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tatiyna
3 years ago
15

How much energy must be absorbed by water with a mass of 0.5 kg in order to raise the temperature from 30°C to 65°C? Note: Water

has a specific heat of 4,190 J/kg °C.
Physics
1 answer:
kvasek [131]3 years ago
6 0

Answer:

73325J

Explanation:

Given parameters:

Mass of water  = 0.5kg

Initial temperature  = 30°C

Final temperature  = 65°C

Specific heat capacity  = 4190J/kg°C

Unknown:

Amount of energy absorbed = ?

Solution:

The amount of energy absorbed can be derived using the expression below;

                H  = m c Δt

H is the amount of energy

m is the mass

c is the specific heat

Δt is the change in temperature

      H  = 0.5 x 4190 x (65 - 30 )

     H  = 73325J

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Answer:

Retina is the part of eye which is used to see things in high details

7 0
2 years ago
What is the wavelength of light that is deviated in the first order through an angle of 13.1 ∘ by a transmission grating having
faltersainse [42]

Answer:

The wavelength of light is 4.53\times10^{-7}\ m

Explanation:

Given that,

Angle = 13.1°

Number of slits = 5000

We need to calculate the wavelength of light

Diffraction of first order is defined as,

d \sin\theta=n\lambda.....(I)

The separation of the slits

d = \dfrac{1}{N}

d=\dfrac{1}{5000}

d=2\times10^{-6}\ m

Now put the value in equation (I)

2\times10^{-6}\sin13.1^{\circ}=\lambda

Here, n = 1

\lambda=4.53\times10^{-7}\ m

Hence, The wavelength of light is 4.53\times10^{-7}\ m

7 0
3 years ago
The resistivity of gold is 2.44 x 10-8 ohms.m at room temperature. A gold wire that is 0.9 mm in diameter and 14 cm long carries
harkovskaia [24]

Answer:

E=0.036 V/m

Explanation:

Given that

Resistivity ,ρ=2.44 x 10⁻⁸ ohms.m

d= 0.9 mm

L= 14 cm

I = 940 m A = 0.94 A

We know that electric field E

E= V/L

V= I R

R=ρL/A

So we can say that

E= ρI/A

Now by putting the values

E=\dfrac{ 2.44\times 10^{-8}\times 0.94}{\dfrac{\pi}{4}(0.9\times 10^{-3})^2}

E=0.036 V/m

6 0
3 years ago
Consider a car travelling at 60 km/hr. If the radius of a tire is 25 cm, calculate the angular speed of a point on the outer edg
vlabodo [156]

To solve this problem it is necessary to apply the concepts given in the kinematic equations of movement description.

From the perspective of angular movement, we find the relationship with the tangential movement of velocity through

\omega = \frac{v}{R}

Where,

\omega =Angular velocity

v = Lineal Velocity

R = Radius

At the same time we know that the acceleration is given as the change of speed in a fraction of the time, that is

\alpha = \frac{\omega}{t}

Where

\alpha =Angular acceleration

\omega = Angular velocity

t = Time

Our values are

v = 60\frac{km}{h} (\frac{1h}{3600s})(\frac{1000m}{1km})

v = 16.67m/s

r = 0.25m

t=6s

Replacing at the previous equation we have that the angular velocity is

\omega = \frac{v}{R}

\omega = \frac{ 16.67}{0.25}

\omega = 66.67rad/s

Therefore the angular speed of a point on the outer edge of the tires is 66.67rad/s

At the same time the angular acceleration would be

\alpha = \frac{\omega}{t}

\alpha = \frac{66.67}{6}

\alpha = 11.11rad/s^2

Therefore the angular acceleration of a point on the outer edge of the tires is 11.11rad/s^2

5 0
2 years ago
Help meeeeeeeeeeeee ill mark brainlist to people
kvasek [131]

Answer: question 1 , would be one question 2 , would be 1 joule and number three would be number one and number four would be , power and last one would be, number two

Explanation: sorry if its wrong

5 0
2 years ago
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