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3 years ago

Which is produced in a synthesis reaction?

Physics

1 answer:

yulyashka [42]3 years ago

3 0

Answer:

Single Compound

Explanation:

A synthesis reaction takes two or more reactants and reacts chemically to turn them all into one substance.

An example of a synthesis reaction is the reaction that takes place between sodium (Na) and chlorine (Cl) to create table salt.

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A 55 kg person is in a head-on collision. The car's speed at impact is 12 m/s. Estimate the net force on the person if he or she

ollegr [7]

1.244 m per second the person driving will go

4 0

3 years ago

A merry-go-round with a a radius of R = 1.63 m and moment of inertia I = 196 kg-m2 is spinning with an initial angular speed of

kondor19780726 [428]

Answer:

1) L = 299.88 kg-m²/s

2) L = 613.2 kg-m²/s

3) L = 499.758 kg-m²/s

4) ω₁ = 0.769 rad/s

5) Fc = 70.3686 N

6) v = 1.2535 m/s

7) ω₀ = 1.53 rad/s

Explanation:

Given

R = 1.63 m

I₀ = 196 kg-m²

ω₀ = 1.53 rad/s

m = 73 kg

v = 4.2 m/s

1) What is the magnitude of the initial angular momentum of the merry-go-round?

We use the equation

L = I₀*ω₀ = 196 kg-m²*1.53 rad/s = 299.88 kg-m²/s

2) What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round?

We use the equation

L = m*v*Rp = 73 kg*4.2 m/s*2.00 m = 613.2 kg-m²/s

3) What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round?

We use the equation

L = m*v*R = 73 kg*4.2 m/s*1.63 m = 499.758 kg-m²/s

4) What is the angular speed of the merry-go-round after the person jumps on?

We can apply The Principle of Conservation of Angular Momentum

L in = L fin

⇒ I₀*ω₀ = I₁*ω₁

where

I₁ = I₀ + m*R²

⇒ I₀*ω₀ = (I₀ + m*R²)*ω₁

Now, we can get ω₁

⇒ ω₁ = I₀*ω₀ / (I₀ + m*R²)

⇒ ω₁ = 196 kg-m²*1.53 rad/s / (196 kg-m² + 73 kg*(1.63 m)²)

⇒ ω₁ = 0.769 rad/s

5) Once the merry-go-round travels at this new angular speed, with what force does the person need to hold on?

We have to get the centripetal force as follows

Fc = m*ω²*R

⇒ Fc = 73 kg*(0.769 rad/s)²*1.63 m = 70.3686 N

6) Once the person gets half way around, they decide to simply let go of the merry-go-round to exit the ride.

What is the linear velocity of the person right as they leave the merry-go-round?

we can use the equation

v = ω₁*R = 0.769 rad/s*1.63 m = 1.2535 m/s

7) What is the angular speed of the merry-go-round after the person lets go?

ω₀ = 1.53 rad/s

It comes back to its initial angular speed

8 0

3 years ago

As part of an interview for a summer job with the Coast Guard, you are asked to help determine the search area for two sunken sh

MrMuchimi

Answer: $37^{\circ}$ North of west

Explanation:

Given

40,000-ton luxury line traveling 20 knots towards west and

60,000 ton freighter traveling towards North with 10 knots

suppose v is the common velocity after collision

conserving momentum in west direction

$20\times 40,000=(20,000+60,000)v\cos \theta$

suppose the final velocity makes \theta angle with x axis

$v\cos \theta =10----1$

Conserving Momentum in North direction

$10\times 60,000=80,000\times v\sin\theta$

$v\sin \theta =\frac{15}{2}----2$

divide 1 and 2

$\tan \theta =\frac{3}{4}$

$\theta =37^{\circ}$

so search in the area $37^{\circ}$ North of west to find the ship

3 0

3 years ago

An object is thrown straight up into the air with an initial speed of 8 m/s, and reaches a greatest height of 15 m before it fal

Ugo [173]

Answer:

The value is $T_t = 2.5659 \ s$

Explanation:

From the we are told that

The initial speed of the object is $u = 8 \ m/s$

The greatest height it reached is $h = 15 \ m$

Generally from kinematic equation we have that

$v^2 = u^2 + 2gH$

At maximum height v = 0 m/s

$0^2 = 8^2 + 2 * - 9.8 * H$

=> $H = 3.27 \ m$

Here H is the height from the initial height to the maximum height

So the initial height is mathematically represented as

$s = h - H$

=> $s = 15 - 3.27$

=> $s = 11.73 \ m$

Generally the time taken for the object to reach maximum height is mathematically evaluated using kinematic equation as follows

$v = u + (-g) t$

At maximum height v = 0 m/s

$0 = 8 - 9.8t$

=> $t = 0.8163 \ s$

Generally the time taken for the object to move from the maximum height to the ground is mathematically using kinematic equation as follows

$h = ut_1 + \frac{1}{2} gt_1^2$

Here the initial velocity is 0 m/s given that its the velocity at maximum height

Also g is positive because we are moving in the direction of gravity

$15 = 0* t + 4.9 t^2$

=> $t_1 = 1.7496$

Generally the total time taken is mathematically represented as

$T_t = t + t_1$

=> $T_t = 0.8163 + 1.7496$

=> $T_t = 2.5659 \ s$

6 0

3 years ago

Which best explains satellites? The moon is a satellite of the Earth, and the Earth is a satellite of the Sun. Mercury and Pluto

expeople1 [14]

A satellite is a body that revolves around another larger body. Where the larger body has a stronger gravitational pull on the smaller body, keeping it in orbit. Thus the moon is the satellite of the Earth is a good example of this.

4 0

3 years ago