All categories

3 years ago

How fast will a runner be going if he accelerated from rest at 3.0 m/s^2 for a distance of 2.5 m?

Physics

1 answer:

KonstantinChe [14]3 years ago

7 0

Answer:

Vf =3.87[m/s]

Explanation:

To solve this problem we must use the following equation of kinematics.

$v_{f}^{2}=v_{o}^{2} +2*a*x$

where:

Vf = final velocity [m/s]

Vo = initial velocity = 0 (from the rest)

a = acceleration = 3 [m/s²]

x = distance = 2.5 [m]

Now replacing:

$v_{f}^{2}=0+2*3*2.5\\v_{f}^{2} =15\\v_{f}=\sqrt{15} \\v_{f}=3.87[m/s]$

You might be interested in

Is the study of muscle movement in the human body and example of biology and physics

Thepotemich [5.8K]

Yes, in broader terms for people who actually study this, this is called biomechanics. Bio refers to the human body and it moving, while the physics part comes in when a ball is thrown for example and you want to know how far it is thrown.

3 0

4 years ago

One object is at rest, and another is moving. The two collide in a one-dimensional, completely inelastic collision. In other wor

zhannawk [14.2K]

Answer:

Part a)

v = 16.52 m/s

Part b)

v = 7.47 m/s

Explanation:

Part a)

(a) when the large-mass object is the one moving initially

So here we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v$

since this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

$v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}$

$v = \frac{(8.4\times 24 + 3.8\times 0)}{3.8 + 8.4}$

$v = 16.52 m/s$

Part b)

(b) when the small-mass object is the one moving initially

here also we can use momentum conservation as the net force on the system of two masses will be zero

so here we can say

$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v$

Again this is a perfect inelastic collision so after collision both balls will move together with same speed

so here we can say

$v = \frac{(m_1v_{1i} + m_2v_{2i})}{(m_1 + m_2)}$

$v = \frac{(8.4\times 0 + 3.8\times 24)}{3.8 + 8.4}$

$v = 7.47 m/s$

4 0

3 years ago

The legs of a weight lifter must ultimately support the weights he has lifted. A human tibia (shinbone) has a circular cross sec

Sergio039 [100]

Answer:

249003822308.05008 N

Explanation:

F = Force

$\sigma$ = Breaking stress of bone = 150 MPa

$d_2$ = Outer diameter = 3.6 cm

$d_1$ = Inner diameter = 2.3 cm

Area of the bone is assumed to be a hollow cylinder

$A=\dfrac{\pi}{4}(d_2^2-d_1^2)$

Stress is given by

$\sigma=\dfrac{F}{A}\\\Rightarrow F=\sigma A\\\Rightarrow F=\dfrac{150\times 10^6}{\dfrac{\pi}{4}(0.036^2-0.023^2)}\\\Rightarrow F=249003822308.05008\ N$