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kipiarov [429]
3 years ago
13

The device shown below contains 2 kg of water. When the cylinder is allowed to fall 250 m, the temperature of the water increase

s by 1.4°C. Suppose 2 kg of water are added to the container and the cylinder is allowed to fall 750 m. What would the increase in temperature be in this case? Gizmo image A. 0.7°C B. 1.4°C C. 2.1°C D. 2.8°C

Physics
1 answer:
Gnesinka [82]3 years ago
4 0

Answer:c. 2.1°C

Explanation:

I just did it

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A 0.25-kg ball attached to a string is rotating in a horizontal circle of radius 0.5 m twice every second, what is the tension i
Svetradugi [14.3K]

Answer:

T = 19.75 N

Explanation:

given,

mass of ball = 0.25 Kg

radius = 0.5 m

frequency = 2 s⁻¹

tension in the string = ?

angular velocity

ω = 2 π f

ω = 2 π x 2

ω = 12.57 rad/s

tension on the string is equal to the centripetal force

T = m ω² r

T = 0.25 x 12.57² x 0.5

T = 19.75 N

Tension in the string is equal to T = 19.75 N

3 0
4 years ago
A 1500 W electric heater is plugged into the outlet of a 120 V circuit that has a 20 A circuit breaker. You plug an electric hai
Mila [183]

Answer:

Explanation:

Current drawn by electric heater = power/volt =1500/120 = 12.5 A.

current drawn by hair drier at 600 watt = 600/120 =5 A

current drawn by hair drier at 900 watt = 900/120 = 7.5 A.

Total current drawn by heater and hair drier used at 900 watt

= 12.5 + 7.5 = 20 A

Breaking current =20 A

So fuse will trip at this point .

3 0
3 years ago
A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.15 m2 and whose thickn
KengaRu [80]

Answer:

88 %

Explanation:

The rate of heat loss by a conducting material of thermal conductivity K, cross-sectional area,A and thickness d with a temperature gradient ΔT is given by

P = KAΔT/d

The total heat lost by the styrofoam wall is P₁ = K₁A₁ΔT₁/d₁ where K₁ =thermal conductivity of styrofoam wall 0.033 W/m-K, A₁ = area of styrofoam wall = 17 m², ΔT₁ = temperature gradient between inside and outside of the wall and d₁ = thickness of styrofoam wall = 0.20 m

The total heat lost by the glass window is P₂ = K₂A₂ΔT₂/d₂ where K₂ =thermal conductivity of glass window pane wall 0.96 W/m-K, A₂ = area of glass window pane = 0.15 m², ΔT₂ = temperature gradient between inside and outside of the window and d₂ = thickness of glass window pane = 7 mm = 0.007 m

The total heat lost is P = P₁ + P₂ = K₁A₁ΔT₁/d₁ + K₂A₂ΔT₂/d₂

Now, since the temperatures of both inside and outside of both window and wall are the same, ΔT₁ = ΔT₂ = ΔT

So, P = K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂

Since P₂ = K₂A₂ΔT₂/d₂ = K₂A₂ΔT/d₂is the heat lost by the window, the fraction of the heat lost by the window from the total heat lost is

P₂/P = K₂A₂ΔT/d₂ ÷ (K₁A₁ΔT/d₁ + K₂A₂ΔT/d₂)

= 1/(K₁A₁ΔT/d₁÷K₂A₂ΔT/d₂ + 1)

= 1/(K₁A₁d₂÷K₂A₂d₁ + 1)

= 1/[(0.033 W/m-K × 17 m² × 0.007 m ÷ 0.96 W/m-K × 0.15 m² × 0.20 m) + 1]

= 1/(0.003927/0.0288 + 1)

= 1/(0.1364 + 1)

= 1/1.1364

= 0.88.

The percentage is thus P₂/P × 100 % = 0.88 × 100 % = 88 %

The percentage of heat lost by window of the total heat is 88 %

6 0
3 years ago
What is the magnification of an astronomical telescope whose objective lens has a focal length of 74 cm and whose eyepiece has a
Novay_Z [31]

Answer:

The magnification of an astronomical telescope is -30.83.

Explanation:

The expression for the magnification of an astronomical telescope is as follows;

M=-\frac{f_o}{f_e}

Here, M is the magnification of an astronomical telescope, f_e is the focal length of the eyepiece lens and f_o is the focal length of the objective lens.

It is given in the problem that an astronomical telescope having a focal length of objective lens 74 cm and whose eyepiece has a focal length of 2.4 cm.

Put f_o=74 cm and f_e=2.4 cm in the above expression.

M=-\frac{74}{2.4}

M=-30.83

Therefore, the magnification of an astronomical telescope is -30.83.

5 0
4 years ago
The human eye is sensitive to yellow-green light having a frequency of about 5.5x 10^14 Hz (a wavelength of about 550 nm) what i
SVEN [57.7K]

Answer:The human eye is sensitive to yellow-green light having a frequency of about 5.5*10^{14} ... What is the energy in joules of the photons associated with this light? ... As the wavelength and frequency of a wave are related, we can find the energy ... In order to find this value, we need Planck's Constant, h=6.626×10−34 J⋅s h ...

Explanation:

5 0
4 years ago
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