All of the waves in the electromagnetic spectrum are ______

On a touchdown attempt, 95.00 kg running back runs toward the end zone at 3.750 m/s. A 113.0 kg line-backer moving at 5.380 m/s

dsp73

Answer:

(a) 1.21 m/s

(b) 2303.33 J, 152.27 J

Explanation:

m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s

(a) Let their velocity after striking is v.

By use of conservation of momentum

Momentum before collision = momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) x v

- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v

v = ( - 356.25 + 607.94) / 208 = 1.21 m /s

(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2

= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)

= 0.5 (1335.94 + 3270.7) = 2303.33 J

Kinetic energy after collision = 1/2 (m1 + m2) v^2

= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J

4 0

2 years ago

A car is initially moving at 10.5 m/s and accelerates uniformly to reach a speed of 21.7 m/s in 4.34 s. How far did the car move

Zarrin [17]

Answer:

A. 69.9m

Explanation:

Given parameters:

Initial velocity = 10.5m/s

Final velocity = 21.7m/s

Time = 4.34s

Unknown:

Distance traveled = ?

Solution:

Let us first find the acceleration of the car;

Acceleration = $\frac{v - u}{t}$

v is final velocity

u is initial velocity

t is the time

Acceleration = $\frac{21.7 - 10.5}{4.34}$ = 2.58m/s²

Distance traveled;

V² = U² + 2aS

21.7² = 10.5² + 2 x 2.58 x S

360.64 = 2 x 2.58 x S

S = 69.9m

3 0

3 years ago

Which is an example of electrical energy being converted or changed into light energy?

Lera25 [3.4K]

When I see the word "which" at the beginning of your question,
I just KNOW that there's a list of choices printed right there
next to he part that you copied, and for some mysterious
reason, you decided not to let us see the choices.

Any flashlight, light bulb, laser, or spark ... like lightning ...
converts some electrical energy into some light energy.

8 0

3 years ago

Read 2 more answers

V=xf-xi/t solve for t

irina [24]

Answer:

$t=\frac{x_f-x_i}{v}$

Explanation:

Starting from the equation:

$v=\frac{x_f-x_i}{t}$

First of all, let's multiply by t on both sides:

$v\cdot t = \frac{x_f-x_i}{t}\cdot t \\vt = x_f - x_i$

And then, let's divide by v on both sides:

$\frac{vt}{v}=\frac{x_f-x_i}{v}\\t=\frac{x_f-x_i}{v}$

So, finally

$t=\frac{x_f-x_i}{v}$

7 0

2 years ago

At a certain instant a particle is moving in the +x direction with momentum +8 kg m/s. During the next 0.13 seconds a constant f

jeka94

Answer:

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

Explanation:

The momentum of the particle is related to force by the following equation:

Δp = F · Δt

Where:

Δp = change in momentum = final momentum - initial momentum

F = constant force.

Δt = time interval.

Let´s calculate the x-component of the momentum after the 0.13 s:

final momentum - 8 kg m/s = -7 N · 0.13 s

final momentum = -7 kg m/s² · 0.13 s + 8 kg m/s

final momentum = 7.09 kg m/s

Now let´s calculate the y-component of the momentum vector after the 0.13 s. Since the particle wasn´t moving in the y-direction, the initial momentum in this direction is zero:

final momentum = 5 kg m/s² · 0.13 s

final momentum = 0.65 kg m/s

Then, the mometum vector will be as follows:

p = (7.09 kg m/s, 0.65 kg m/s)

The magnitude of this vector is calculated as follows:

$|p| = \sqrt{(7.09 kg m/s)^{2} + (0.65 kg m/s)^{2}} = 7.12 kg m/s$

The momentum of the particle at the end of the 0.13 s time interval is 7.12 kg m/s

4 0

3 years ago

All of the waves in the electromagnetic spectrum are _______ waves.