Light travelling in one material enters another material in which it travels faster. The

A 2.40 kg stone is sliding in the +x-direction on a horizontal, frictionless surface. It collides with a 4.00 kg stone at rest.

tester [92]

Answer:

4.32 kg.m/s

Explanation:

SInce the 2.4 kg stone is mobing at speed of 3.6m/s at an angle of 30 degree measured from the x+ direction toward y+ direction. Its y-component momentum is product of its velocity in the y-direction and its mass

$P_y = m*v_y = m*v*sin(\theta)$

$P_y = 2.4*3.6*sin(30)$

$P_y = 2.4*3.6*0.5 = 4.32 kg.m/s$

5 0

3 years ago

A 1.0 kg ball and a 2.0 kg ball are connected by a 1.1-m-long rigid, massless rod. The rod is rotating cw about its center of ma

katen-ka-za [31]

The torque that will bring the balls to halt, t = 0.34 Nm

What is rotating?

In physics and mechanics, torque is the rotational equivalent of linear force. It is also referred to as the moment, moment of force, rotational force, or turning effect, depending on the field of study. It represents the capability of a force to produce a change in the rotational motion of the body.

Center of Mass of Rod = (1 * 0 + 2.0 * 1.1) /(1.0 + 2.0)

Center of Mass of Rod = 0.73 m

ω = 21 rpm

ω = 21 * (2*pi)/60 rad/s

ω = 2.2 rad/s

Time to bring Halt = 5.2 s

Angular De-acceleration = 2.2/5.2 rad/s^2

α = 0.423 rad/s^2

We know,

torque = (moment of inertia) * (angular acceleration)

t = I * α

Where

I = m1*r1^2 + m2*r2^2

I = 1kg * (0.73)^2 + 2kg * (1.1-0.73)^2

I = 0.8067 kg*m^2

Substituting Values -

torque = 0.8067 Kg*m^2 * 0.423 rad/s^2

torque = 0.34 N*m

The torque that will bring the balls to halt, t = 0.34 Nm

To learn more about the torque the link is given below:

brainly.com/question/18883167?

#SPJ4

7 0

1 year ago

A thin 14.7 m long copper rod in a uniform magnetic field has a mass of 49.0 g. When the rod carries a current of 0.317 A, it fl

MatroZZZ [7]

$F_B = ILBsin \theta$

F force
I current
L length of wire in magnetic field
B magnetic field strength
θ angle between wire and magnetic field

if θ = 90°

$B = \frac{F_B}{IL} \\ \\ F_B = F_G = mg \\ \\ B = \frac{mg}{IL} = \frac{0.049* 9.81 }{0.317*14.7}$

3 0

3 years ago

On average, both arms and hands together account for 13% of a person's mass, while the head is 7.0% and the trunk and legs accou

BabaBlast [244]

Answer:

176.38 rpm

Explanation:

mass percentage of arms and legs = 13%

mass percentage of legs and trunk = 80%

mass percentage of head = 7%

Total mass of the skater = 74.0 kg

length of arms = 70 cm = 0.7 m

height of skater = 1.8 m

diameter of trunk = 35 cm = 0.35 m

Initial angular momentum = 68 rpm

We assume:

The spinning skater with her arms outstretched as a vertical cylinder (head, trunk, and legs) with two solid uniform rods (arms and hands) extended horizontally.
friction between the skater and the ice is negligible.

We split her body into two systems, the spinning hands as spinning rods

1. Each rod has moment of inertia = $\frac{1}{3} mL^{2}$