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3 years ago

7

Can someone help with this???

2 answers:

Rus_ich [418]3 years ago

8 0

no BECQUSE POSUM BROOB SHSHSJ

Alina [70]3 years ago

4 0

Answer:

Help with what?

Explanation:

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A bag of blood with a density of 1050 kg/m3 is raised about 1.00 m higher than the level of a patient's arm. How much greater is

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A bag of blood with a density of 1050 kg/m4 is raised about 1.00 m higher than the level of a patient's arm. How much greater is the blood pressure at the patient's arm than it would be if the bag were at the same height as the army is the solisan of kind

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3 years ago

after 2 hours the displacement of a car is 160km north the initial displacement of the car is 0. what is the average velocity of

gulaghasi [49]

Answer:

D

Explanation:

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3 years ago

A gull is flying horizontally 10.80 m above the ground at 6.00 m/s. The bird is carrying a clam in its beak and plans to crack t

UkoKoshka [18]

Answer:

v_y = 14.55 m/s

Explanation:

given,

height at which gull is flying = 10.80 m

speed of the gull = 6 m/s

acceleration due to gravity = 9.8 m/s²

Relative to the seagull, the x-speed is 0,

because the seagull has the same x-speed.

Only the y-speed counts:

$v_y^2 = u^2 + 2 g h$

$v_y^2 = 0^2 + 2 g h$

$v_y = \sqrt{2gh}$

$v_y = \sqrt(2\times 9.8 \times 10.8)$

$v_y = \sqrt(211.68)$

v_y = 14.55 m/s

hence, the speed at which the clam smash the rock is v_y = 14.55 m/s

8 0

3 years ago

skad [1K]

Answer:

64q

Explanation:

Note that

Charges are addaptive quantities

Here

Capacitance C and potential V is given

We know

Q=CV

So charge is also same

Now

there are 64 drops so 64q charge

5 0

2 years ago

Read 4 more answers

A block weighing 3.7 kg is suspended from the ceiling of a truck trailer by a hanging bungee cord. The cord has a cross-sectiona

sweet [91]

Answer:

$Y = 2.27 \times 10^{10} N/m^2$

Explanation:

Natural length of the string is given as

$L_o = 43 cm$

length of the string while block is hanging on it

$L = 53 cm$

extension in length is given as

$\Delta L = 10 cm$

now we have strain in the string is given as

$strain = \frac{\Delta L}{L}$

$strain = \frac{{10 cm}{43 cm}$

$strain = 0.23$

similarly we will have cross-sectional area of the string is given as

$A = 40 \times 10^{-6} m^2$

now the stress in the string is given as

$Stress = \frac{T}{A}$

$Stress = \frac{mg}{A}$

$Stress = \frac{3.7 \times 9.81}{40 \times 10^{-6}}$

$stress = 9.07 \times 10^5 N/m^2$

Now Young's Modulus is given as

$Y = \frac{stress}{strain}$

$Y = \frac{9.07 \times 10^5}{40\times 10^{-6}}$

$Y = 2.27 \times 10^{10} N/m^2$

5 0

3 years ago