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3 years ago

What can you conclude about this product?

Physics

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aksik [14]3 years ago

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A puck of mass 0.70 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the movi

natali 33 [55]

Answer:

$v_1 = \ 5.196 \frac{m}{s}$

$v_2 = 3 \frac{m}{s}$

Explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

$\vec{p}_i = \vec{p}_f$

where the suffix i means initial, and the suffix f means final.

The initial momentum will be:

$\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}$

as the second puck is initially at rest:

$\vec{v}_{2_i} = 0$

Using the unit vector $\vec{i}$ pointing in the original line of motion:

$\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}$

$\vec{p}_i = 0.70 \ kg \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0$

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

So:

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f$

$\vec{p}_f = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

$\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

So, our velocity vectors will be:

$\vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )$

$\vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

We got

$\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}$

$4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \ v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ ) + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

So, we got the equations:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ v_1 \ cos(30 \°) + 0.7 \ kg \ v_2 \ cos(-60 \°)$

and

$0 = 0.7 \ kg \ v_1 \ sin(30 \°) + 0.7 \ kg \ v_2 \ sin(-60 \°)$ .

From the last one, we get:

$0 = 0.7 \ kg \ ( v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°) )$

$0 = v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°)$

$v_1 \ sin(30 \°) = - \ v_2 \ sin(-60 \°)$

$v_1 = \ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) }$

and, for the first one:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ ( v_1 \ cos(30 \°) + v_2 \ cos(60 \°) )$

$\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} = v_1 \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = (\ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = v_2 (\ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + cos(60 \°)$

$6 \ \frac{m}{s} = v_2 * 2$

so:

$v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s}$

and

$v_1 = \ 3 \frac{m}{s} \ \frac{sin(60 \°)}{ sin(30 \°) }$

$v_1 = \ 5.196 \frac{m}{s}$

3 0

3 years ago

Compare the energy consumption of two commonly used items in the household. Calculate the energy used by a 1.40 kW toaster oven,

Nikolay [14]

Explanation:

It is given that,

Power consumed by toaster oven, $P=1.4\ kW$

Time taken, t = 6 minutes = 0.1 hour

Energy used by toaster, $W_{toaster}=P\times t$

$W_{toaster}=1.4\ kW\times 0.1\ h=0.14\ kWh$

Similarly,

Power consumed by toaster oven, $P=11\ W$

Time taken, t = 9 minutes = 0.15 hour

Energy used by toaster, $W_{toaster}=P\times t$

$W_{toaster}=11\ W\times 0.15\ h=1.65\ Wh=0.0016\ kWh$

Hence, this is the required solution.

3 0

3 years ago

qual o nome da posição que o planeta tem maior velocidade? este ponto é mais próximo ou mais afastado do sol?

klio [65]

Em inglês, esse ponto é 'perihelion' ... 'periélio'. É o mais próximo que a Terra ou o planeta chega do sol.

7 0

3 years ago

Current is produced when charges are accelerated by an electric field to move to a position of lower

Vaselesa [24]

Current is created when charges are quickened by an electric field to move where the position of lower temperature. An electric current is a stream of electric charge. In electric circuits, this charge is regularly conveyed by moving electrons in a wire.

8 0

3 years ago

Examine the images of the Grand Canyon below. Notice that most of the canyon consists of layers of sedimentary rocks, but if you

Tomtit [17]

Intense temperature and pressure of regional metamorphism

Explanation:

The process that cause the formation of the Vishnu Schist is the intense temperature and pressure as a result of regional metamorphism.

Regional metamorphism is an extensive metamorphism of an area as a result of temperature and pressure changes.
The schist is a foliated metamorphic rock usually found in areas of moderate to high grade temperature and pressure.
The Vishnu schist must have been metamorphosed before the new sediments were deposited on top.

Learn more:

Contact metamorphism brainly.com/question/1970623

#learnwithBrainly

7 0

3 years ago