All categories

3 years ago

(this is somehow part of my science unit, dont ask why)

Physics

1 answer:

nirvana33 [79]3 years ago

6 0

Answer:

Height of coffee mug = 5 cm

Explanation:

Changing all the data given in to unit cm

50 mm = 5 cm

20 cm = 20 cm

8 m = 800 cm

0.1 km = 100 m = 10000 cm

Since we need to find the height of coffee mug, it's height will not be more than 10 cm.

So 5 cm is the correct choice.

You might be interested in

Calculating average speed

djverab [1.8K]

Average speed =

(distance covered during some period of time)
divided by
(length of time to cover that distance).

7 0

3 years ago

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago

A gun has a muzzle speed of 80 meters per second. What angle of elevation should be used to hit an object 180 meters away? Negle

olga55 [171]

Knowing the initial velocity and angle, the horizontal range formula is given by R= V^2sin(2teta) / g, so we can get
sin(2teta)=Rg/V^2
sin(2teta)= (180 x 9.8)/ 80^2= 0.27, sin(2teta)=0.27, 2teta=arcsin(0.27)=15.66, so teta=15.66/2

teta=7.83°

5 0

3 years ago

Read 2 more answers

Uses of concave lens

natita [175]

Answer:

Concave Lens Uses. Telescope and Binoculars Spectacles Lasers Cameras FlashlightsPeepholes. ...

Used in telescope and binoculars. ...

Concave lens used in glasses. ...

Uses of concave lens in lasers. ...

Use of concave lens in cameras. ...

Used in flashlights. ...

Concave lens used in peepholes.

3 0

2 years ago

Suppose Galileo dropped a lead ball (100 kilograms) and a glass ball (1 kilogram) from the Leaning Tower of Pisa. Which one hit

pogonyaev

If he dropped them both at the same time, then as close as anyone could tell, they both hit the ground at the same time.

8 0

3 years ago