Our eyes can detect light only within a range of ____________ called visible light. frequencies speeds mediums periods

Jim is driving a 2268-kg pickup truck at 19 m/s and releases his foot from the accelerator pedal. The car eventually stops due t

lord [1]

Answer:

The initial kinetic energy of the truck is 409374 J

Explanation:

This problem can be solved in two ways. Let´s solve it first in the easiest way.

The kinetic energy is calculated using this equation:

E = 1/2 · m · v²

Where:

E = kinetic energy

m = mass

v = velocity

Then, the kinetic energy of the truck will be:

E = 1/2 · 2268 kg · (19 m/s)² = 409374 J

And that´s it.

But we can complicate it a bit:

The kinetic energy is the work needed to move an object from rest to a desired velocity. If the object is moving, the work needed to stop it must be of the same magnitude as its kinetic energy (in the opposite direction to the movement).

The equation for work is:

W = F · d

Where:

W= work

F = force

d = distance

We know the magnitude of the force applied to the truck, but we do not know for how much distance that force was applied. The distance can be calculated using the equation for the position of an object moving in a straight line:

x = x0 + v0 · t + 1/2 · a · t²

where

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

But we still do not know the time nor the acceleration.

The acceleration can be obtained from the equation of force:

F = m · a

Where

F = force

m = mass

a = acceleration

Then:

900 N = 2268 kg ·a

a = 900 N /2268 kg = 0.397 m/s²

Now, we can calculate the time needed for the truck to stop. We know that at the final time, the velocity is 0. Then, we can use the equation for velocity to obtain that time:

v = v0 + a · t

Where:

v = velocity at time t

v0 = initial velocity

a = acceleration

t = time

Then:

v = 19 m/s - 0.397 m/s² · t

0 = 19 m/s - 0.397 m/s² · t

-19 m/s / -0.397 m/s² = t (acceleration is negative because it is opposite to the direction of the movement)

t = 47.86 s (The truck stoped at 47.86 s after releasing the foot from the accelerator pedal)

With the time and acceleration, we can calculate the traveled distance.

x = 0 m + 19 m/s · 47.86 s - 1/2 · 0.397 m/s² · (47.86s)²

x = 454.66 m (without rounding the acceleration nor the time, the value will be 454.86 m)

Now, we can calculate the work done to stop the truck which will be of the same magnitude as the kinetic energy:

W = 900 N · 454.66 m = 409194 J

(if you do all the calculations without rounding, you will get the same value as we calculated above using the equation of kinetic energy, 409374 J).

3 0

3 years ago

a ball of mass 15 kg moving at +20 m/s collides head-on with another ball of mass 10 kg moving at -15 Ms. if after collision the

Karo-lina-s [1.5K]

Given

m1 = 15kg

vi1 = +20 m/s

vf1 = +5 m/s

m2 = 10 kg

vi2= -15 m/s

vf2 = ?

Procedure

Conservation of momentum, general law of physics according to which the quantity called momentum that characterizes motion never changes in an isolated collection of objects; that is, the total momentum of a system remains constant.

$\begin{gathered} m_1v_{i1}+m_2v_{i2}=m_1v_{f2}+m_2v_{f2} \\ 15kg\cdot20m/s+10\operatorname{kg}\cdot-15m/s=15\operatorname{kg}\cdot5m/s+10\operatorname{kg}\cdot v_{f2} \\ 150-75=10v_{f2} \\ v_{f2}=75/10 \\ v_{f2}=7.5\text{ m/s} \end{gathered}$

The speed of the 10kg Ball is 7.5m/s

8 0

1 year ago

The magnitude of the force of the bottom block on the top block is _____ the magnitude of the force of the earth on the top bloc

ra1l [238]

Hello. This question is incomplete. The full question is:

Two blocks are stacked on top of each other on the floor of an elevator. For each of the following situations, select the correct relationship between the magnitudes of the two forces given.

The elevator is moving downward at a constant speed.

The magnitude of the force of the bottom block on the top block is _____ the magnitude of the force of the earth on the top block.

Answer:

The magnitude of the force of the bottom block on top block is equal to the magnitude of the force of the top block on bottom block.

Explanation:

As the elevator is descending, there is only a normal force being applied to the lower surface of the block. This force has a magnitude equal to the force of the upper block, because the only acceleration that is acting in this case is the force of gravity. From that force, the resulting force is zero.

6 0

3 years ago

2

Xelga [282]

Answer:

About 7.67 m/s.

Explanation:

Mechanical energy is always conserved. Hence:

$\displaystyle \begin{aligned} E_i & = E_f \\ \\ U_i + K_i &= U_f + K_f\end{aligned}$

Where <em>U</em> is potential energy and <em>K</em> is kinetic energy.

Let the bottom of the slide be where potential energy equals zero. As a result, the final potential energy is zero. Additionally, because the child starts from rest, the initial kinetic energy is zero. Thus:

$\displaystyle U_i = K_f$

Substitute and solve for final velocity:
$\displaystyle \begin{aligned} mgh_i &= \frac{1}{2}mv_f^2 \\ \\ 2gh_i &= v^2_f \\ \\ v_f &= \sqrt{2gh_i} \\ \\ & =\sqrt{2(9.8\text{ m/s$^2$})(3.00\text{ m})} \\ \\ & \approx 7.67\text{ m/s} \end{aligned}$