True or False: An electric circuit has to be complete for the load to receive electricity.

An object moves with a positive acceleration. Could the object be moving with increasing speed, decreasing speed or constant spe

Masja [62]

Answer:

Increasing speed.

Explanation:

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

In this scenario, an object moves with a positive acceleration. Thus, the object is moving with an increasing speed and as such it has acceleration in the same direction as its velocity with respect to time.

3 0

3 years ago

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

initial temperature of metals, = $T_m$
initial temperature of water, = $T_i$
specific heat capacity of copper, $C_p$ = 0.385 J/g.K
specific heat capacity of aluminum, $C_A$ = 0.9 J/g.K
both metals have equal mass = m

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

The change in heat energy for copper metal;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

For aluminum metal, the quantity of heat transferred is calculated as;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

The change in heat energy for aluminum metal ;

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0

3 years ago

Space satellites in the inner solar system (such as those in orbit around Earth) are often powered by solar panels. Satellites t

erastova [34]

Answer:

D

Explanation:

See attached file

7 0

4 years ago

Urgent help for a physics exercise!

Zigmanuir [339]

a)

Amplitude of wave is given as maximum displacement from mean position

So here amplitude is 1.25 cm

b)

Wavelength is the length of the wave that it travels in one time period

From graph we can say the wavelength is given as 3cm

PART C)

Time period of wave is the time after which it repeats its shape

Speed of the wave = 21 cm/s

time period = wavelength / speed

$T = \frac{3}{21}$

$T = \frac{1}{7} s$

Now frequency is

$f = \frac{1}{T}$

$f = 7 Hz$

PART D)

Time period = \frac{1}{f}[/tex]

$T = 0.143 s$

6 0

4 years ago

Consider a room that is initially at the outdoor temperature of 208C. The room contains a 40-W lightbulb, a 110-W TV set, a 300-

Lostsunrise [7]

Answer:

The rate of increase of the energy content of the room when all the eletrical devices are on is 1650 W.

Explanation:

From conservation of energy,

Rate of energy transfer ( $\dfrac{dE}{dt}$ ) = Rate of change of energy within the room ( $E_{in} - E_{out}$ ).

As according to the problem no heat transfer occurs through the walls of the room we can assume that there is no energy transferred to the outside of the room, i.e., $E_{out} = 0$ .

If all the electric devices are on inside the room, then the rate of increase of energy is equal to the power ( $P_{in}$ ) consumed bt the electrical devices inside the room. Therefore we can write,

$\dfrac{dE}{dt} = \dfrac{dE_{in}}{dt} = P_{in} = P_{bulb} + P_{TV} + P_{refrigerator} + P_{iron} = (40 + 110 + 300 + 1200)~W = 1650~W$

7 0

3 years ago