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3 years ago

Which one of the following is a decomposition reaction?

Physics

2 answers:

lara [203]3 years ago

7 0

I dont know the answer to the problem sorry and plus it is not middle schools level

dexar [7]3 years ago

6 0

That is not middle school grade level.

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How do the forces acting on a boat compare to the force acting on a plane ?

erica [24]

Answer:

on there boat there will two main forces act 1,upthrust which comes from water and air and other side on the plane air and gravity.

Explanation:

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3 years ago

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Which SPF level should a person look for in sunscreen as part of planning a day at the beach? pls i need it now

slega [8]

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4 years ago

A 240-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope a

erastovalidia [21]

Explanation:

Below are attachments containing the solution.

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3 years ago

A closely wound rectangular coil of 80 turns has dimensions 25.0 \rm cm by 40.0 \rm cm. The plane of the coil is rotated from a

Pepsi [2]

Answer:

88.3

Explanation:

Emf in a rotating coil is given by rate of change of flux:

E= dФ/dt=(NABcos∅)/ dt

N: number of turns in the coil= 80

A: area of the coil= 0.25×0.40= 0.1

B: magnetic field strength= 1.1

Ф: angle of rotation= 90- 37= 53

dt= 0.06s

E= (80 × 0.4× 0.25×1.10 × cos53)/0.06= 88.3V

4 0

3 years ago

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3.1 * Consider a gun of mass M (when unloaded) that fires a shell of mass m with muzzle speed v. (That is, the shell's speed rel

kramer

Answer:

$v_s=\frac{v}{1+\frac{m}{M} }$

Explanation:

Let the shells speed with respect to ground be $v_s$

Let the shells speed with respect to ground be $v_g$

mass of shell, $m$

mass of gun, $M$

The relative velocity of shell with respect to a free unconstrained gun, $v$

According to the Newton's third law of motion the direction of the velocity of gun and shell will be in the opposite direction.

So, the relation between the relative velocity and their individual velocity will be:

$v=v_s+v_g$ ......................(1)

<u>And according to the conservation of momentum (as the condition is very close to the elastic collision):</u>

$M.v_g-m.v_s=0$

substitute the value of $v_g$ from equation (1)

$M\times (v-v_s)=m\times v_s$

$M.v-M.v_s=m.v_s$

$M.v=M.v_s+m.v_s$

$v_s=\frac{M.v}{(M+m)}$

$v_s=\frac{v}{(\frac{(M+m)}{M}) }$

$v_s=\frac{v}{1+\frac{m}{M} }$

5 0

3 years ago