Using the periodic table, choose the more reactive nonmetal. Si or N

A car traveling at 27 m/s slams on its brakes to come to a stop. It decelerates at a rate of 8 m/s2 . What is the stopping dista

qaws [65]

v² - u² = 2 a ∆x

where u = initial velocity (27 m/s), v = final velocity (0), a = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆x = stopping distance.

So

0² - (27 m/s)² = 2 (-8 m/s²) ∆x

∆x = (27 m/s)² / (16 m/s²)

∆x ≈ 45.6 m

5 0

3 years ago

A straight line can represent a _____-dimensional coordinate system.

NISA [10]

A. 1 dimensional coordinate system

5 0

2 years ago

A small 12.00g plastic ball is suspended by a string in a uniform, horizontal electric field with a magnitude of 10^3 N/C. If th

tensa zangetsu [6.8K]

Answer:

The net charge is $67.89 \mu C$

Solution:

As per the question:

Mass of the plastic bag, m = 12.0 g = $12.0\times 10^{-3}\ kg$

Magnitude of electric field, E = $10^{3}\ N/C$

Angle made by the string, $\theta = 30^{\circ}$

Now,

To calculate the net charge, Q on the ball:

Vertical component of the tension in the string, $T = Tcos\theta$

Horizontal component of the tension in the string, $T = Tsin\theta$

Now,

Balancing the forces in the x-direction:

$Tsin\theta = QE$

$Q = {Tsin\theta}{E}$ (1)

Balancing the forces in the y-direction:

$Tcos\theta = mg$

where

g = acceleration due to gravity = $9.8\ m/s^{2}$

Thus

$T = \frac{mg}{cos\theta }$

$T = \frac{12.0\times 10^{-3}\times 9.8}{cos30^{\circ}} = 0.1357\ N$

Use T = 0.1357 N in eqn (1):

$Q = {0.1357\times sin30^{\circ}}{10^{3}} = 6.789\times 10^{- 5}\ C$

$Q = 67.89\times 10^{- 5}\ C = 67.89\mu C$

7 0

3 years ago

We often use a spoon to cool coffee by stirring. Some of the heat is conducted from the coffee to the air by this method, but th

Ilia_Sergeevich [38]

Answer:

Explanation:

heat lost by coffee = heat gain by aluminum spoon

heat lost by coffee = mc ( 85° C - T)

175 ml of water = 175 grams

175 grams = (175 / 1000) kg = 0.175 kg

heat loss by coffee = 0.175 × 4182 J/Kgk ( 85 - T) = 731.85 ( 85-T) = 62207.25 - 731.85 T

heat gain by aluminum spoon = 0.012kg × 897 J/kgK ( T - 22) = 10.764 T - 236.808

10.764 T - 236.808 = 62207.25 - 731.85 T

10.764 T + 731.85 T = 62207.25 + 236.808

T = 84.086 ° C approx 84°C

3 0

2 years ago

A 4.6-kg block of ice originally at 263 K is placed in thermal contact with a 15.7-kg block of silver (cAg = 233 J/kg-K) which i

Viktor [21]

Answer:

temperature at 326.44 K system achieve equilibrium

Explanation:

given data

mass of block of ice = 4.6 kg

temperature = 263 K

thermal contact = 15.7-kg

specific heat of silver cAg = 233 J/kg-K

initially temperature = 1052 K

to find out

what temperature will the system achieve equilibrium

solution

first we consider final temperature of the system is T

we know that specific heat of water (C w) = 4186 J(kg K)

and

specific heat of ice ( C i ) = 2030 J/(kg K)

and

latent heat of fusion of ice ( Lf ) = 3.33 × $10^{5}$ J/kg

and we know that system is insulated

so heat lost by silver = heat generated by ice .................1

so we can say

mass of silver × specific heat of silver × ( initial temp - final temp ) = mass of ice × specific heat of ice × ( ice temp ) + mass of ice × latent heat of fusion of ice + mass of ice × specific heat of water × (final temperature )

put here value we get

mass of silver × specific heat of silver × ( initial temp - final temp ) = mass of ice × specific heat of ice × ( ice temp ) + mass of ice × latent heat of fusion of ice + mass of ice × specific heat of water × (final temperature )

15.7 × 233 × ( 1052 - T ) = 4.6 × 2030 × 10 + 4.6 × 3.33 × $10^{5}$ + 4.6 × 4286 × ( T - 273 )

solve we get

T = 326.44 K

so temperature at 326.44 K system achieve equilibrium

7 0

2 years ago