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6 days ago

6

A well-greased, essentially frictionless, metal bowl has the shape of a hemisphere of radius 0.150 m. You place a pat of butter

of mass 5.00×10−3kg at the rim of the bowl and let it slide to the bottom of the bowl. What is the speed of the pat of butter when it reaches the bottom of the bowl?

1 answer:

ValentinkaMS [2.4K]6 days ago

8 0

Since there is no friction in the bowl, the total mechanical energy remains constant. Thus, we can conclude that the initial potential energy of the butter is equal to its final kinetic energy at the bowl's bottom, allowing us to calculate the speed v.

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Two projectiles are launched with the same initial speed from the same location, one at a 30° angle and the other at a 60° angle

Ostrovityanka [2208]

Answer:

Explanation:

Let us denote the launch angle as $\theta _1=30^{\circ}$ .

$\theta _2=60^{\circ}$ , $\theta _1$ , and $\theta _2$ are complementary angles which means their ranges are identical.

$H_{max}=\frac{u^2(\sin \theta )^2}{2g}$

$H_{30}=\frac{u^2(\sin 30)^2}{2g}$

$H_{30}=\frac{u^2}{8g}$

Time of flight is indicated as $=\frac{2u\sin \theta }{g}$ .

$t_{30}=\frac{u}{g}$

For $\theta =60^{\circ}$

$H_{60}=\frac{u^2(\sin 60)^2}{2g}$

$H_{60}=\frac{3u^2}{8g}$

$t_{60}=\frac{2u\sin 60}{g}=\frac{\sqrt{3}u}{g}$

$H_{60}=3\times H_{30}$

$t_{60}=\sqrt{3}\times t_{30}$

7 0

18 days ago

What is the sources of error and suggestion on how to overcome it in the hooke's law experiment?

Yuliya22 [2446]

In the study of physics, Hooke's law can be expressed as:

F = kx

This law indicates that the spring force F is proportional to the extension x, with k being the spring constant.

In experiments, this is often examined using the setup illustrated in the included figure. The spring is tested, and a known weight is applied underneath it. This weight exerts a gravitational pull, essentially its weight, on the spring. While the spring elongates, the displacement can be measured using a ruler.

Several potential errors can arise during this experiment. Firstly, the person's measurement reading may be faulty. Digital scales offer greater accuracy as they reduce human error, while ruler readings can be subjective, especially if not viewed at eye level. Additionally, the object's weight may be inaccurately measured if the scale is untrustworthy. Lastly, the measuring equipment may not be correctly calibrated.

6 0

21 day ago

A box sliding on a horizontal frictionless surface runs into a fixed spring, compressing it a distance x1 from its relaxed posit

serg [2598]

Answer:

x₂=2×1

Explanation:

According to the work-energy theorem, we can assume that the gravitational potential energy at the lowest point of compression is zero since the kinetic energy change is 0;

mgx-(kx)²/2 =0 where m refers to the object's mass, g indicates the acceleration due to gravity, k denotes spring constant, and x represents the spring's compression.

mgx=(kx)²/2

x=2mg/k----------------compression when the object is at rest

However, ΔK.E =-1/2mv²⇒kx²=mv² -----------where v symbolizes the object's velocity and K.E signifies kinetic energy

Thus, if kx²=mv² then

v=x *√(k/m) ----------------where v=0

<pDoubling v results in multiplying x *√(k/m) by 2, leading to x₂ being double x₁

7 0

1 month ago

Can pockets of vacuum persist in an ideal gas? Assume that a room is filled with air at 20∘C and that somehow a small spherical

kicyunya [2264]

Answer:

The time required is 20 μs

Explanation:

Here is the data provided:

temperature = 20°C = 293 K

radius = 1 cm

atomic mass of air = 29 u

To determine

the duration for air to refill the vacuum space

solution:

We calculate the root mean square velocity of air particles. This can be expressed as:

$\frac{1}{2}mv^2 = \frac{3}{2}RT$

where m indicates mass, t is temperature, v is speed, and R is the ideal gas constant, which is approximately 8.3145 (kg·m²/s²)/K·mol.

v = $\sqrt{\frac{3RT}{M} }$ ............................1

v = $\sqrt{\frac{3(8.314)293}{29*10{-3}kg} }$

Resulting in v = 501.99 m/s.

<pNow, to cover the distance of 1 cm,<pThe duration needed for air is calculated as:

time taken = $\frac{r}{v}$

which gives us:

time taken = $\frac{1*10^{-2}m}{501.99}$

so, time taken = 19.92 × $10^{6}$ seconds = 20μs.

Thus, the required time is 20 μs.

3 0

1 month ago

Starting with only the Balmer series light (visible light), how could we ensure that the solar panels generate a current that Ma

ValentinkaMS [2433]

The right answer is (a).

Solar panels create electric current through the photoelectric effect, which describes how photons strike certain material surfaces, resulting in the release of electrons when light with the correct frequency hits them. A photon will interact with an electron on the panel, causing it to be ejected from the panel's surface.

As the illumination on the panel becomes brighter, the intensity of the light rises, indicating an increase in the number of photons. Each photon has the potential to liberate an electron; thus, as the number of incoming photons rises, so does the quantity of freed electrons. Given that the photoelectric current reflects the rate at which these electrons flow, an increase in light intensity leads to a corresponding rise in the photoelectric current.

If the frequency of the light is increased without a change in brightness, the photoelectric current remains the same because the total number of photons does not increase. Yet, the electrons that are ejected do escape with higher kinetic energy. However, since the total number of electrons liberated stays unchanged, the current remains constant regardless of the electrons' increased energy. Thus, option b is incorrect.

Increasing the wavelength of the light means the energy of the photons decreases. This would cause the emitted electrons to have lower energy. However, if the brightness is consistent, the number of electrons remains the same, and as a result, there would be no change in the photoelectric current. Therefore, choice (c) is also incorrect.

The correct answer is (a). To generate the needed current, the brightness of the incident light must be increased.

8 0

27 days ago

Read 2 more answers