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lilavasa [31]
2 years ago
10

What is this.best answer gets brainiest​

Physics
2 answers:
dexar [7]2 years ago
6 0
It’s a Black hole clearly
Afina-wow [57]2 years ago
5 0

Answer:

the Thanksgiving turkey

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A 20 m high filled water tank develops a 0.50 cm hole in the vertical wall near the base. With what speed does the water shoot o
Rina8888 [55]

Answer:

The speed of the water shoot out of the hole is 20 m/s.

(d) is correct option.

Explanation:

Given that,

Height = 20 m

We need to calculate the velocity

Using formula Bernoulli equation

\dfrac{1}{2}\rho v_{1}^2+\rho gh_{1}=\dfrac{1}{2}\rho v_{2}^2+\rho gh_{2}

Where,

v₁= initial velocity

v₂=final velocity

h₁=total height

h₂=height of the hole from the base

Put the value into the formula

v_{1}^2=2g(h_{2}-h_{1})

v_{1}=\sqrt{2g(h_{2}-h_{1})}

v_{1}=\sqrt{2\times9.8\times(20-0.005)}

v_{1}=19.7\ m/s= approximate\ 20\ m/s

Hence, The speed of the water shoot out of the hole is 20 m/s.

7 0
3 years ago
How many water (in kg) can raise its temperature from 23°C to 100°C by adding 21500 Joules of energy?
igomit [66]

Answer: Approximately 8.0g of water

Explanation:

4 0
3 years ago
The smallest known galaxy, Segue 2, has an approximate radius of 1.05 × 1015 kilometers. Use the conversion factors 1 light-year
scoundrel [369]

( 1.05 x 10¹⁵ km ) x ( 1 LY / 9.5 x 10¹² km ) x ( 1 psc / 3.262 LY ) =

(1.05) / (9.5 x 3.262) x (km · LY · psc) / (km · LY) x (10¹⁵⁻¹²) =

(0.03388) x (psc) x (10³) =

33.88 parsecs


5 0
3 years ago
As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted o
astraxan [27]

Answer:

3.6 x 10⁶ Pa

Explanation:

A = Area of the heel = 1.50 cm² = 1.50 x 10⁻⁴ m²

m = mass of the woman = 55.0 kg

g = acceleration due to gravity = 9.8 m/s²

Force of gravity on the heel is given as

F = mg

Inserting the values

F = (55) (9.8)

F = 539 N

Pressure exerted on the floor is given as

P = \frac{F}{A}

P = \frac{539}{1.5\times 10^{-4}}

P = 3.6 x 10⁶ Pa

6 0
3 years ago
An initially motionless test car is accelerated to 115 km/h in 8.58 s before striking a simulated deer. The car is in contact wi
hoa [83]

Answer:

a)       a = 3.72 m / s², b)    a = -18.75 m / s²

Explanation:

a) Let's use kinematics to find the acceleration before the collision

             v = v₀ + at

as part of rest the v₀ = 0

             a = v / t

Let's reduce the magnitudes to the SI system

              v = 115 km / h (1000 m / 1km) (1h / 3600s)

              v = 31.94 m / s

              v₂ = 60 km / h = 16.66 m / s

l

et's calculate

             a = 31.94 / 8.58

             a = 3.72 m / s²

b) For the operational average during the collision let's use the relationship between momentum and momentum

            I = Δp

            F Δt = m v_f - m v₀

            F = \frac{m ( v_f - v_o)}{t}

            F = m [16.66 - 31.94] / 0.815

            F = m (-18.75)

Having the force let's use Newton's second law

            F = m a

            -18.75 m = m a

             a = -18.75 m / s²

4 0
2 years ago
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