Energy a substance or system has because of its motion Is what?

You are standing next to your friend, ready to race across a 100 meter field. When the go signal is given, you take 5 seconds to

zhuklara [117]

1) 1.08 m/s^2

Explanation:

Acceleration is equal to the change in velocity divided by the time taken:

$a=\frac{v-u}{\Delta t}$

where

v is the final velocity

u is the initial velocity

$\Delta t$ is the time taken

In this problem, we have:

- initial velocity: u = 0 (you start from rest)

- final velocity: v = 5.4 m/s

- time taken: $\Delta t = 5 s$

Therefore, the acceleration is

$a=\frac{v-u}{\Delta t}=\frac{5.4 m/s-0}{5 s}=1.08 m/s^2$

2) -0.54 m/s^2

We can calculate the acceleration to slow down using the same formula as before, but this time the data are as follows:

- initial velocity : u = 5.4 m/s

- final velocity : v = 0 (you come to a stop)

- time taken : $\Delta t = 10 s$

using the same formula, we find

$a=\frac{v-u}{\Delta t}=\frac{0-5.4 m/s}{10 s}=-0.54 m/s^2$

And the negative sign means it is a deceleration.

5 0

3 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

(c) $x_f=2.6m$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

(c) Using the kinematics equation:

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$