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3 years ago

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HELP IS APPRECIATED!!

1 answer:

kobusy [5.1K]3 years ago

8 0

Yeah i believe it is going to be c

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You are camping with two friends, Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close toget

aleksklad [387]

Answer:

35.7 m

Explanation:

Let

$\mid A\mid=18.5 m$

$\mid B\mid=41 m$

We have to find the distance between Joe's and Karl'e tent.

$A_x=Acos\theta$

$A_y=Asin\theta$

Substitute the values then we get

$A_x=18.5cos23^{\circ}=17 m$

$A_y=18.5sin 23^{\circ}=7.2 m$

$B_x=41cos37.5^{\circ}=32.5 m$

$B_y=41sin37.5^{\circ}=-24.96 m$

Because vertical component of B lie in IV quadrant and y-inIV quadrant is negative.

By triangle addition of vector

$B=A+C$

$C=B-A$

$C_x=B_x-A_x=32.5-17=15.5 m$

$C_y=B_y-A_y=-24.96-7.2=-32.16\approx=-32.2 m$

$\mid C\mid=\sqrt{C^2_x+C^2_y}$

$\mid C\mid=\sqrt{(15.5)^2+(-32.2)^2}=35.7 m$

Hence, the distance between Joe's and Karl's tent=35.7 m

6 0

2 years ago

ABCD next 3 letters??????

jasenka [17]

Answer:EFG

Explanation:AYEEE

8 0

3 years ago

Read 2 more answers

A photon has 3.4 × 10–18 joules of energy. Planck’s constant is 6.63 × 10–34 J•s.

hodyreva [135]

Ok i will answer for real this time. Please give me brainliest.
<span>The Answerr is:
5.12*10^15. Since e=h*f, f=e/h. 3.4*10^(-18)/h.
</span>i am so sorry i was doing a challenge and i needed answers to get 100 pts.
Hope I Helped
~TeenOlafLover <3

5 0

2 years ago

In which situation is a person doing work on an object?

Marianna [84]

Answer : When a person applies force and covers distance then the person is doing work on an object.

6 0

2 years ago

Read 2 more answers

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 50.9 c

seraphim [82]

Answer:

Flow rate 2.34 m3/s

Diameter 0.754 m

Explanation:

Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.

The area at the well head is

$A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2$

So the volume flow rate along the pipe is

$\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s$

We can use the similar logic to find the cross-section area at the refinery

$A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2$

The radius of the pipe at the refinery is:

$A_r = \pi r^2$

$r^2 =A_r/\pi = 0.446/\pi = 0.141$

$r = \sqrt{0.141} = 0.377m$

So the diameter is twice the radius = 0.38*2 = 0.754m

6 0

2 years ago