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2 years ago

The product side of a chemical reaction is shown. → 7Ti2(SO4)3 Which number represents a coefficient?

Physics

2 answers:

AnnZ [28]2 years ago

5 0

D) 7 is the right answer

solniwko [45]2 years ago

3 0

7 is the answer.................................

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Which parts of the electric circuit considered as fuse

Marizza181 [45]

Answer:

Rewirable or Kit – Kat Type Fuses are a type of Low Voltage (LV) Fuses. They are most commonly used in house wiring, small industries and other small current applications. Rewirable Fuses consists of two main parts: a Fuse Base, which contains the in and out terminal, and a Fuse Carrier, which holds the Fuse Element.

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1 year ago

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The Back Saver Sit and Reach from the Fitnessgram measure this fitness component.

MAVERICK [17]

The correc answer is B.

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2 years ago

Consider a system to be two train cars traveling toward each other. What is the total momentum of the system before the train ca

Brut [27]

Let say the two train cars are of masses $m_1$ and $m_2$

now if the speed of two cars are $v_1$ and $v_2$

then we can say that the momentum of two cars before they collide is given by

$P = m_1v_1 - m_2v_2$

here two cars are moving in opposite direction so we can say that the net momentum is subtraction of two cars momentum.

Now since in these two car motion there is no external force on them while they collide

So the momentum of two cars are always conserved.

hence we can say that the final momentum of two cars will be same after collision as it is before collision

$P = m_1v_1 - m_2v_2$

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2 years ago

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If a person weighs 818 N on earth and 5320 N on the surface of a nearby planet, what is the acceleration due to gravity on that

alexandr402 [8]

Answer:

$g'=63.74\ m/s^2$

Explanation:

It is given that,

Weight of the person on Earth, W = 818 N

Weight of a person is given by the following formula as :

$W=mg$

g is the acceleration due to gravity on earth

$m=\dfrac{W}{g}$

$m=\dfrac{818\ N}{9.8\ m/s^2}$

m = 83.46 kg

The mass of an object is same everywhere. It does not depend on the location.

Let W' is the weight of the person on the surface of a nearby planet, W' = 5320 N

g' is the acceleration due to gravity on that planet. So,

$g'=\dfrac{W'}{m}$

$g'=\dfrac{5320\ N}{83.46\ kg}$

$g'=63.74\ m/s^2$

So, the acceleration due to gravity on that planet is $63.74\ m/s^2$ . Hence, this is the required solution.

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2 years ago

a projectile is launched at an angle of 30 degrees and lands later at the same level. if it's initial speed is 50 m/s, solve for

Mrrafil [7]

$using \: the \: formula \\ t = \frac{2u \sin( \alpha ) }{g} where \: u = initial \: speed \: \\ \alpha = angle \: of \: projection \\ g = acceleration \: due \: to \: gravity \\ \frac{2 \times 50 \times \sin(30) }{10} \\ \frac{100 \times 0.5}{10} = \frac{50}{10} = 5seconds$

Maximum height
= (Usinα)^2/2g
(50*0.5)^2/20
25^2/20
625/20
=31.25metres
horizontal distance = Range= [U^2 * sin2α]/g
[50^2 * sin60]/10
2500 * 0.8660/10
2165/10=216.5metres

3 0

2 years ago