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3 years ago

Whích phase change is the result of particles moving more quickly? (2 points)

Physics

1 answer:

sdas [7]3 years ago

8 0

Answer:

liquid to gas

Explanation:

let me know if this helps

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When Sweden is playing Denmark, it is SWE-DEN. The remaining letters, not used, is DEN-MARK

iren [92.7K]

Yes it is! It would be SWE vs. DEN

4 0

4 years ago

1.60 kg frictionless block is attached to an ideal spring with force constant 315 N/m . Initially the spring is neither stretche

Tatiana [17]

Answer:

(a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

Explanation:

Given that,

Mass of block =1.60 kg

Force constant = 315 N/m

Speed = 13.0 m/s

(a). We need to calculate the amplitude of the motion

Using conservation of energy

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2$

$A^2=\dfrac{mv^2}{k}$

Put the value into the relation

$A^2=\dfrac{1.60\times13.0^2}{315}$

$A=\sqrt{0.858}$

$A=0.926\ m$

(b). We need to calculate the block’s maximum acceleration

Using formula of acceleration

$a=A\omega^2$

$a=A\times\dfrac{k}{m}$

Put the value into the formula

$a=0.926\times\dfrac{315}{1.60}$

$a=182.31\ m/s^2$

(c). We need to calculate the maximum force the spring exerts on the block

Using formula of force

$F=ma$

Put the value into the formula

$F= 1.60\times182.31$

$F=291.69\ N$

Hence, (a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

7 0

3 years ago

How much is 100 kg in Newtons?

tatyana61 [14]

Answer:980.67

Explanation:

8 0

3 years ago

g If you keep the launch angle fixed, but double the initial launch speed, what happens to the range?

kiruha [24]

Answer:

Range will become 4 times of initial range

Explanation:

Let the velocity of projection is u

And angle at which projectile is projected is $\Theta$

And acceleration due to gravity is $g\ m/sec^2$

So range of projectile is equal to $R=\frac{u^2sin2\Theta }{g}$ ........eqn 1

Now in second case it is given that velocity of launching is doubled

So new velocity $u_{new}=2u$

So new range will be equal to $R_{new}=\frac{(2u)^2sin2\Theta }{g}=\frac{4u^2sin2\Theta }{g}$ .....eqn 2

Now dividing eqn 2 by eqn 1

$\frac{R_{new}}{R}=\frac{4u^2sin2\Theta }{g}\times \frac{g}{u^2sin2\Theta }$

$R_{new}=4R$

So if we double the initial launch speed then range will become 4 times

4 0

4 years ago

Calculate the GPE of a rock with a mass of 55kg on a cliff that is 27m high

GaryK [48]

Answer:

P=mgh

P=mghm=55

P=mghm=55g=9.8 or ~10

P=mghm=55g=9.8 or ~10h=27

Explanation:

hope it will help you have a great day bye and Mark brainlist if the answer is correct

 $kai6417$ 

 #carry on learning

4 0

2 years ago