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Marizza181 [45]
3 years ago
8

1.) Waves transfer ______ without moving ______

Physics
1 answer:
kondor19780726 [428]3 years ago
6 0
I believe it goes as the following::

1) Waves transfer energy without moving particles

2) The two types of waves are longitudinal and transverse waves OR The two types of waves are mechanical and electromagnetic waves. Both are applicable and should be correct!

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A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre
vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}

v_1=v_{01}+a_1t

We must consider that it's launched from the ground (y_{01}=0m) and from rest (v_{01}=0m/s), with an upwards acceleration a_{1}=28m/s^2 that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m

And the velocity achieved in part 1:

v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 (y_{02}=1317.26m) and its initial velocity is the one achieved in part 1 (v_{02}=271.6m/s), now in free fall, which means with a downwards acceleration a_{2}=-9,8m/s^2. For the data we have it's faster to use the formula v_f^2=v_0^2+2ad, where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s

Then, to get y_2, we do:

2a_2(y_2-y_{02})=-v_{02}^2

y_2-y_{02}=-\frac{v_{02}^2}{2a_2}

y_2=y_{02}-\frac{v_{02}^2}{2a_2}

And we substitute the values:

y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m

3 0
3 years ago
PLEASE HELP ASAP!!!!!
Igoryamba

While falling, both the sheet of paper and the paper ball experience air resistance. But the surface area of the sheet is much more than that of the spherical ball. And air resistance varies directly with surface area. Hence the sheet experiences more air resistance than the ball and it falls more slowly than the paper ball.

Hope that helps!

8 0
2 years ago
Explain why free fall acceleration near earths surface is constant
Naya [18.7K]
Because gravity is constant 
<span>the only force acting in free-fall is gravity which points downward at 9.8 m/s</span>
4 0
2 years ago
FREE BRAINLEST JUST COMMENT MRBEAST
Pepsi [2]

Answer:

MRBEAST-

Explanation:

6 0
3 years ago
Read 2 more answers
Jada is rowing a boat across a river that has a current of 5 m/s in the ˆ j direction. Leanne, standing on the shore, observes J
olchik [2.2K]

Answer: d. 8.25 m/s

Explanation:

We are given that Current= 5 m/s in j direction

Velocity= 8 m/s i + 3 m/s j

Now, we have to find Jada's speed with respect to the water.

First we find Jada's velocity with respect to water

v= (8 i + 3 j) - (5 j)

v= 8i - 2 j

To find the speed, we take the magnitude of this velocity vector we have

|v|= \sqrt{(8)^2+(-2)^2}

|v|= \sqrt{68} = 8.246 m/s

which comes out to be around = 8.25 m/s

So option d is correct.

5 0
2 years ago
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