Ask question

Ask question

All categories

3 years ago

12

The diagram below shows a transverse wave moving toward the right along a rope. At the instant shown, point P on the rope is mov

ing toward the:
Please explain the answer as well, thanks so much

1 answer:

polet [3.4K]3 years ago

8 0

Answer:

2. i think

Explanation:

If the p were to keep moving it would go down then it would go up.

You might be interested in

. You are in an airplane flying over the ocean. How could you most accurately determine your distance from the water?

pshichka [43]

Answer:

i dont nobhhjkkkkkkuii

Explanation:

hjko0fraer

8 0

2 years ago

Relativistic velocity is of the order of _____ of velocity of light A- 1/15th of the velocity of light B-1/20th of the velocity

ratelena [41]

Answer:

Relativistic velocity is of the order of 1/10th of the velocity of light

Explanation:

We define relativistic speed (or velocity) as a speed that is a significant fraction of the speed of light: c = 3*10^8 m/s

Such that for these speeds, the special relativity theory starts to apply (the relativity effects starts to apply).

Usually, we define relativistic speeds as those that are of the order (or larger) of c/10, which is one-tenth of the speed of light.

Then the correct option is C:

Relativistic velocity is of the order of 1/10th of the velocity of light

4 0

3 years ago

An American Red Cross plane needs to drop emergency supplies to victims of a hurricane in a remote part of the world. The plane

Andreas93 [3]

Refer to the diagram shown below.

Assume that air resistance is ignored.

Note:
The distance, h, of a falling object with initial vertical velocity of zero at time t is
h = (1/2)gt²
where
g = 9.8 m/s²

The initial vertical velocity of the supplies is 0 m/s.
It the time taken for the supplies to reach the ground is t, then
(50 m) = (1/2)*(9.8 m/s²)*(t s)²

Hence obtain
t² = 50/4.9 = 10.2041
t = 3.1944 s

The horizontal distance traveled at a speed of 100 m/s is
d = (100 m/s)*(3.1944 s) = 319.44 m

Answer: 319.4 m (nearest tenth)

3 0

4 years ago

A force of 32.5N stretches a spring of 0.500cm. What force will stretch a similar spring 1.60 cm?

Vlad [161]

Answer:

104 N

Explanation:

Calculate the spring stiffness:

F = kx

32.5 N = k (0.500 cm)

k = 65 N/cm

Find the force for the new length:

F = kx

F = (65 N/cm) (1.60 cm)

F = 104 N

3 0

3 years ago

A model rocket blasts off from the ground, rising straight upward with a constant acceleration that has a magnitude of 85.0 m/s2

sergey [27]

Answer:

Maximum height attained by the model rocket is 2172.87 m

Explanation:

Given,

Initial speed of the model rocket = u = 0
acceleration of the model rocket = $a\ =\ 85.0 m/s^2$
time during the acceleration = t = 2.30 s

We have to consider the whole motion into two parts

In first part the rocket is moving with an acceleration of a = 85.0 $m/s^2$ for the time t = 2.30 s before the fuel abruptly runs out.

Let $s_1$ be the height attained by the rocket during this time intervel,

$s_1\ =\ ut\ +\ \dfrac{1}{2}at^2\\\Rightarrow s_1\ =\ 0\ +\ 0.5\times 85\times 2.30^2\\\Rightarrow s_1\ =\ 224.825\ m$

And Final velocity at that point be v

$\therefore v\ =\ u\ +\ at\\\Rightarrow v\ =\ 0\ +\ 85.0\times 2.3\\\Rightarrow v\ =\ 195.5\ m/s.$

Now, in second part, after reaching the altitude of 224.825 m the fuel abruptly runs out. Therefore rocket is moving upward under the effect of gravitational acceleration,

Let ' $s_2$ ' be the altitude attained by the rocket to reach at the maximum point after the rocket's fuel runs out,

At that insitant,

initial velocity of the rocket = v = 195.5 m/s.

a = $-g\ =\ -9.81\ m/s^2$
Final velocity of the rocket at the maximum altitude = $v_f\ =\ 0$

From the kinematics,

$v^2\ =\ u^2\ +\ 2as\\\Rightarrow 0\ =\ u^2\ -\ 2gs_2\\\Rightarrow s_2\ =\ \dfrac{u^2}{2g}\\\Rightarrow s_2\ =\ \dfrac{195.5^2}{2\times 9.81}\\\Rightarrow s_2\ =\ 1948.02\ m$

Hence the maximum altitude attained by the rocket from the ground is

$s\ =\ s_1\ +\ s_2\ =\ 224.85\ +\ 1948.02\ =\ 2172.87\ m$

6 0

3 years ago