.8x10^10N. (Attractive).
