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astra-53 [7]
3 years ago
5

How much force is required to accelerate a 2 kg mass at 3 m/s2

Physics
2 answers:
nadya68 [22]3 years ago
7 0
Force = mass x acceleration
= 2 x 3 = 6 Newtons
Lesechka [4]3 years ago
4 0
F = ma
F = 2 *3
F = 6 N
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Water flows through a horizontal pipe of varying cross-section. In the first section, the cross-sectional area is 10 cm2 and flo
Stels [109]

Answer:

(a) the flow speed of the second section is 11 m/s

(b) the pressure of the second section is 6.33 x 10⁴ Pa

Explanation:

Given;

flow rate in the first section, Q₁ = 2750 cm³/sec

area of the first cross section, A₁ = 10 cm²

pressure in the first cross section, P₁ = 1.2 x 10⁵ Pa

area of the second section, A₂ = 2.5 cm²

(a) the flow speed of the second section (V₂)

Apply continuity equation;

Q₁ = Q₂

Q₁ = A₂V₂

V₂ = Q₁ / A₂

V₂ = (2750) / (2.5)

V₂ = 1100 cm/s = 11 m/s

(b) the pressure of the second section (P₂)

Apply Bernoulli's equation;

P₁ + ¹/₂ρV₁² = P₂ + ¹/₂ρV₂²

where;

ρ is density of water = 1000 kg/m³

V₁ is the speed of water in the first section;

Q₁ = A₁V₁

V₁ = Q₁ / A₁

V₁ = (2750) / (10)

V₁ = 275 cm/s = 2.75 m/s

P₂ = P₁ + ¹/₂ρV₁² - ¹/₂ρV₂²

P₂ = P₁ + ¹/₂ρ(V₁² - V₂²)

P₂ = 1.2 x 10⁵ Pa + ¹/₂ x 1000 (2.75² - 11²)

P₂ = 1.2 x 10⁵ Pa + 500(-113.438)

P₂ = 1.2 x 10⁵ Pa - 0.567  x 10⁵ Pa

P₂ = 0.633 x 10⁵ Pa

P₂ = 6.33 x 10⁴ Pa

8 0
2 years ago
You add 500 mL of water at 10°C to 100 mL of water at 70°C. What is the
Sphinxa [80]

Answer:

Option (c) : 20°C

Explanation:

t(final) =  \frac{w1 \times t1 + w2 \times t2}{w1 + w2}

T(final) = 500* 10 + 100*70/600 = 20°C

4 0
3 years ago
Two types of energy can be either kinetic or potential energy. Identify those two types. For each type, explain why it can be ei
love history [14]

Answer:

All forms of energy are either kinetic or potential. The energy associated with motion is called kinetic energy . The energy associated with position is called potential energy . Potential energy is not "stored energy".

Explanation:

8 0
3 years ago
Read 2 more answers
(20) A rocket is launched vertically. At time t = 0 seconds, the rocket’s engine shuts down. At the time, the rocket has reached
Diano4ka-milaya [45]

Answer:

Explanation:

Given that,

h(t) = -9.8t² / 2 + 125t + 500

for t > 0.

At t = 0, the rocket is at height h = 500m, at a velocity of Vo = 125m/s.

We want to find the maximum height reached by rocket

Using mathematics maxima and minima

let find the turning point when dh/dt = 0

dh/dt = -9.8t + 125

dh / dt = 0 = -9.8t + 125

9.8t = 125

t = 125 / 9.8

t = 12.76s

Let find the turning point to know if this time t = 12.76 is maximum or minimum point

Let find d²h / dt²

d²h / dt² = -9.8

Since, d²h/dt² < 0, then, at t = 12.76s is the maximum points.

Then, the maximum height reached is

h =  -9.8t² / 2 + 125t + 500

h =  -9.8(12.76)² / 2 + 125(12.76) + 500

h = -797.80 + 1595 + 500

h = 1297.2 m

The maximum height reached is 1297.2 m

From the attachment, the maximum height is 1297.2m at t = 12.76sec.

Comment, the result are the same for both the analysis aspect and the graphical aspect.

3 0
3 years ago
What are the steps to extract metal from the earths crust
Pani-rosa [81]
<span>Methods of extraction include: extract by electrolysis, extract by reaction with carbon or carbon monoxide, and extracted by various chemical reactions.</span>
4 0
3 years ago
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