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2 years ago

Part of a neutralization reaction between an acid and a base is shown below. What is the second product of the reaction?

Physics

2 answers:

icang [17]2 years ago

8 0

Salt compound. it is a double displacement

stira [4]2 years ago

4 0

Answer:

Salt Compound

Explanation:

acid + base →→→ water +salt compound

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If a car has an initial velocity of 20 m/s and accelerates at 2.0 m/s2 for 100 m, what is its final velocity?

rodikova [14]

Answer:

28.28 ms-1

Explanation:

v² = u² + 2as

v² = 20² + 2×2×100

v²= 800

v = 28.28 ms-1

3 0

2 years ago

1. A 59 kg person is in a vehicle travelling at 41 m/s. The vehicle runs into a telephone pole. At impact, it

Tems11 [23]

Answer:

a. 12,600 N

b. 1290 kg

Explanation:

a. Impulse = change in momentum

F Δt = m Δv

F (0.192 s) = (59 kg) (0 m/s − 41 m/s)

F = -12,600 N

b. F = mg

12,600 N = m (9.8 m/s²)

m = 1290 kg (or 2,830 lbs)

5 0

3 years ago

a metal of length 15.04cm is heated until its temperature rises by 65°c. if its new length is 16.00m calculate its linear expans

spin [16.1K]

Answer:

ΔL = L0 C ΔT

We need to find C the constant of expansivity

C = ΔL / (L0 ΔT)

C = .96 / (15.04 * 65) = 9.82 * 10^-4 / deg C

5 0

1 year ago

A satellite is in circular orbit at an altitude of 1500 km above the surface of a nonrotating planet with an orbital speed of 9.

Ksju [112]

To solve this problem we will use the Newtonian theory about the speed of a body in space for which the speed of a body in the orbit of a planet is summarized as:

$v = \sqrt{\frac{2GM}{R}}$

Where,

G = Gravitational Universal Constant

M = Mass of Planet

r = Radius of the planet ('h' would be the orbit from the surface)

The escape velocity is

$v = 14.9km/h = 14900m/s$

Through this equation we can find the mass of the Planet in function of the distance, therefore

$M = \frac{v^2R}{2G}$

$M = \frac{14900^2R}{2(6.67*10^{-11})}$

$M = 16.64*10^{17}R$

The orbital velocity is

$v_o = \sqrt{\frac{GM}{R+h}}$

$9200^2 = \frac{(6.67*10^{-11})(16.64*10^{17})R}{R+1500*10^3}$

$11.1*10^7R = (R+15000*10^3)(9200)^2$

$2.64*10^7R = 12.69*10^{13}$

$R = 4.81*10^6m$

The time period of revolution is,

$T = \frac{2\pi(R+h)}{v_o}$

$T = \frac{2\pi(4.81*10^6+1.5*10^6)}{9200}$

$T = 4307s$

$T = 72min = 1hour12min$

Therefore the orbital period of the satellite is closes to 1 hour and 12 min

3 0

3 years ago

What are simple machines please?

lubasha [3.4K]

A basic mechanical device that is used for applying a force.

Examples are: inclined plane, lever, wedge, wheel and axle, pulley, and screw...Hope this helps, have a BLESSED and wonderful day!

5 0

3 years ago

Read 2 more answers