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2 years ago

What is the momentum of a 3 kg bowling ball moving at 3 m/s?

Physics

1 answer:

nadezda [96]2 years ago

6 0

Answer: 9kgm/s

Explanation:

Given that,

Mass of ball = 3 kg

Speed by which ball moves = 3 m/s. Linear momentum of the ball = ?

Since momentum refers to the quantity of motion of the moving ball,

Linear momentum = Mass x Speed

= 3kg x 3m/s

= 9 kgm/s

Thus, the linear momentum of the ball is 9kgm/s

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N2 + O2 → 2NO N-N triple bond: 941 kJ/mol O-O double bond: 495 kJ/mol N-O bond: 201 kJ/mol

kiruha [24]

Answer:

$\large \boxed{\text{761 kJ}}$

Explanation:

You calculate the energy required to break all the bonds in the reactants.

Then you subtract the energy needed to break all the bonds in the products.

N₂ + O₂ ⟶ 2NO

N≡N + O=O ⟶ 2O-N=O

Bonds: 2N≡N 1O=O 2N-O + 2N=O

D/kJ·mol⁻¹: 941 495 201 607

$\begin{array}{rcl}\Delta H & = & \sum{D_{\text{reactants}}} - \sum{D_{\text{products}}}\\& = & 2 \times 941 +1 \times 495 - (2 \times 201 + 2\times 607)\\&=& 2377 - 1616\\&=&\textbf{761 kJ}\\\end{array}\\\text{The enthalpy of reaction is $\large \boxed{\textbf{761 kJ}}$}.$

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2 years ago

A motor does 8000j of work in 20 seconds. What is the power of the motor

lozanna [386]

Power = work / time = 8000J / 20s = 400W

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3 years ago

An object is projected horizontally at 14.1 m/s from the top of a 195.0 meter cliff.

Roman55 [17]

What does a physical map show?

the names of countries, states, and cities

the history of an area

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3 years ago

How is energy sent from Earth to space?

Nostrana [21]

It might be radiation and reflection but I’m not sure

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3 years ago

Read 2 more answers

A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500

defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

$M = \frac{\mu_0 N_1 N_2A_1}{l_1}$

Where,

$\mu =$ permeability of free space

$N_1 =$ Number of turns in solenoid 1

$N_2 =$ Number of turns in solenoid 2

$A_1=$ Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

$\mu_0 = 4\pi *10^{-7}H/m$

$N_1 = 500$

$N_2 = 40$

$A = 7.5*10^{-4}m^2$

$l = 0.5m$

Substituting,

$M = \frac{\mu_0 N_1 N_2A_2}{l_1}$

$M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}$

$M = 3.77*10^{-4}H$

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

8 0

3 years ago