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Lina20 [59]
3 years ago
10

You are standing on a scale in an elevator. suddenly you notice your weight decreases. what do you conclude? a. the elevator is

accelerating downwards. b. the elevator is accelerating upwards. c. the elevator is moving at a constant velocity upwards. d. your diet is working. e. the elevator is moving at a constant velocity downwards.
Physics
1 answer:
solong [7]3 years ago
3 0

Answer:

The elevator is accelerating downwards.

Explanation:

When an elevator accelerating downward it moves together with the person inside and the weight of the person does not change, but if the person is standing on a scale the contact force between the person and the scale is reduced.  The scale therefore has to push upward with less force on the person to support the person's weight. Therefore the Normal Force is smaller, so the reading on the scale becomes less than the true weight.

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Explain why saucepans are much more common with plastic handles than metal handles.
horrorfan [7]
Because it’s cheaper and efficient
3 0
3 years ago
Read 2 more answers
A 10 g bullet is fired into, and embeds itself in, a 2 kg block attached to a spring with a force constant of 19.6 n/m and whose
dmitriy555 [2]

Answer:

x=0.478\ m is the compression in the spring

Explanation:

Given:

  • mass of the bullet, m=10\ g=0.01\ kg
  • mass of block, M=2\ kg
  • stiffness constant of the spring, k=19.6\ N.m^{-1}
  • initial velocity of the spring just before it hits the block, u=300\ m.s^{-1}

<u>Now since the bullet-mass gets embed into the block, we apply the conservation of momentum as:</u>

m.u=(M+m).v

0.01\times 300=(2+0.01)\times v

v=1.4925\ m.s^{-1}

Now this kinetic energy of the combined mass gets converted into potential energy of the spring.

\rm Kinetic\ energy=Spring\ potential\ energy

\frac{1}{2} (M+m).v^2=\frac{1}{2} k.x^2

\frac{1}{2} \times (2+0.01)\times 1.4925^2=\frac{1}{2} \times 19.6\times x^2

x=0.478\ m is the compression in the spring

4 0
3 years ago
Read 2 more answers
The mass of a ship before launch is 55,000 metric tons. The ship is launched down a ramp and drops a total of 10 vertical meters
skelet666 [1.2K]

Answer:

ΔT = 17.11 °C

Explanation:

In this case, we have a ship standing on a place with a given mass and it's about to be launched to a lock containing water.

At first, before launch, the ship has a potential energy, and when the ship hits the water after being launched, this potential energy is transformed into kinetic energy.

So, let's calculate first the potential energy of the ship:

E = mgh   (1)

We have the mass, gravity and height, so we need to replace the given data here. Before we do that, let's remember to use the correct units. A ton is 1000 kg, so replacing and converting we have:

E = (55000 ton * 1000 kg/ton) * (9.8 m/s²) * 10 m

E = 5.39x10⁹ J

Now this energy will be the same when the ship hits the water, only that is kinetic energy that will result in the rise of temperature. To get this rise we use the following expression:

E = m * C * ΔT   (2)

We have the energy, the mass of water (assuming density of water as 1 kg/m³) and the specific heat, so, replacing in (2) and solving for ΔT we have:

ΔT = E / m * C    (3)

ΔT = 5.39x10⁹ / 4200 * 75000

<h2>ΔT = 17.11 °C</h2>

Hope this helps

5 0
3 years ago
A student drops two metallic objects into a 120-g steel con- tainer holding 150 g of water at 25°C. One object is a 200-g cube o
marissa [1.9K]

Answer:

The mass of the aluminum chunk is 258 g

Explanation:

Given;

mass of steel container =  120-g

mass of water = 150 g

initial temperature of water, = 25°C

mass of copper cube, M_{cu} = 200 g

initial temperature of the copper cube, T_c_u = 85°C

initial temperature of the aluminum chunk T_A_l = 5.0°C

Neglecting heat loss, heat exchanged by the two metallic objects is the same since initial temperature is equal to final temperature of water.

M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu}

where;

C_{AL} is specific heat capacity of aluminum

\delta T_{Al} is change in temperature of aluminum

C_c_u is the specific heat capacity of copper

\delta T_c_u is the change in temperature of copper

M_{Al}C_{Al} \delta T_{Al} = M_{cu}C_{cu} \delta T_{cu} \\\\M_{Al} = \frac{M_{cu}C_{cu} \delta T_{cu}}{C_{Al} \delta T_{Al}} \\\\M_{Al} = \frac{0.2*387*60}{900*20} = 0.258 \ kg

Therefore, the mass of the aluminum chunk is 258 g

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