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2 years ago

A tightly sealed glass jar is an example of which type of system?

2 answers:

7 0

Thermal expansion and contraction, specifically the different coefficients of thermal expansion between glass & metal is a closed system ; D

noname [10]2 years ago

4 0

D. hope it helps. :D

You might be interested in

Short-range forecasts tends to ________ longer-range forecasts.

spin [16.1K]

Short-range forecasts are more accurate than longer range ones. Short term forecasts may use mathematical techniques such as moving averages and, exponential smoothing. Longer term forecasts not only use different methodologies, such as qualitative vs. quantitative, they also tend to consider different issues.

5 0

2 years ago

How much does a person weigh if it takes 700 kg*m/s to move them 10 m/s NEED ASAP

madreJ [45]

Answer:

$\huge\boxed{m = 70 \ kg}$

Explanation:

Given Data:

Momentum = P = 700 kg m/s

Velocity = v = 10 m/s

Required:

Mass = m = ?

Formula:

P = mv

Solution:

m = P / v

m = 700 / 10

m = 70 kg

$\rule[225]{225}{2}$

Hope this helped!

<h3>~AnonymousHelper1807</h3>

5 0

1 year ago

Two roads intersect at right angles, one going north-south, the other east-west. an observer stands on the road 60 meters south

Sloan [31]

observer is standing at distance d = 60 m south from the intersection

cyclist is travelling at speed v = 10 m/s

now after t = 8 s its displacement from intersection is given by

$x = 10*8 = 80 m$

so the position of cyclist makes an angle with the observer

$\theta = tan^{-1}\frac{80}{60} = 53 degree$

now the component of velocity of cyclist along the line joining its position with the observer is given as

$v = v_o cos\phi$

here

$\phi = 90 -\theta$

$\phi = 90 - 53 = 37 degree$

$v = 10 cos37 = 8 m/s$

so at this instant cyclist is moving away with speed 8 m/s

7 0

2 years ago

Read 2 more answers

A 0.0140 kg bullet traveling at 205 m/s east hits a motionless 1.80 kg block and bounces off it, retracing its original path wit

makvit [3.9K]

Answer:

Final velocity of the block = 2.40 m/s east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Mass of bullet = 0.0140 kg

Consider east as positive.

Initial velocity of bullet = 205 m/s

Mass of Block = 1.8 kg

Initial velocity of block = 0 m/s

Initial momentum = 0.014 x 205 + 1.8 x 0 = 2.87 kg m/s

Final velocity of bullet = -103 m/s

We need to find final velocity of the block( u )

Final momentum = 0.014 x -103+ 1.8 x u = -1.442 + 1.8 u

We have

2.87 = -1.442 + 1.8 u

u = 2.40 m/s

Final velocity of the block = 2.40 m/s east.

7 0

2 years ago

A 1500 kg car traveling at 20 m/s suddenly runs out of gas while approaching the valley shown in the figure. The alert driver im

geniusboy [140]

Answer:

v_f = 17.4 m / s

Explanation:

For this exercise we can use conservation of energy

starting point. On the hill when running out of gas

Em₀ = K + U = ½ m v₀² + m g y₁

final point. Arriving at the gas station

Em_f = K + U = ½ m v_f ² + m g y₂

energy is conserved

Em₀ = Em_f

½ m v₀ ² + m g y₁ = ½ m v_f ² + m g y₂

v_f ² = v₀² + 2g (y₁ -y₂)

we calculate

v_f ² = 20² + 2 9.8 (10 -15)

v_f = √302

v_f = 17.4 m / s

8 0

2 years ago