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3 years ago

Most metals are in the liquid state at room temperature.

Physics

2 answers:

Veseljchak [2.6K]3 years ago

6 0

Answer:

The answer to this question is FALSE

aivan3 [116]3 years ago

4 0

Answer:

Most metals are in the liquid state at room temperature.<u>False</u><u>.</u>

Explanation:

it needs excess heat except mercury and lithium.

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When a falling meteoroid is at a distance above the Earth's surface of 3.40 times the Earth's radius, what is its acceleration d

Alchen [17]

Answer:

g = 0.85 m $s^{-2}$

Explanation:

g = $\frac{GM}{h^{2} }$

were; g is the acceleration due to Earth's gravity, G is Newton's gravitation constant (6.674 x $10^{-11}$ N $m^{2}$ $kg^{-2}$ ), M is the mass of the earth (5.972 x $10^{24}$ kg), and h is the distance of meteoroid to the earth.

h = 3.40 x R

= 3.40 x 6371 km

h = 21661.4 km

= 21661400 m

Thus,

g = $\frac{6.674*10^{-11}*5.972*10^{24} }{(21661400)^{2} }$

= $\frac{3.9857 *10^{14} }{4.6922*10^{14} }$

= 0.84944

g = 0.85 m $s^{-2}$

The acceleration due to the Earth's gravitation is 0.85 m $s^{-2}$ .

6 0

3 years ago

Lexi is balancing equations. She is finding one equation to be very difficult to balance. Which explains how to balance the equa

Rina8888 [55]

Answer: ZnSO4 + Li2CO3 = ZnCO3 + Li2SO4 - Chemical Equation Balancer

Equation is already balanced.

Explanation: ZnSO4 + Li2CO3 = ZnCO3 + Li2SO4

6 0

3 years ago

6. A car, 1110 kg, is traveling down a horizontal road at 20.0 m/s when it locks up its brakes. The coefficient of friction betw

USPshnik [31]

Answer:

$x=22.65m$

Explanation:

We have an uniformly accelerated motion, with a negative acceleration. Thus, we use the kinematic equations to calculate the distance will it take to bring the car to a stop:

$v_f^2=v_0^2-2ax\\\frac{0^2-v_0^2}{-2a}=x\\x=\frac{v_0^2}{2a}$

The acceleration can be calculated using Newton's second law:

$\sum F_x:F_f=ma\\\sum F_y:N=mg$

Recall that the maximum force of friction is defined as $F_f=\mu N$ . So, replacing this:

$\mu N=ma\\\mu mg=ma\\a=\mu g\\a=0.901(9.8\frac{m}{s^2})\\a=8.83\frac{m}{s^2}$

Now, we calculate the distance:

$x=\frac{v_0^2}{2a}\\x=\frac{(20\frac{m}{s})^2}{2(8.83\frac{m}{s^2})}\\x=22.65m$

7 0

3 years ago

A 0.30-kg object connected to a light spring with a force constant of 22.6 N/m oscillates on a frictionless horizontal surface.

gtnhenbr [62]

Answer:

(a) vmax = 0.34m/s

(b) v = 0.13m/s

(d) x = 0.039m

Explanation:

Given information about the spring-mass system:

m: mass of the object = 0.30kg

k: spring constant = 22.6 N/m

A: amplitude of the motion = 4.0cm = 0.04m

(a) The maximum speed of the object is given by the following formula:

$v_{max}=\omega A$ (1)

w: angular frequency of the motion.

The angular frequency is calculated with the following relation:

$\omega=\sqrt{\frac{k}{m}}$ (2)

You replace the expression (2) into the equation (1) and replace the values of the parameters:

$v_{max}=\sqrt{\frac{k}{m}}A=\sqrt{\frac{22.6N/m}{0.30kg}}(0.04m)=0.34\frac{m}{s}$

The maximum speed of the object is 0.34 m/s

(b) If the object is compressed 1.5cm the amplitude of its motion is A = 0.015m, and the maximum speed is:

$v_{max}=\sqrt{\frac{22.6N/m}{0.30kg}}(0.015m)=0.13\frac{m}{s}$

The speed is 0.13m/s

(c) To find the speed of the object when it passes the point x=1.5cm, you first take into account the equation of motion:

$x=Acos(\omega t)$

You solve the previous equation for t:

$t=\frac{1}{\omega}cos^{-1}(\frac{x}{A})\\\\\omega=\sqrt{\frac{22.6N/m}{0.30kg}}=8.67\frac{rad}{s}\\\\t=\frac{1}{8.67}cos^{-1}(\frac{1.5cm}{4.0cm})=0.13s$

With this value of t, you can calculate the speed of the object with the following formula:

$v=\omega Asin(\omega t)\\\\v=(8.67rad/s)(0.04m)sin((8.67rad/s)(0.13s))=0.31\frac{m}{s}$

The speed of the object for x = 1.5cm is v = 0.31 m/s

(d) To calculate the values of x on which v is one-half the maximum speed, you first calculate the time t:

$\frac{v_{max}}{2}=\omega A sin(\omega t)\\\\t=\frac{1}{\omega}sin^{-1}(\frac{v_{max}}{2\omega A})\\\\t=\frac{1}{8.67rad/s}sin^{-1}(\frac{0.13m/s}{2(8.67rad/s)(0.04m)})=0.021s$

The position will be:

$x=Acos(\omega t)=0.04mcos((8.67rad/s)(0.021s))=0.039m$

The position of the object on which its speed is one-half its maximum velocity is 0.039

5 0

4 years ago

When making a budget, you should prioritize ___________ before "__________." A. Needs/wants B. Wants/needs C. Electricity/Water

Blizzard [7]

Answer A

Explanation: because u need to make sure you know the difference between a need and a want

7 0

3 years ago