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3 years ago

Please help I need help I can’t fail .

Physics

1 answer:

Evgen [1.6K]3 years ago

6 0

Answer:

Explanation:

Gas particles move fastest.

Plasma lacks fixed volume and shape.

Liquid lacks fixed shape, but has fixed volume.

Solid has both fixed volume and shape.

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A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t

mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 $\frac{m}{s^{2}}$

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

$K=U$

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}$

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}$

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=$

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}$

part 3

Acceleration can be find using Newton's second law:

$F=ma$

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

$-kx=ma$

$a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}$

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

$T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s$

So between 0 and 4.5 cm we have half a period:

$t=\frac{T}{2}=0.66s$

7 0

3 years ago

In the periodic table, similar elements are arranged in vertical columns called

tester [92]

Answer:

The vertical columns on the periodic table are called groups or families because of their similar chemical behavior.

Explanation:

3 0

3 years ago

Read 2 more answers

An athlete executing a long jump leaves the ground at a 28.0º angle and travels 7.80 m. (A) What was the takeoff speed?

Naddika [18.5K]

Answer:

9.6 m/s

Explanation:

Angle of projection, θ = 28°

Horizontal distance, R = 7.8 m

Let the velocity of projection is given by u.

The formula used to find the velocity of projection is given by

$R=\frac{u^{2}Sin2\theta }{g}$

$7.8=\frac{u^{2}Sin\left (2\times 28 \right ) }{9.8}$

$7.8=\frac{u^{2}\times 0.829}{9.8}$

u = 9.6 m/s

Thus, the velocity of projection is 9.6 m/s.

7 0

2 years ago

A soccer ball is kicked At 8m/s. It lands on the ground after being in the air for .85 seconds. At what angle is it kicked?

SpyIntel [72]

If ball remains in air for total time T = 0.85 s

this is also known as time of flight

In order to find the time of flight we can use kinematics

$\Delta Y = v_y*t + \frac{1}{2}at^2$

so for complete motion its displacement in y direction will be zero

$0 = v_y* 0.85 + \frac{1}{2}(-9.8)(0.85^2)$

$0 = v_y*0.85 - 3.54$

$v_y = 4.165 m/s$

now we know that net velocity of the ball is 8 m/s

while is y direction component we got is vy = 4.165 m/s

now by component method we can say

$v_y = v sin\theta$

$4.165 = 8 sin\theta$

$\theta = sin^{-1}\frac{4.165}{8}$

$\theta = 31.4^0$

so it is projected at an angle of 31.4 degree above horizontal

8 0

3 years ago

A particle with a charge of +4.20nC is in a uniform electric field E⃗ directed to the left. It is released from rest and moves t

Morgarella [4.7K]

Answer:

(A). The work done is $1.50\times10^{-6}\ J$ .

(B). The potential of the starting point with respect to the endpoint is 357.14 V.

(C). The magnitude of E is 5952.38 N/C.

Explanation:

Given that,

Charge = 4.20 nC

Distance = 6.00 cm

Kinetic energy $K.E=1.50\times10^{-6}\ J$

The particle start from rest.

So, the initial kinetic energy i zero.

(A). We need to calculate the work by the electric force

Using formula of work done

$W = \Delta K.E$

$W=K.E_{f}-K.E_{i}$

Put the value into the formula

$W= 1.50\times10^{-6}-0$

$W=1.50\times10^{-6}\ J$

The work done is $1.50\times10^{-6}\ J$ .

(B). We need to calculate the potential of the starting point with respect to the endpoint

We know that.

Change in potential energy = change in kinetic energy

$\Delta P.E=\Delta K.E$

So, $U = 1.50\times10^{-6}$

Using formula of potential

$V=\dfrac{U}{q}$

Put the value into the formula

$V=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}}$

$V=357.14\ V$

The potential of the starting point with respect to the endpoint is 357.14 V.

(C). We need to calculate the magnitude of E

Using formula of work done

$W=F\times r$ ....(I)

Using formula of force

$F=qE$

Put the value in the equation (I)

$W=qE\times r$

$E=\dfrac{W}{q\times r}$

Put the value into the formula

$E=\dfrac{1.50\times10^{-6}}{4.20\times10^{-9}\times6.00\times10^{-2}}$

$E=5952.38\ N/C$

The magnitude of E is 5952.38 N/C.

Hence, This is the required solution.

7 0

3 years ago