All categories

3 years ago

Calculate the total momentum of the three-sphere system if the three spheres are swung at −1.0 msto the left.

Physics

1 answer:

Advocard [28]3 years ago

6 0

Answer:

, pfx = pix + Jx.

Explanation:

The momentum principle tells us that impulse transfers momentum to an object.

If an object has 2 kgm/s of momentum, a 1 kgm/s impulse delivered to the object

increases its momentum to 3 kgm/s. That is, pfx = pix + Jx.

Just as we did with energy, we can represent this “momentum accounting” with a

momentum bar chart. For example, the bar chart of FIGURE 11.6 represents the ball

colliding with a wall in Figure 11.4. Momentum bar charts are a tool for visualizing

an interaction

You might be interested in

What is meant by fundamental unit? write any two difference between mass and weight.

yulyashka [42]

Answer:

Fundamental unit is any unit that is not dependent on other units and other units can be derived from them

Explanation:

Units such as Kilogram, Mass and Time are said to be fundamental units because they are independent.

Differences between Mass and weight;

1. Mass is the measure of the amount of matter in a body while weight is a measure of how the force of gravity acts upon that mass.

2. Mass is a scalar quantity while weight is a vector quantity

8 0

3 years ago

A 37 N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block ar

Ne4ueva [31]

Answer:

The frictional force acting on the block is 14.8 N.

Explanation:

Given that,

Weight of block = 37 N

Coefficients of static = 0.8

Kinetic friction = 0.4

Tension = 24 N

We need to calculate the maximum friction force

Using formula of friction force

$f=\mu mg$

Put the value into the formula

$f=0.8\times37$

$f=29.6\ N$

So, the tension must exceeds 29.6 N for the block to move

We need to calculate the frictional force acting on the block

Using formula of frictional force

$f = \mu N$

Put the value in to the formula

$f=0.4\times37$

$f=14.8\ N$

Hence, The frictional force acting on the block is 14.8 N.

6 0

3 years ago

Your heart pumps 80 g of blood with each beat. The blood starts from rest and reaches a speed of 0.60 m/s in the aorta. If each

Gala2k [10]

Answer:

d. 0.3 N

Explanation:

Force: This is defined as the product of mass of a body and its acceleration.

The S.I unit of Force is Newton (N).

Mathematically Force can be represented as,

F = ma .................. Equation 1

Where F = force, m = mass, a = acceleration.

also

a = (v-u)/t............... Equation 2

Where v = final velocity, u = initial velocity, t = time.

Given: v = 0.60 m/s, u = 0 m/s ( From rest), t= 0.16 s.

Substitute into equation 2

a = (0.60-0)/0.16

a = 3.75 m/s²

Also given: m = 80 g = 0.08 kg.

Substitute into equation 1,

F = 0.08(3.75)

F = 0.3 N.

Hence he average force = 0.3 N

The right option is d. 0.3 N

6 0

3 years ago

The inductance of a closely packed coil of 560 turns is 8.9 mH. Calculate the magnetic flux through the coil when the current is

Y_Kistochka [10]

Answer:

$0.11\times 10^{-6}weber$

Explanation:

We have given number of turns N = 560

Inductance L = 8.9 mH

Current through the coil = 7 mA

Inductance of the coil is given as $L=\frac{N\Phi }{I}$

Where N is number of turns I is current and $\Phi$ is flux

So $\Phi =\frac{LI}{N}=\frac{8.9\times 10^{-3}\times 7\times 10^{-3}}{560}=0.11\times 10^{-6}weber$

6 0

3 years ago

Read 2 more answers

A wheel rotates with a constant angular acceleration of  rad/s2. During a certain time interval its angular displacement is  r

Hatshy [7]

Answer:

The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

Explanation:

Given that,

Angular acceleration $\alpha=\pi\ rad/s^2$

Angular displacement $\theta=\pi\ rad$

Angular velocity $\omega =2\pi\ rad/s$

We need to calculate the angular velocity at the beginning

Using formula of angular velocity

$\alpha =\dfrac{\omega^2-\omega_{0}^2}{2\theta}$

$\omega_{0}^2=\omega^2-2\alpha\theta$

Where, $\alpha$ = angular acceleration

$\omega$ = angular velocity

Put the value into the formula

$\omega_{0}^2=(2\pi)^2-2\times\pi\times\pi$

$\omega=\sqrt{2\pi^2}$

$\omega_{0}=\pi\sqrt{2}\ rad/s$

Hence, The angular velocity at the beginning of the interval is $\pi\sqrt{2}\ rad/s$ .

3 0

3 years ago