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kenny6666 [7]
2 years ago
8

Which statement is a hypothesis?

Physics
2 answers:
olga_2 [115]2 years ago
3 0
B, do earthworms prefer bright light or darkness!
jarptica [38.1K]2 years ago
3 0

B IS THE ANSWER. I JUST TOOK THE QUIZ.

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In the future, an experimental spacecraft leaves space dock for a test flight. After flying in a very large circle at constant s
Fantom [35]

Answer:

2.58 x 10⁸ m/s

Explanation:

Time dilation fomula will be applicable here, which is given below.

t = \frac{T}{\left ( 1-\frac{v^2}{c^2} \right )^\frac{1}{2}}

Where T is dilated time or time observed by clock in motion , t is stationary time , v is velocity of clock in motion and c is velocity of light .

c is 3 times 10⁸ ms⁻¹ , T is 7.24 h , t is 3.69 h. Put these values in the formula

7.24 = \frac{3.69}{\left ( 1-\frac{v^2}{c^2} \right )^\frac{1}{2}}\\

\frac{v^2}{c^2}=0.744\\\\

v=2.58\times 10^8

5 0
3 years ago
You put a 3 kg block in the box, so the total mass is now 9 kg, and you launch this heavier box with an initial speed of 5 m/s.
OlgaM077 [116]

Answer:

Δt=0.85 seconds

Explanation:

In this chase the speed does not change as the mass change.So we can use the follow equation to find the required time

Δt=Δv/gμ

To stop the final speed will be zero therefore the change in speed will be

Δv= vf-vi

Δv=0-5 m/s

Δv= -5 m/s

Now we plug our values for  Δv,g and μ to find time

Δt=Δv/gμ

Δt=(-5m/s) ÷(9.8m/s² × 0.6)

Δt=0.85 seconds

8 0
3 years ago
An unknown source plays a pitch of middle C (262 Hz). How fast would the sound wave from this source have to travel to raise
Viefleur [7K]

Answer:

The sound travels at v_{s}=11.4 \mathrm{m} / \mathrm{s}

Option: c

Explanation:

Unknown source plays of middle C (fs) = 262 Hz

The sound wave from this source have to travel to raise the pitch to C sharp is (fd) = 272 Hz

\begin{array}{l}{velocity of sound in air(v)=343 \mathrm{m} / \mathrm{s}, f_{\mathrm{s}}=262 \mathrm{Hz}, f_{\mathrm{d}}=271 \mathrm{Hz}} \\ {velocity of receiver(v_{\mathrm{d}})=0 \mathrm{m} / \mathrm{s},velocity of source( v_{\mathrm{s}}) \text { is unknown }}\end{array}

\text { Speed of sound } \mathrm{V}_{\mathrm{S}}=343 \mathrm{m} / \mathrm{s}

f_{\mathrm{d}}=f_{\mathrm{s}}\left(\frac{v-v_{\mathrm{d}}}{v-v_{\mathrm{s}}}\right)

\frac{f_{d}}{f_{s}}=\left(\frac{v-v_{d}}{v-v_{s}}\right)

\left(v-v_{s}\right)=\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)

v_{s}=v-\frac{f_{s}}{f_{d}}\left(v-v_{d}\right)

Substitute the given values in the formula,

v_{s}=343+\frac{262}{271}(343-0)

v_{s}=343+0.966(343)

v_{s}=343-331.33

v_{s}=11.4 \mathrm{m} / \mathrm{s}

Therefore, The sound travels at v_{s}=11.4 \mathrm{m} / \mathrm{s}

4 0
3 years ago
The bands on Jupiter are ultimately caused by...
Margaret [11]
Precision because it’s the answer
7 0
3 years ago
A motorcycle accelerates from 10. m/s to 25 m/s in 5.0 seconds. What is the average acceleration of the bike?
valentinak56 [21]

If a motorcycle accelerates from 10 m/s to 15 m/s in 5 seconds, the average acceleration of the bike is 3 m/s/s. This only means that the motorcycle’<span>s velocity will increase 3 m/s every second.  You need to divide 15 m/s which is the difference of 10 and 25 to 5 seconds.</span>

3 0
4 years ago
Read 2 more answers
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