all of the following answer choices are ideas from daltons atomic theory. which one of them do we now know is not true?

A transformer's secondary coil has twice as many turns as its primary. If the primary is connected to 6 V of DC, how many volts

Aleksandr-060686 [28]

Zero is induced in the secondary, and the total output of the transformer consists of smoke.
Transformers only work with AC.

6 0

4 years ago

a stone is thrown horizonttaly from a cliff of a hill with an initial velocity of 30m/s it hits the ground at a horizontal dista

ELEN [110]

Answer:

a) Time = 2.67 s

b) Height = 35.0 m

Explanation:

a) The time of flight can be found using the following equation:

$x_{f} = x_{0} + v_{0_{x}}t + \frac{1}{2}at^{2}$ (1)

Where:

$x_{f}$ : is the final position in the horizontal direction = 80 m

$x_{0}$ : is the initial position in the horizontal direction = 0

$v_{0_{x}}$ : is the initial velocity in the horizontal direction = 30 m/s

a: is the acceleration in the horizontal direction = 0 (the stone is only accelerated by gravity)

t: is the time =?

By entering the above values into equation (1) and solving for "t", we can find the time of flight of the stone:

$t = \frac{x_{f}}{v_{0}} = \frac{80 m}{30 m/s} = 2.67 s$

b) The height of the hill is given by:

$y_{f} = y_{0} + v_{0_{y}}t - \frac{1}{2}gt^{2}$

Where:

$y_{f}$ : is the final position in the vertical direction = 0

$y_{0}$ : is the initial position in the vertical direction =?

$v_{0_{y}}$ : is the initial velocity in the vertical direction =0 (the stone is thrown horizontally)

g: is the acceleration due to gravity = 9.81 m/s²

Hence, the height of the hill is:

$y_{0} = \frac{1}{2}gt^{2} = \frac{1}{2}9.81 m/s^{2}*(2.67 s)^{2} = 35.0 m$

I hope it helps you!

5 0

3 years ago

An aluminum calorimeter with a mass of 100 g contains 250 g of water. The calorimeter and water are in thermal equilibrium at 10

Alexeev081 [22]

Answer:

a) $c=1822.3214\ J.kg^{-1}.K^{-1}$

b) This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c) The material is peat, possibly.

d) The material cannot be ice because ice doesn't exists at a temperature of 100°C.

Explanation:

Given:

mass of aluminium, $m_a=0.1\ kg$

mass of water, $m_w=0.25\ kg$

initial temperature of the system, $T_i=10^{\circ}C$

mass of copper block, $m_c=0.1\ kg$

temperature of copper block, $T_c=50^{\circ}C$

mass of the other block, $m=0.07\ kg$

temperature of the other block, $T=100^{\circ}C$

final equilibrium temperature, $T_f=20^{\circ}C$

We have,

specific heat of aluminium, $c_a=910\ J.kg^{-1}.K^{-1}$

specific heat of copper, $c_c=390\ J.kg^{-1}.K^{-1}$

specific heat of water, $c_w=4186\ J.kg^{-1}.K^{-1}$

Using the heat energy conservation equation.

The heat absorbed by the system of the calorie-meter to reach the final temperature.

$Q_{in}=m_a.c_a.(T_f-T_i)+m_w.c_w.(T_f-T_i)$

$Q_{in}=0.1\times 910\times (20-10)+0.25\times 4186\times (20-10)$

$Q_{in}=11375\ J$

The heat released by the blocks when dipped into water:

$Q_{out}=m_c.c_c.(T_c-T_f)+m.c.(T-T_f)$

where

$c=$ specific heat of the unknown material

For the conservation of energy : $Q_{in}=Q_{out}$

so,

$11375=0.1\times 390\times (50-20)+0.07\times c\times (100-20)$

$c=1822.3214\ J.kg^{-1}.K^{-1}$

b)

This value of specific heat is close to the specific heat of ice at -40° C and the specific heat of peat (a variety of coal).

c)

The material is peat, possibly.

d)

The material cannot be ice because ice doesn't exists at a temperature of 100°C.

7 0

3 years ago

Identify two types of simple machines in this compund machine.

sleet_krkn [62]

the answer would be "screw and lever!", thats the answer because the middle is made into a rod of metal rapped in metal to screw into the cork, and the lever part is the two handles on the side to lift it out of the bottle.

4 0

3 years ago

Read 2 more answers

Two charges, Q1 and Q2, are separated by 6·cm. The repulsive force between them is 25·N. In each case below, find the force betw

Misha Larkins [42]

Answer:

a) 5 N b) 225 N c) 5 N

Explanation:

a) Per Coulomb's Law the repulsive force between 2 equal sign charges, is directly proportional to the product of the charges, and inversely proportional to the square of the distance between them, acting along the line that joins the charges, as follows:

F₁₂ = K Q₁ Q₂ / r₁₂²

So, if we make Q1 = Q1/5, the net effect will be to reduce the force in the same factor, i.e. F₁₂ = 25 N / 5 = 5 N

b) If we reduce the distance, from r, to r/3, as the factor is squared, the net effect will be to increase the force in a factor equal to 3² = 9.

So, we will have F₁₂ = 9. 25 N = 225 N

c) If we make Q2 = 5Q2, the force would be increased 5 times, but if at the same , we increase the distance 5 times, as the factor is squared, the net factor will be 5/25 = 1/5, so we will have:

F₁₂ = 25 N .1/5 = 5 N

3 0

3 years ago