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solniwko [45]
3 years ago
11

Which object has more heat energy , a hot cup or a cold cup? why

Physics
2 answers:
NARA [144]3 years ago
8 0

a hot cup because if a hot cup is hotter it's been in the sun too long

olga55 [171]3 years ago
7 0

hot water molecules are moving faster than cold water molecules. So hot water has more energy than cold one.

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The 0.8-Mg car travels over the hill having the shape of a parabola. When the car is at point A, it is traveling at 9 m/s and in
Kay [80]

The image is missing, so i have attached it.

Answer:

resultant normal force; N= 6727.9 N

resultant frictional force;F_f = -1144.33 N

Explanation:

From the image,

y = 20(1 - x²/6400)

Expanding, we have;

y = 20 - 20x²/6400)

dy/dx = -40x/6400

From the diagram, x = 80

At x = 80,

dy/dx = -40(80)/6400

dy/dx = -0.5

Also, d²y/dx² = -40/6400

d²y/dx² = -1/160

Now,

The radius of curvature is;

R = [(1 + (dy/dx)²)^(3/2)]/(d²y/dx²)

Plugging in the relevant values;

R = [(1 + (-0.5)²)^(3/2)]/(-1/160)

R = -223.61m

But we'll take the absolute value as radius cannot be negative.

Thus;

R = 223.61m

We know that acceleration (a_n) = v²/R

Thus, a_n = 9²/223.61

a_n = 81/223.61

a_n = 0.3622 m/s²

Now, to get the resultant normal force. From the diagram, resolving forces, gives;

N = W•cosθ - m•a_n

We are given; m = 0.8 x 10³ kg

Now, tan θ = dy/dx

And dy/dx at the distance of 80 = -1.5

Thus,tanθ = - 1.5

θ = tan^(-1)(-1.5)

θ = 26.6°

(the negative sign was ignored)

Thus;

ΣF_n;

N = mg•cos26.6 - m•a_n

N = (0.8 x 10³ x 9.81 x 0.8942) - (0.8 x 10³ x 0.3622)

N = 6727.9 N

Now for the resultant frictional force;

ΣF_t;

F_f + Wsin26.6 = m(v/t)

Where;

F_f is resultant frictional force

v/t is rate of change of velocity which is given as 3 m/s²

Thus;

F_f = m(v/t) - Wsin26.6

F_f = (0.8 x 10³ x 3) - (0.8 x 10³ x 9.81 x 0.4478)

F_f = 2400 - 3514.33

F_f = -1144.33 N

5 0
4 years ago
A laser emits light of frequency 4.74 x 1014 hz. what is the wavelength of the light in nm?
Ulleksa [173]
<span>Use this formula:Wavelength=c/v, where c is the speed of light, and v the frequency. 6.33 X 10^-7=3 X 10^8/v v=3 X 10^8/6.33 X 10^-7 v=4.74 X 10^14 Hertz,</span>
5 0
3 years ago
3. An electric motor is used to lift a 6.0 kg of mass through a height of 1 metre. The energy it uses is measure on an energy me
olga55 [171]

Answer:

Climbing stairs and lifting objects is work in both the scientific and everyday sense—it is work done against the gravitational force. When there is work, there is a transformation of energy. The work done against the gravitational force goes into an important form of stored energy that we will explore in this section.



Figure 1. (a) The work done to lift the weight is stored in the mass-Earth system as gravitational potential energy. (b) As the weight moves downward, this gravitational potential energy is transferred to the cuckoo clock.

Let us calculate the work done in lifting an object of mass m through a height h, such as in Figure 1. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg. The work done on the mass is then W = Fd = mgh. We define this to be the gravitational potential energy (PEg) put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force. For convenience, we refer to this as the PEg gained by the object, recognizing that this is energy stored in the gravitational field of Earth. Why do we use the word “system”? Potential energy is a property of a system rather than of a single object—due to its physical position. An object’s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth’s surface, but this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs.

Converting Between Potential Energy and Kinetic Energy

Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass, gravitational force will do an amount of work equal to mgh on it, thereby increasing its kinetic energy by that same amount (by the work-energy theorem). We will find it more useful to consider just the conversion of PEg to KE without explicitly considering the intermediate step of work. (See Example 2.) This shortcut makes it is easier to solve problems using energy (if possible) rather than explicitly using forces.

More precisely, we define the change in gravitational potential energy ΔPEg to be ΔPEg = mgh, where, for simplicity, we denote the change in height by h rather than the usual Δh. Note that h is positive when the final height is greater than the initial height, and vice versa. For example, if a 0.500-kg mass hung from a cuckoo clock is raised 1.00 m, then its change in gravitational potential energy is

mgh=(0.500 kg)(9.80 m/s2)(1.00 m) =4.90 kg⋅m2/s2=4.90 Jmgh=(0.500 kg)(9.80 m/s2)(1.00 m) =4.90 kg⋅m2/s2=4.90 J

Note that the units of gravitational potential energy turn out to be joules, the same as for work and other forms of energy. As the clock runs, the mass is lowered. We can think of the mass as gradually giving up its 4.90 J of gravitational potential energy, without directly considering the force of gravity that does the

5 0
3 years ago
A small metal bead, labeled A, has a charge of 28 nC .It is touched to metal bead B, initially neutral, so that the two beads sh
Dmitry_Shevchenko [17]

Answer:

Explanation:

Let the charge on bead A be q nC  and the charge on bead B be 28nC - qnC

Force F between them

4.8\times10^{-4} = \frac{9\times10^9\times q\times(28-q)\times10^{-18}}{(5\times10^{-2})^2}

=120 x 10⁻⁸ = 9 x q(28 - q ) x 10⁻⁹

133.33 = 28q - q²

q²- 28q +133.33 = 0

It is a quadratic equation , which has two solution

q_A = 21.91 x 10⁻⁹C or q_B = 6.09 x 10⁻⁹ C

3 0
3 years ago
Where would you expect the oceanic crust to be the hottest
Dimas [21]
It is the hottest at the mid ocean ridge because new crust is formed from the lava.
8 0
4 years ago
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