Answer:
A 50 kg ball traveling at 20 m/s would have 4 times more kinetic energy.
A 50 kg ball traveling at 5 m/s would have 4 times less kinetic energy.
A 50 kg person falling at 10 m/s would have the same kinetic energy.
Explanation:
hope this helps:)
The short answer is that the displacement is equal tothe area under the curve in the velocity-time graph. The region under the curve in the first 4.0 s is a triangle with height 10.0 m/s and length 4.0 s, so its area - and hence the displacement - is
1/2 • (10.0 m/s) • (4.0 s) = 20.00 m
Another way to derive this: since velocity is linear over the first 4.0 s, that means acceleration is constant. Recall that average velocity is defined as
<em>v</em> (ave) = ∆<em>x</em> / ∆<em>t</em>
and under constant acceleration,
<em>v</em> (ave) = (<em>v</em> (final) + <em>v</em> (initial)) / 2
According to the plot, with ∆<em>t</em> = 4.0 s, we have <em>v</em> (initial) = 0 and <em>v</em> (final) = 10.0 m/s, so
∆<em>x</em> / (4.0 s) = (10.0 m/s) / 2
∆<em>x</em> = ((4.0 s) • (10.0 m/s)) / 2
∆<em>x</em> = 20.00 m
The correct answer for this question is this one: "C. Neither Natalie nor Will." Natalie and Will are discussing socialization. Natalie says that socialization occurs when an animal becomes accustomed to the people in the household. <span>Will says that socialization is easily attained if the animal is first exposed to humans after 12 weeks of age.</span>
Answer:
4.96×10¯¹⁰ N
Explanation:
The following data were obtained from the question:
Mass 1 (M1) = 300 Kg
Mass 2 (M2) = 300 Kg
Separating distance (r) = 110 m
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Gravitational force (F) =?
The gravitational force between the two goal posts can be obtained as follow:
F = GM1M2 / r²
F = 6.67×10¯¹¹ × 300 × 300 / 110²
F = 6.003×10¯⁶ / 12100
F = 4.96×10¯¹⁰ N
Therefore the gravitational force between the two goal posts is 4.96×10¯¹⁰ N
Answer:
s = 0.9689 m
Explanation:
given,
Height of fall of paratroopers = 362 m
speed of impact = 52 m/s
mass of paratrooper = 86 Kg
From from snow on him = 1.2 ✕ 10⁵ N
now using formula
F = m a
a = F/m


Using equation of motion
v² = u² + 2 a s


s = 0.9689 m
The minimum depth of snow that would have stooped him is s = 0.9689 m