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3 years ago

5

About how much gravitational potential energy is stored in an object with a mass of 4 kg that is at rest at a height of 8 meters

above the ground? (Use 10 m/s/s as the gravitational acceleration).
Question 3 options:

128 J

320 J

32 J

0 J

2 answers:

kirza4 [7]3 years ago

7 0

Answer:

it should be C I have has this question before

Advocard [28]3 years ago

4 0

Answer: Gravitational potential energy = mgh

m=mass of the body

g= acceleration due to gravity

h=height

P.E=4*8*10=320

P.E=320J

Explanation:

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50 grams of ice cubes at -15°C are used to chill a water at 30°C with mass mH20 = 200 g. Assume that the water is kept in a foam

Arada [10]

Answer : The final temperature is, $25.0^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ice = $2.09J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ice = 50 g

$m_2$ = mass of water = 200 g

$T_f$ = final temperature = ?

$T_1$ = initial temperature of ice = $-15^oC$

$T_2$ = initial temperature of water = $30^oC$

Now put all the given values in the above formula, we get:

$50g\times 2.09J/g^oC\times (T_f-(-15))^oC=-200g\times 4.184J/g^oC\times (T_f-30)^oC$

$T_f=25.0^oC$

Therefore, the final temperature is, $25.0^oC$

5 0

4 years ago

Hallar la energía mecánica de un pendulo , en un punto de su oscilación donde la energía cinética es 0,39 J y en ese mismo punto

iris [78.8K]

Answer:

E = 0,39 + 0,59 = 0,98 J

Explanation:

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3 years ago

A curent-carrying 2.0-cm long segment of wire is inside a long solenoid adius - 40 cm - 800 turns/m, current - 50 mA). The wire

klasskru [66]

Answer:

The value is $F_{max} = 12 \muN$

Explanation:

From the question we are told that

The length is $l = 2.0 \ cm = 0.02 \ m$

The radius of the the solenoid is $r =40 \ cm$

The number of turns per meter is $N = 800 \ turns / m$

The current through the solenoid is $I = 50 \ mA = 50*10^{-3} \ A$

The current through the segment is $I_s = 12 \ A$

Generally the magnetic force is mathematically represented as

$F = B * I_s l sin(\theta)$

At maximum $\theta = 90^o$

So

$F_{max} = B * I_s l$

Here B is the magnetic field is mathematically represented as

$B = \mu_o * N * I$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} N/A^2$

So

$B = 4\pi * 10^{-7} * 800 *50*10^{-3}$

=> $B = 5.0272 *10^{-5} \ T$

So

$F_{max} = 5.0272 *10^{-5} * 12 * 0.02$

$F_{max} = 1.2 *10^{-5}$

$F_{max} = 12 \muN$

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3 years ago

Your friend comes back to school after a week of being sick.

Serhud [2]

That Organ is be percussion.

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3 years ago

An object weighs 100 newtons on earth.What is its weight on the moon?

Pachacha [2.7K]

Answer:

a) The gravitational acceleration at the surface of the Moon is g moon=1.67 m/s

2

The ratio of weights (for a given mass ) is the ratio of g-values, so

W

moon

=(100N)(1.67/9.8)=17N.

(b) For the force on that object caused by Earth's gravity to equal 17 N, then the free fall acceleration at its location must be

ag

=1.67m/s

2

. Thus , .

ag

= r 2

Gm

E

⇒

a

g

Gm

=1.5×10

7

m

So the object would need to be a distance of r/R

E

=2.4 "radii" from Earth's center.

4 0

3 years ago

Read 2 more answers