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3 years ago

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4. A metal of mass 1.55kg was heated from 300K to 320K in 6 minutes by a boiling ring of 85 W rating, calculate the specific hea

t capacity of the metal. {Neglect heat losses to the surrounding.}

1 answer:

Rufina [12.5K]3 years ago

7 0

Can't say anything 'bout it.....

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1.An elevator is ascending with constant speed of 10 m/s. A boy in the elevator throws a ball upward at 20 m/ a from a height of

laiz [17]

(a) The maximum height reached by the ball from the ground level is 75.87m

(b) The time taken for the ball to return to the elevator floor is 2.21 s

<u>The given parameters include:</u>

constant velocity of the elevator, u₁ = 10 m/s

initial velocity of the ball, u₂ = 20 m/s

height of the boy above the elevator floor, h₁ = 2 m

height of the elevator above the ground, h₂ = 28 m

To calculate:

(a) the maximum height of the projectile

total initial velocity of the projectile = 10 m/s + 20 m/s = 30 m/s (since the elevator is ascending at a constant speed)

at maximum height the final velocity of the projectile (ball), v = 0

Apply the following kinematic equation to determine the maximum height of the projectile.

$v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\ gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m$

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection

h = h₁ + h₂ + h₃

h = 28 m + 2 m + 45.87 m

h = 75.87 m

(b) The time taken for the ball to return to the elevator floor

Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m

Apply the following kinematic equation to determine the time to return to the elevator floor.

$h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s$

To learn more about projectile calculations please visit: brainly.com/question/14083704

6 0

3 years ago

Ships A and B leave port together. For the next two hours, ship A travels at 40.0 mph in a direction 35.0° west of north while t

kompoz [17]

Answer:

Explanation:

Given

Ship A velocity is 40 mph and is traveling 35 west of north

Therefore in 2 hours it will travel $40\times 2=80 miles$

thus its position vector after two hours is

$r_A=-80sin35\hat{i}+80cos35\hat{j}$

similarly B travels with 20 mph and in 2 hours

$=20\times 2=40 miles Its position vector[tex]r_B=40sin80\hat{i}+40cos80\hat{j}$

Thus distance between A and B is

$r_{AB}=\left ( -40sin80-80sin35\right )\hat{i}+\left ( 80cos35-40cos80\right )\hat{j}$

$|r_{AB}|=\sqrt{\left ( -40sin80-80sin35\right )^2+\left ( 80cos35-40cos80\right )^2}$

$|r_{AB}|=103.45 miles$

Velocity of A

$v_A=-40sin35\hat{i}+40cos35\hat{j}$

Velocity of B

$v_B=20sin80\hat{i}+20cos80\hat{j}$

Velocity of A w.r.t B

$v_{AB}=v_A-v_B$

$v_{AB}=\left ( -20sin80-40sin35\right )\hat{i}+\left ( 40cos35-20cos80\right )\hat{j}$

4 0

3 years ago

In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phas

Anni [7]

Answer:

The phase difference is $\Delta \phi = 180^o$

Explanation:

From the question we are told that

The distance between the slits is $d = 0.2 \ mm = \frac{0.2}{1000} = 0.2 *10^{-3} \ m$

The distance to the screen is $D = 100 cm = \frac{100}{100} = 1 \ m$

The wavelength is $\lambda = 400nm$

The distance of the wave from the central maximum is $L = 5mm = 5*10^{-3} m$

Generally the path difference of this waves is mathematically represented as

$y = d sin \theta$

Here $\theta$ is the angle between the the line connecting the mid-point of the slits with the screen and the line connecting the mid-point of the slits to the central maximum

This implies that

$tan \theta = \frac{L}{D}$

=> $\theta = tan ^{-1} \frac{L}{D}$

$\theta = tan ^{-1} [\frac{5*10^{-3}}{1}]$

$\theta =0.2865$

Substituting values into the formula for path difference

$y = 0.2 *10^{-3} sin(0.2864)$

$y = 9.997*10^{-7} \ m$

The phase difference is mathematically represented as

$\Delta \phi = \frac{2 \pi }{\lambda } * y$

Substituting values

$\Delta \phi = \frac{2 \pi }{400 *10^{-9} } \ * 9.997*10^{-7}$

$\Delta \phi =5 \pi$

Converting to degree

$\Delta \phi =5 \pi radians = 5 (180^o) = 180^o$

the solution is subtracted by 360° in order to get the actual angle

4 0

3 years ago

While watching a movie a spaceship explodes and there is a loud bang and flash of light. What is wrong with this scene? Explain

Ksenya-84 [330]

Well if the ship was in space their shouldn’t be a loud bang. Because you can’t hear anything in space

4 0

3 years ago

irius, the brightest star in the sky, is 2.6 parsecs (8.6 light-years) from Earth, giving it a parallax of 0.379 arcseconds. Ano

Norma-Jean [14]

The actual distance of Regulus from Earth is 23.81 parsecs.

Given:

Parallax of Regulus, p = 0.042 arc seconds

Calculation:

When an observer changes their position, an apparent change in the object's position takes place. This change can be calculated using the angle ( or semi-angle) made by the observer and object i.e. the angle made between the two lines of observation from the object to the observer.

Thus from the relation of parallax of a celestial body we get:

S = 1/ tan p ≈ 1 / p

where S is the actual distance between the object and the observer

p is the parallax angle observed

Here for Regulus, we get:

S = 1 / p

= 1 / (0.042) [ 1 parsecs = 1 arcseconds ]

= 23.81 parsecs

We know that,

1 parsecs = 3.26 light-years = 206,000 AU

Converting the actual distance into light years we get:

23.81 parsecs = 23.81 × (3.26 light yrs) = 77.658 light-years

Therefore, the actual distance of Regulus from Earth is 23.81 parsecs which is 77.658 in light years.

Learn more about astronomical units here:

<u>brainly.com/question/16471213</u>

#SPJ4

6 0

1 year ago