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3 years ago

Which is a chemical property of iron?

Physics

1 answer:

Free_Kalibri [48]3 years ago

6 0

Im pretty sure its the first option

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During what stage of engine operation does the piston move upward in the cylinder and force the burned gases out of the cylinder

trasher [3.6K]

*l Take in air and fuel (Intake)
*l Compress (squeeze) the air and fuel (Compression)
*l Ignite and burn the air-and-fuel mixture (Power)
*l Get rid of the burned fuel gases (Exhaust)The Answer is C.Exhaust

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3 years ago

A student woke up in the morning and checked the rain gauge in her backyard. The rain gauge was empty, but the grass was moist.

faust18 [17]

It would most likely be C
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3 years ago

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To completely describe the motion of an object, you need

notsponge [240]

Answer:

Explanation:

7 0

3 years ago

A projectile of mass 1.800 kg approaches a stationary target body at 4.800 m/s. The projectile is deflected through an angle of

Artyom0805 [142]

Answer:

P = 5.22 Kg.m/s

Explanation:

given,

mass of the projectile = 1.8 Kg

speed of the target = 4.8 m/s

angle of deflection = 60°

Speed after collision = 2.9 m/s

magnitude of momentum after collision = ?

initial momentum of the body = m x v

= 1.8 x 4.8 = 8.64 kg.m/s

final momentum after collision

momentum along x-direction

P_x = m v cos θ

P_x = 1.8 x 2.9 x cos 60°

P_x = 2.61 kg.m/s

momentum along y-direction

P_y = m v sin θ

P_y = 1.8 x 2.9 x sin 60°

P_y = 4.52 kg.m/s

net momentum of the body

$P = \sqrt{P_x^2 + P_y^2}$

$P = \sqrt{4.52^2 + 2.61^2}$

P = 5.22 Kg.m/s

momentum magnitude after collision is equal to P = 5.22 Kg.m/s

5 0

3 years ago

A person of mass 70 kg stands at the center of a rotating merry-go-round platform of radius 2.9 m and moment of inertia 900 kg⋅m

Cloud [144]

Explanation:

It is given that,

Mass of person, m = 70 kg

Radius of merry go round, r = 2.9 m

The moment of inertia, $I_1=900\ kg.m^2$

Initial angular velocity of the platform, $\omega=0.95\ rad/s$

Part A,

Let $\omega_2$ is the angular velocity when the person reaches the edge. We need to find it. It can be calculated using the conservation of angular momentum as :