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4 years ago

To move a heavy couch across a carpeted floor what could be used to decrease the frictional force

Physics

2 answers:

vitfil [10]4 years ago

7 0

Use a piece of cardboard

aleksklad [387]4 years ago

3 0

Put a lubricant between the surface of the object and the floor. use round objects, like pencils, to decrease the friction and push the refrigerator over the pencils

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glass shattering in to pieces bubblegum being stretched a piece of metal expanding due to heat These are all examples of Questio

kogti [31]

I think it’s physical change, because the objects aren’t changing into another substance, just changing their physical shape. Hope this helps!

4 0

3 years ago

Read 2 more answers

The charge entering the positive terminal of an element is q = 5 sin 4πt mC while the voltage across the element (plus to minus)

Artemon [7]

Answer:

(a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

Explanation:

Given that,

Charge $q=5\sin4\pi t\ mC$

Voltage $v=3\cos4\pi t\ V$

Time t = 0.3 sec

We need to calculate the current

Using formula of current

$i(t)=\dfrac{dq}{dt}$

Put the value of charge

$i(t)=\dfrac{d}{dt}(5\sin4\pi t)$

$i(t)=5\times4\pi\cos4\pi t$

$i(t)=20\pi\cos4\pi t$

(a).We need to calculate the power delivered to the element

Using formula of power

$p(t)=v(t)\times i(t)$

Put the value into the formula

$p(t)=3\cos4\pi t\times20\pi\cos4\pi t$

$p(t)=60\pi\times10^{-3}\cos^2(4\pi t)$

$p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi t}{2})$

Put the value of t

$p(t)=60\pi\times10^{-3}(\dfrac{1+\cos8\pi\times0.3}{2})$

$p(t)=30\pi\times10^{-3}(1+\cos8\pi \times0.3)$

$p(t)=187.68\ mW$

(b). We need to calculate the energy delivered to the element between 0 and 0.6 s

Using formula of energy

$E(t)=\int_{0}^{t}{p(t)dt}$

Put the value into the formula

$E(t)=\int_{0}^{0.6}{30\pi\times10^{-3}(1+\cos8\pi \times t)}$

$E(t)=30\pi\times10^{-3}\int_{0}^{0.6}{1+\cos8\pi \times t}$

$E(t)=30\pi\times10^{-3}(t+\dfrac{\sin8\pi t}{8\pi})_{0}^{0.6}$

$E(t)=30\pi\times10^{-3}(0.6+\dfrac{\sin8\pi\times0.6}{8\pi}-0-0)$

$E(t)=57.52\ mJ$

Hence, (a). The power delivered to the element is 187.68 mW

(b). The energy delivered to the element is 57.52 mJ.

6 0

3 years ago

A 3.62 g bullet moving at 270 m/s enters and stops in an initially stationary 2.30 kg wooden block on a horizontal frictionless

scZoUnD [109]

Momentum before collision must be equal to momentum after collision

(m1 + m2)v = m1v1 + m2v2

3.62g*270 m/s=2.30kg* (x)
convert the units to be the same that is convert the kg mass to grams
1kg=1000g
2.30kg=y
230/100*1000=2300g
3.62*270=2300X
977.4g/m/s=2300X
977.4/2300=0.4250M/S
Finding a velocity after the gun is embedded on the block of wood
3.62*270+2300g*0.425=(3.62+2300)*V
977.5+977.5=2303V
1955=2303V
V=1955/2303
V=0.8489M/S

5 0

3 years ago

2. A gas with a constant pressure of 270 kPa does 36,000 J of work as it expands.

liberstina [14]

C sorry if wrong ,,,

5 0

3 years ago

Two carts collide on a low-friction track. Cart 1 has an initial momentum of +10 kg⋅m/sι^ and a final momentum of -5.0 kg⋅m/sι^

Basile [38]

According to the saving of the momentum, the total momentum before collision is equal to the total momentum after collision

in symbols:

$P_{i1}+P_{i2}=P_{f1}+P_{f2}$