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3 years ago

Which sI units are combined to describe force?

Physics

2 answers:

aalyn [17]3 years ago

7 0

The units of mass, length, and time are.

The combination is the unit of force, called the "Newton".

1 Newton is the force that accelerates 1 kilogram of mass at the rate of 1 meter per second every second.

Reptile [31]3 years ago

3 0

Answer:

Symbol N

Explanation:

The SI unit of force is the newton, symbol N

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Elect the correct answer. Two identical projectiles, A and B, are launched with the same initial velocity, but the angle of laun

allsm [11]

Answer:

A. The range of A and B are equal.

Explanation:

Let u be the initial speed of both the projectile.

The gravitational force acting in the projectile is in the downward direction, sp the speed of the projectile in the horizontal direction remains constant and equals to the initial horizontal speed.

For projectile, a projectile having initial velocity u at an angle \theta with the horizontal direction,

The speed in the horizontal direction $= u\cos\theta$

and the speed in the vertical direction is $= u\sin\theta$ upward.

For A:

The speed in the horizontal direction $= u\cos75^{\circ}$

and the speed in the vertical direction is $= u\sin75^{\circ}$ upward.

For B:

The speed in the horizontal direction $= u\cos15^{\circ}$

and the speed in the vertical direction is $= u\sin15^{\circ}$ upward.

Let $t_A$ and $t_B$ are the time of flight for projectile A and B respectively.

As the range is the horizontal distance traveled by the projectile, so

The range for the projectile A $= u\cos75^{\circ}\times t_A\cdots(i)$

The range for the projectile B = $u\cos15^{\circ}\times t_B\cdots(ii)$

At the highest point, the vertical velocity is 0.

Bu using the equation of motion $v^2=u^2 +2a s.$

Here, the final velocity v=0, the initial velocity $u = u \sin \theta$ , h= vertical distance up to the highest point, and $a= -g$ (as per sign convention).

So, $s= \frac{u^2\sin^2 \theta}{2g}$

For projectile A: The maximum height attained.

$s_A= \frac{u^2\sin^2 75^{\circ}}{2g}$

For projectile B: The maximum height attained.

$s_B= \frac{u^2\sin^2 15^{\circ}}{2g}$

As $\sin^2 75^{\circ} > \sin^2 15^{\circ}$ , the height of A is attained by A is more than the heigHt attained by B.

Now, the times required to reach the highest point from the ground and again form the highest point to the ground are the same.

So, the total time of flight = 2 x (Time to reach the highest point)

In a similar way, by using the equation of motion v=u+at,

The time to reach the highest point $=\frac {u\sin\theta}{g}$

where g is the acceleration due to gravity.

So, the total time of flight

$= 2 \times \frac {u\sin\theta}{g}$

The total time of flight for A

$=2 \times \frac {u\sin75^{\circ}}{g}$

The total time of flight for A

$=2 \times \frac {u\sin15^{\circ}}{g}$

Now, from equations (i) and (ii),

The range for the projectile A =

$u\cos75^{\circ}\times \frac {2u\sin75^{\circ}}{g}=\frac {u^2 \sin 150^{\circ}}{g}= \frac {u^2 \sin 30^{\circ}}{g}$

The range for the projectile B =

$u\cos15^{\circ}\times \frac {2u\sin15^{\circ}}{g}=\frac {u^2 \sin 30^{\circ}}{g}.$

Both the projectile have the same range.

Hence, option (A) is correct.

7 0

3 years ago

A concave lens has a focal length of 20 cm. A real object is 30 cm from the lens. Where is the image? What is the magnification?

Pani-rosa [81]

Answer:

12 cm and 0.4

Explanation:

f = - 20 cm, u = - 30 cm

Let v be the position of image and m be the magnification.

Use lens equation

1 / f = 1 / v - 1 / u

- 1 / 20 = 1 / v + 1 / 30

1 / v = - 5 / 60

v = - 12 cm

m = v / u = - 12 / (-30) = 0.4

6 0

3 years ago

What is the root mean square speed of an As4 particle as it is sublimed? (Assume at high temperatures arsenic acts like an ideal

Norma-Jean [14]

Answer: $v_{rms} =$ 273m/s

Explanation: Root Mean Square Speed of an atom or molecule is the speed of a particle in a gas. It is the average speed a particle in a gas can have.

It can be calculated:

$\frac{1}{2}mv^{2} = \frac{3}{2} k_{B}T$