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3 years ago

If Cell fragments are found in a rock sample it was most likely that The Rock formed

Physics

1 answer:

Ira Lisetskai [31]3 years ago

7 0

5 is 4) shallow water, 6 is 2)sedimentary i think but I'm not 100% sure and 7 is 21%

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What is the potential energy of a 3 kg- ball that is on the ground?

hammer [34]

With respect to the ground, the 3 kg object has no or 0 potential energy, as one can compute the potential energy as Ep = mgh. Since the object is on the ground it posses no height and thus no potential energy.

4 0

3 years ago

Spymaster Paul, flying a constant 215km/h horizontally in a low-flying helicopter, want to drop secret documents into his contac

aleksandrvk [35]

Answer:

$\theta = 49.56 degree$

Explanation:

Relative horizontal velocity of plane with respect to the car is given as

$v_r = v_p - v_c$

so we have

$v_p = 215 km/h$

$v_c = 155 km/h$

now we have

$v_r = 215 - 155$

$v_r = 60 km/h$

$v_r = 16.67 m/s$

Now time taken by the object to drop the vertical height is given as

$y = \frac{1}{2}gt^2$

$78 = \frac{1}{2}(9.81)t^2$

$t = 3.98 s$

so the distance of the car must be

$d = v_r\times t$

$d = 16.67 \times 3.98$

$d = 66.47 m$

Angle of the car with horizontal is given as

$tan\theta = \frac{y}{x}$

$tan\theta = \frac{78}{66.47}$

$\theta = tan^{-1}(1.17)$

$\theta = 49.56 degree$

5 0

2 years ago

Brizan has just lost his job. He is proactive in trying to resolve this source of stress: He immediately uses the Internet to lo

ratelena [41]

The answer probably would be A

3 0

2 years ago

Read 2 more answers

Water flows through a 4.50-cm inside diameter pipe with a speed of 12.5 m/s. At a later position, the pipe has a 6.25-cm inside

jek_recluse [69]

Given,

The initial inside diameter of the pipe, d₁=4.50 cm=0.045 m

The initial speed of the water, v₁=12.5 m/s

The diameter of the pipe at a later position, d₂=6.25 cm=0.065 m

From the continuity equation,

$\begin{gathered} A_1v_1=A_2v_2 \\ \pi(\frac{d_1}{2})^2v_1=\pi(\frac{d_2}{2})^2v_2 \\ \Rightarrow d^2_1v_1=d^2_2v_2 \end{gathered}$

Where A₁ is the area of the cross-section at the initial position, A₂ is the area of the cross-section of the pipe at a later position, and v₂ is the flow rate of the water at the later position.

On substituting the known values,

$\begin{gathered} 0.045^2\times12.5=0.065^2\times v_2 \\ \Rightarrow v_2=\frac{0.045^2\times12.5}{0.065^2} \\ =5.99\text{ m/s} \end{gathered}$

Thus, the flow rate of the water at the later position is 5.99 m/s

4 0

1 year ago

what can you say about the motion of an object if its speed time graph is a straight line parallel to the time axis ? Also draw

MaRussiya [10]

Answer:

It says about the motion and the graph of the object is stationary, basically travelling at the same speed at any time of the graph. It will never change.

Explanation:

To draw a diagram:

1. Draw an object and represent the speed as stationary and constant at any time.

5 0

2 years ago