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3 years ago

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With all his gear, Neil Armstrong weighed 360 pounds on Earth. When he landed on the Moon, he weighed 60 pounds. Why? A. His mas

s changed while traveling to the Moon. B. The gravity on the Moon is six times the gravity on Earth. C. The gravity on the Moon is less than the gravity on Earth. D. He experienced the effects of a gravity slingshot.

2 answers:

Neporo4naja [7]3 years ago

7 0

Answer:

C

Explanation:

Lyrx [107]3 years ago

6 0

Answer:

C) The gravity on the Moon is less than the gravity on Earth.

Explanation:

Since Moon is smaller than Earth, any mass on Moon will experience less gravitational force and so that mass will weigh less on Moon.

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Sasha lifts a couch 8.2 meters from the ground floor of her house to the attic. If the couch has a mass of 120 kg, what is the g

gavmur [86]

As we know that gravitational potential energy is given by

$U = mgH$

here we have

m = mass = 120 kg

$g = 9.81 m/s^2$

h = height = 8.2 m

now from above formula

$U = 120kg (9.81 m/s^2) (8.2 m)$

$U = 9653.04 J$

so above is the gravitational potential energy of the couch

4 0

2 years ago

Read 2 more answers

An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su

s2008m [1.1K]

Answer: A. 23.59 minutes.

B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

mass of ice, m= 0.555 kg

temperature of ice, T= -16.6°C

rate of heat transfer, $q=820 J.min^{-1}$

specific heat of ice, $c_{i}= 2100 J.kg^{-1}.K^{-1}$

latent heat of fusion of ice, $L_{i}=334\times10^{3}J.kg^{-1}$

<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

$Q=mc\Delta T$

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

$Q= 0.555\times2100\times16.6$

$Q= 19347.3 J$

Now, we require 19347.3 joules of heat to bring the ice to 0°C and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

$t=\frac{Q}{q}$

$=\frac{19347.3}{820}$

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, $Q_{L}= mL_{i}$

$Q_{L}= 0.555\times334\times10^{3}= 185370 J$

<u>Now the time required for supply of 185370 J:</u>

$t=\frac{Q_{L}}{q}$

$t=\frac{185370}{820}$

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

8 0

3 years ago

A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round

Svetllana [295]

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

$u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s$

$v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s$

Now we find the centripetal acceleration which is given by

$a_c=\frac{v^2}{r}$

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

$a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2$

we also have a tangential acceleration, which is given by

$a_t = \frac{v-u}{t}$

where

t = 17.0 s

Substituting values,

$a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2$

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

$a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2$

6 0

2 years ago

Read 2 more answers

How much energy is needed to raise a 50kg block up from the ground to a height of 5 meters?

miss Akunina [59]

Answer:

The answer to your question is Pe = 2452.5 J

Explanation:

Data

mass = 50 kg

height = 5 m

gravity = 9.81 m/s²

Process

The energy of this process is Potential energy which is proportional to the mass of the body, the gravity and the height of the body.

Pe = mgh

Substitution

Pe = (50)(5)(9.81)

Simplification

Pe = 2452.5 J

8 0

2 years ago

Anika asks Eva to roll a basketball and then a bowling

Alex73 [517]

Answer:

the bowling ball, because it has more mass and therefore more inertia

Explanation:

As per law of inertia we know that if an object is having more inertia then it is difficult to change state of motion.

Inertia is the property of mass of an object which always resist to change the state of motion of the object.

If an object has more inertia then it is more difficult to change the state of motion.

Now we know that we have one bowling ball and one basket ball, since bowling ball is having more mass then it must have more inertia so it is difficult to start the motion in bowling ball.

So correct answer will be

the bowling ball, because it has more mass and therefore more inertia

8 0

2 years ago