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3 years ago

12

If you answe my question ill get you 50 pionts The smallest piece of matter that still has the properties of an element is a(n)

1.compound
2.electron

3.atom

4.nucleus

1 answer:

timurjin [86]3 years ago

8 0

Answer:

I think its number 3, correct me if I'm wrong.

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A volleyball player bumps a ball across a net with the velocity and angle shown below. What is the maximum height of the ball?

Marrrta [24]

Answer:

D. 12.4 m

Explanation:

Given that,

The initial velocity of the ball, u = 18 m/s

The angle at which the ball is projected, θ = 60°

The maximum height of the ball is given by the formula

h = u² sin²θ/2g m

Where,

g - acceleration due to gravity. (9.8 m/s)

Substituting the values in the above equation

h = 18² · sin²60 / 2 x 9.8

= 18² x 0.75 / 2 x 9.8

= 12.4 m

Hence, the maximum height of the ball attained, h = 12.4 m

6 0

3 years ago

Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

$m_{T}=2384.5$

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u> $m_{1}$ <u>:</u>

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

<u>For </u> $m_{2}$ <u>:</u>

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

<u>For </u> $m_{3}$ <u>:</u>

$m_{3} = m_{T} - (m_{1}+m_{2})$

$m_{3} = 2384.5-1520.0$

$m_{3}$ = 864.5kg

8 0

3 years ago

Five hundred joules of heat are added to a closed system. The initial internal energy of the system is 87 J, and the final inter

Aleonysh [2.5K]

We can solve the problem by using the first law of thermodynamics:

$\Delta U= Q-W$

where

$\Delta U$ is the variation of internal energy of the system

Q is the heat added to the system

W is the work done by the system

In this problem, the variation of internal energy of the system is

$\Delta U=U_f-U_i=134 J-87 J=47 J$

While the heat added to the system is

$Q=500 J$

therefore, the work done by the system is

$W=Q-\Delta U=500 J-47 J=453 J$

5 0

3 years ago

Read 2 more answers

A physicist performing a sensitive measurement wants to limit the magnetic force on a moving charge in her equipment to less tha

lawyer [7]

Answer:

Maximum charge will be $6.66\times 10^{-9}C$

Explanation:

We have given force ion the moving charge $F=10^{-12}N$

Maximum speed of the moving charge v = 30 m /sec

Magnetic field $B=5\times 10^{-5}T$

We have to fond the charge

Force on moving charge is given by

$F=qvB$

So $10^{-12}=q\times 30\times 5\times 10^{-5}$

$q=6.66\times 10^{-9}C$

Maximum charge will be $6.66\times 10^{-9}C$

5 0

3 years ago

By what factor must we increase the amplitude of vibration of an object at the end of a spring in order to double its maximum sp

strojnjashka [21]

Answer:

$A'=2A$

Explanation:

According to the law of conservation of energy, the total energy of the system can be expresed as the sum of the potential energy and kinetic energy:

$E=U+K=\frac{kA^2}{2}\\E=\frac{kx^2}{2}+\frac{mv^2}{2}=\frac{kA^2}{2}$

When the spring is in its equilibrium position, that is $x=0$ , the object speed its maximum. So, we have:

$\frac{k(0)^2}{2}+\frac{mv_{max}^2}{2}=\frac{kA^2}{2}\\A^2=\frac{mv_{max}^2}{k}\\A=\sqrt{\frac{mv_{max}^2}{k}}$

In order to double its maximum speed, that is $v'{max}=2v_{max}$ . We have:

$A'=\sqrt{\frac{m(v'_{max})^2}{k}}\\A'=\sqrt{\frac{m(2v_{max})^2}{k}}\\A'=\sqrt{\frac{4mv_{max}^2}{k}}\\A'=2\sqrt{\frac{mv_{max}^2}{k}}\\A'=2A$

6 0

3 years ago