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2 years ago

If you have a density of 100kg/L and a mass of 1000 units, tell me the following: second what is the volume

Physics

1 answer:

adell [148]2 years ago

8 0

Answer:

volume is 0.1 L

Explanation:

you can use the equation density=mass/volume

100 = 1000 / v

divide by 1000 on both sides

0.1 = v

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TEA [102]

Answer:

Friction, normal force, and weight

Explanation:

If the book slows down, it means that there must be friction acting in the opposite direction of the direction the book is moving in.

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2 years ago

A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann

Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

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2 years ago

An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2.3 kg mass of the pulley is concentrated on i

Sergeu [11.5K]

Answer:

$a = 2.77~{\rm m/s^2}$

Explanation:

Since the pulley has a mass concentrated on its rim, the pulley can be considered as a ring.

The moment of inertia of a ring is

$I = mr^2 = (2.3)(23.5\times 10^{-2})^2 = 0.127$

The mass on the left is heavier, that is the pulley is rotating counterclockwise.

By Newton's Second Law, the net torque is equal to moment of inertia times angular acceleration.

$\tau = I \alpha$

Here, the net torque is the sum of the weight on the left and the weight on the right.

$\tau = m_1gR - m_2gR = (1.65)(9.8)(23.5\times 10^{-2}) - (1)(9.8)(23.5\times 10^{-2}) = 1.497~{\rm Nm}$

Applying Newton's Second Law gives the angular acceleration

$\tau = I\alpha\\1.497 = 0.127\alpha\\\alpha = 11.78~{\rm rad/s^2}$

The relation between angular acceleration and linear acceleration is

$a = \alpha R$

Then, the linear acceleration of the masses is

$a = 11.78 \times 23.5\times 10^{-2} = 2.77~{\rm m/s^2}$

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3 years ago