Please Please answer! I will give brainiest and points if it's right!

Federal board paper of class 10 of physics

zzz [600]

Answer:

If you are looking for past papers you can search that up and you will find plenty of resources that will help you out.

8 0

3 years ago

Vector has a magnitude of 6.0 m and points 30° north of east. Vector has a magnitude of 4.0 m and points 30° east of north. The

Brut [27]

Answer:

The resultant vector is $\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j$ .

Explanation:

First, each vector is determined in terms of absolute coordinates:

6-meter vector with direction: 30º north of east.

$\vec A = (6\,m)\cdot (\cos30^{\circ} \,i + \sin 30^{\circ}\,j)$

$\vec A = 5.196\,i + 3\,j$

4-meter vector with direction: 30º east of north.

$\vec B = (4\,m)\cdot (\cos 60^{\circ}\,i + \sin 60^{\circ}\,j)$

$\vec B = 2\,i + 3.464\,j$

The resultant vector is obtaining by sum of components:

$\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j$

The resultant vector is $\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j$ .

5 0

4 years ago

Water evaporating from a pond does so as if it were diffusing across an air film 0.15 cm thick. The diffusion coefficient of wat

QveST [7]

Answer:

The water level will drop by about 1.24 cm in 1 day.

Explanation:

Here Mass flux of water vapour is given as

$j_{H_2O}=\frac{D}{l} \bigtriangleup c$

where

$j_{H_2O}$ is the mass flux of the water which is to be calculated.

D is diffusion coefficient which is given as $0.25 cm^2/s$

l is the thickness of the film which is 0.15 cm thick.
$\bigtriangleup c$ is given as

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT}$

In this

$P_{sat}$ is the saturated water pressure, which is look up from the saturated water property at 20°C and 0.5 saturation given as 2.34 Pa

$P_a$ is the air pressure which is given as 0.5 times of $P_{sat}$

R is the universal gas constant as $8.314 kJ/kmol-K$

T is the temperature in Kelvin scale which is $20+273= 293K$

By substituting values in the equation

$\bigtriangleup c= \frac{P_{sat}-P_a}{RT} \\ \bigtriangleup c= \frac{P_{sat}-0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5P_{sat}}{RT} \\ \bigtriangleup c= \frac{0.5 \times 2.34}{8.314 \times 293} \\\bigtriangleup c= 0.48 mol/m^3$

Converting $\bigtriangleup c$ into $cm^3/cm^3$

As 1 mole of water 18 $cm^3$ so

$\bigtriangleup c= 0.48 mol/m^3 \\ \bigtriangleup c= 0.48 \times 18 \times 10^{-6} cm^3/cm^3 \\ \bigtriangleup c= 8.64 \times 10^{-6} cm^3/cm^3$

Putting this in the equation of mass flux equation gives

$j_{H_2O}=\frac{D}{l} \bigtriangleup c \\ j_{H_2O}=\frac{0.25}{0.15} \times 8.64 \times 10^{-6} \\ j_{H_2O}=14.4 \times 10^{-6} cm/s$

For calculation of water level drop in a day, converting mass flux as

$j_{H_2O}=14.4 \times 10^{-6} \times 24 \times 3600 cm/day\\ j_{H_2O}=1.24 cm/day$

So the water level will drop by about 1.24 cm in 1 day.

7 0

4 years ago

A 1423-kg car is moving along a level highway with a speed of 26.4 m/s. The driver takes the foot off the accelerator and the ca

Leona [35]

Answer:

Final speed, v = 28.81 m/s

Explanation:

Given that,

Mass of the car, m = 1423 kg

Initial speed of the car, u = 26.4 m/s

Force experience by the car, F = 901 N

Distance, d = 106 m

To find,

The speed of the car after traveling this distance.

Solution,

The force experienced by a car is equal to the product of mass and acceleration.

$F=ma$

$a=\dfrac{F}{m}$

$a=\dfrac{901}{1423}$

$a=0.63\ m/s^2$

Let v is the final speed of the car. Using third equation of motion to find it as :

$v^2-u^2=2ad$

$v^2=2ad+u^2$

$v^2=2\times 0.63\times 106+(26.4)^2$

v = 28.81 m/s

So, the final speed of the car is 28.81 m/s.

5 0

4 years ago

A student fires a cannonball diagonally

rodikova [14]

Answer:The cannons total flight time is 2.23seconds

Explanation:

A ball fired diagonally is fired at an angle of 45° to the horizontal. The motion of the fired ball is a projectile motion. A projectile is a motion in which an object fired into space with an initial velocity U is allowed to fall freely under the influence of gravitational force.

To total time of flight T of the cannon ball can be expressed as;

T = 2Usin(theta)/g where;

U is the initial velocity or speed of the ball = 31m/s

theta is the angle that the ball make with the horizontal = 45°

g is the acceleration due to gravity = 9.81m/s²

Substituting the given datas into the formula we have;

T = 31sin45°/9.81

T = 31×0.7071/9.81

T = 21.92/9.81

T = 2.23seconds

8 0

3 years ago

Please Please answer! I will give brainiest and points if it's right!​

Please Please answer! I will give brainiest and points if it's right!