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3 years ago

A body of mass 3 kg and volume 4 x 10^m' is hung from a balance

Physics

1 answer:

motikmotik3 years ago

4 0

Answer:

it's b I guess not sure

Explanation:

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The type of change that occurs when a solid becomes a liquid is_______.

GenaCL600 [577]

That would be melting. From a liquid to a gas is evaporation, and from a solid directly to a gas is sublimation. From a gas to a liquid is condensation, and from a liquid to a solid is freezing.

6 0

3 years ago

Which statement is true about the process shown? Free neutron Product Target nucleus Free neutron Free neutron Product Free neut

Monica [59]

Answer: B

Explanation:

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5 0

3 years ago

A horizontal spring with stiffness 0.5 N/m has a relaxed length of 19 cm (0.19 m). A mass of 22 grams (0.022 kg) is attached and

lawyer [7]

Answer:

v = 0.0147 m / s

Explanation:

For this exercise let's use energy conservation

Starting point. Fully stretched spring

Em₀ = Ke = ½ k (x-x₀)²

Final point. Unstretched position

Emf = K = ½ m v²

Emo = Emf

½ k (x- x₀)² = ½ m v²

v = √m/k (x-x₀)

Let's calculate

v = √(0.022 / 0.5) (0.26-0.19)

v = 0.0147 m / s

6 0

3 years ago

A raft of 180 kg carries two swimmers with masses of 50 kg and 80 kg. The raft is initially floating at rest. The two swimmers s

nirvana33 [79]

Answer:

v=0.5m/s

Explanation:

Given a raft of 180kg, and two simmers of 50kg and 80kg:

The two rafts have an initial velocity of zero:

$\bigtriangleup P_{raft}+\bigtriangleup P_{50}+\bigtriangleup P_{80}=0\\\\m_R\bigtriangleup v_R+m_{50}\bigtriangleup v_{50}+m_{80}\bigtriangleup v_{80}=0\\\\m_R\bigtriangleup v_R=-m_{50}\bigtriangleup v_{50}-m_{80}\bigtriangleup v_{80}\\\\\bigtriangleup v_R=\frac{-m_{50}\bigtriangleup v_{50}-m_{80}\bigtriangleup v_{80}}{m_R}\\\\\bigtriangleup v_R=\frac{-(50kg\times3m/s+80kg\times-3m/s)}{180kg}\\\\\bigtriangleup v_R=0.5m/s$

Hence, the raft starts moving with avelocity of 0.5m/s

-The raft will continue moving in the direction it was initially moving. The raft has a larger mass than the combined mass of the two swimmers, by law of momentum conservation, it's change in velocity will be negligible.

8 0

4 years ago

A rifle is aimed horizontally at a target 49 m away. the bullet hits the target 2.3 cm below the aim point. part a what was the

Nezavi [6.7K]

We use the equation of motion for vertical component,

$s_{y} = u_{y} t+\frac{1}{2} gt^2.$

Here, $s_{y}$ is displacement of bullet, $u_{y}$ is vertical initial velocity of bullet which is equal to zero because bullet was fired horizontally, and t is time of flight.

Therefore,

$s_{y} =\frac{1}{2}gt^2$

Given, $s_{y} =2.3 \ cm = 2.3 \times 10^{-2} \ m$

Substituting the values, we get time of flight

$2.3 \times 10^{-2} \ m = \frac{1}{2} \times 9.8 \ m/s^2 \times t^2 \\\\ t =\sqrt{46.94 \times 10^{-4} \ s } = 0.069 \ s$

4 0

4 years ago