<h2>
Answer:</h2>
0.126m
<h2>
Explanation:</h2>
According to Hooke's law, the force (F) acting on a spring to cause an extension or compression (e) is given by;
F = k x e -------------------(i)
Where;
k = the spring's constant.
From the question, the force acting on the spring is the weight(W) of the mass. i.e
F = W -----------------------(ii)
<em>But;</em>
W = m x g;
where;
m = mass of the object
g = acceleration due to gravity [usually taken as 10m/s²]
<em>From equation (ii), it implies that;</em>
F = W = m x g
<em>Now substitute F = m x g into equation(i) as follows;</em>
F = k x e
m x g = k x e ------------------(iii)
<em>From the question;</em>
m = m1 = 3.5kg
k = 278N/m
<em>Substitute these values into equation (iii) as follows;</em>
3.5 x 10 = 278 x e
35 = 278e
<em>Now solve for e;</em>
e = 35/278
e = 0.126m
Therefore, the distance the spring is stretched from its unstretched length (which is the same as the extension of the spring) is 0.126m
Volume of gold in the phone = 10 cm^3
= 0.<span>00001 m^3 </span>
Density of gold = 19300 kg/m^3
1 kg mass = 2.2 pounds
Mass of 10 cm^3 of gold = 0<span>.00001 m^3 * (19300 kg/m^3)
= 0.193 kg
So
0.193 kg = 0.193 * 2.2 pounds
= 0.43 pounds
I think there is something wrong with the options given in the question.</span>
Answer:
He could jump 2.6 meters high.
Explanation:
Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.
Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.
Answer:
yes ( true)
Explanation:
positive effects on all the body systems.
A = 1.15m/s2, Vf = 80.0km/h --> we need it in m/s, so:
Vf = 80km/h × 1000m/1km × 1h/3600s
= 22.22m/s
Top speed = Vf, initial speed = Vi
time (t) = V(Vf-Vi) ÷ a
t = (22.22-0)m/s ÷ 1.15m/s2
t = 22.22m/s × s2/1.15m
= 19.32 seconds