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luda_lava [24]
3 years ago
14

Asap plz Which statement describes a controlled experiment?

Physics
1 answer:
miss Akunina [59]3 years ago
6 0

Answer:

B, it includes a control group and an experimental group.

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05. The time required to complete one lap around a perfectly circular track having a radius of 1,835 meters is 86
likoan [24]

Answer:

v = 134.06 m/s

Explanation:

Given that,

Radius of a circular track is 1,835 m

Time required to complete one lap around a perfectly circular track is 86 seconds

We need to find the car's velocity. Velocity is equal to,

v=d/t

On circular path,

v=\dfrac{2\pi r}{t}\\\\v=\dfrac{2\pi \times 1835}{86}\\\\v=134.06\ m/s

So, car's velocity is 134.06 m/s.

7 0
4 years ago
~choices below~
mart [117]
A is the correct answer
5 0
3 years ago
Read 2 more answers
A ball rolls horizontally off a table and a height of 1.4 m with a speed of 4 m/s. How long does it take the ball to reach the g
Hitman42 [59]

For vertical motion, use the following kinematics equation:

H(t) = X + Vt + 0.5At²

H(t) is the height of the ball at any point in time t for t ≥ 0s

X is the initial height

V is the initial vertical velocity

A is the constant vertical acceleration

Given values:

X = 1.4m

V = 0m/s (starting from free fall)

A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)

Plug in these values to get H(t):

H(t) = 1.4 + 0t - 4.905t²

H(t) = 1.4 - 4.905t²

We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:

1.4 - 4.905t² = 0

4.905t² = 1.4

t² = 0.2854

t = ±0.5342s

Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))

t = 0.53s

8 0
4 years ago
Read 2 more answers
The product of 14 and a cubed
olya-2409 [2.1K]

Answer:

14 × a^3

Explanation:

Product means multiplication

Cubed means to the power of 3

6 0
4 years ago
Read 2 more answers
A liquid of density 1270 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the f
FinnZ [79.3K]

Answer:

{P_2}-P_1=49.99\ KPa

Explanation:

v_1=9.43\ m/s

d_1=11.7\ cm/s

d_2=17.5\ cm/s

From continuity equation

A_1v_1=A_2v_2

v_1d_1^2=v_2d_2^2

v_2=\dfrac{9.43\times 11.7^2}{17.5^2}

v_2=4.21\ m/s

\dfrac{P_1}{\rho g}+\dfrac{v_1^2}{2g}+y_1=\dfrac{P_2}{\rho g}+\dfrac{v_2^2}{2g}+y_2

{P_1}+\rho\dfrac{v_1^2}{2}+\rho y_1g={P_2}+\rho\dfrac{v_2^2}{2}+\rho y_2g

\rho\dfrac{v_1^2}{2}+\rho y_1g -\rho\dfrac{v_2^2}{2}-\rho y_2g={P_2}-P_1

1270\times \dfrac{9.43^2}{2}+1270\times 0\times 10 -1270\times\dfrac{4.21^2}{2}-1270\times 0.175\times 10={P_2}-P_1

{P_2}-P_1=49.99\ KPa

4 0
3 years ago
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