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3 years ago

If you wanted to produce more power,what could you do to maximize power

2 answers:

6 0

In physics, power is defined as energy per unit time. You will also hear it described as work per unit time. The standard unit of measure for power is the watt, where a watt is defined as joules (energy) per second (time). This is expressed as a fraction as J/s. If you wanted to increase the power in any operation, you can either increase the energy (more joules) or reduce the time (fewer seconds).

Vikentia [17]3 years ago

6 0

Same work in less time, or more work in same time

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A heated piece of metal cools according to the function c(x) = (.5)^(x _ 11), where x is measured in hours. A device is added th

Agata [3.3K]

You just need to replace x with 5 in each function

.5^5 - 11
-5-3

.5 ^-6
-8

64 - 8 = 56 A Celcius

Hope this helps

3 0

2 years ago

Read 2 more answers

Electron kinetic energies are often measured in units of electron-volts (1 eV 1.6 x 10-19 J), which is the kinetic energy of an

PolarNik [594]

Answer:

4.1 eV

Explanation:

Kinetic energy, K = 0.8 eV = 0.8 x 1.6 x 10^-19 J = 1.28 x 10^-19 J

wavelength, λ = 253.5 nm = 253.5 x 10^-9 m

According to the Einstein energy equation

$E = W_{o}+K$

Where, E be the energy incident, Wo is the work function and K is the kinetic energy.

h = 6.634 x 10^-34 Js

c = 3 x 10^8 m/s

$E=\frac{hc}{\lambda }=\frac{6.634 \times 10^{-34} \times 3 \times 10^{8}}{253.5\times 10^{-9}}=7.85 \times 10^{-19} J$

So, the work function, Wo = E - K

Wo = 7.85 x 10^-19 - 1.28 x 10^-19

Wo = 6.57 x 10^-19 J

Wo = 4.1 eV

Thus, the work function of the metal is 4.1 eV.

5 0

2 years ago

Two students walk in the same direction along a straight path at a constant speed. One walks at a speed of 0.90 m/s and the othe

mr Goodwill [35]

Answer:

A. 456 seconds

Explanation:

We are given that two students walk in the same direction along a straight path at a constant speed.

One student walks with a speed=0.90 m/s

second student walks with speed=1.9 m/s

Total distance covered by each students=780 meter

We have to find who is faster and how much time extra taken by slower student than the faster student.

Time taken by one student who travel with speed 0.90 m/s= $\frac{780}{0.90}$

Time= $\frac{distance}{speed}$

Time taken by one student who travel with speed 0.90 m/s

= $\frac{780}{0.90}$

Time taken by one student who travel with speed 0.90 m/s

=866.6 seconds

Time taken by second student who travel with speed 1.9 m/s= $\frac{780}{1.9}$

=410.5 seconds

The second student who travels with speed 1.9 m/s is faster than the student travels with speed 0.90 m/s .

Extra time taken by the student travels with speed 0.90 m/s=866.6-410.5=456.1 seconds

Extra time taken by the student travels with speed 0.90 m/s=456 seconds

Hence, option A is true.

7 0

2 years ago

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The psychologist known for latent learning and cognitive maps is _________. A. Robert Rescorla B. Edward Tolman C. William James

Lubov Fominskaja [6]

Answer:

Explanation:

B. Edward Tolman

4 0

3 years ago

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3) Calculate the kinetic energy of a 7 kg mass traveling at a velocity of 4 m/sec.

murzikaleks [220]

$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

Let's solve ~

Given terms :

Mass (m) = 7 kg

velocity (v)= 4 m/s

The formula to find kinetic Energy is ~

$\boxed{ \boxed{ \sf{ \frac{1}{2} m{v}^{2} }}}$

Now, apply the formula according to given situation

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \: \dfrac{1}{2} \times 7 \times ( {4)}^{2}$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \: \dfrac{1}{2} \times 7 \times 16$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:7 \times 8$

${ \qquad{ \sf{ \dashrightarrow}}} \: \: \sf \:56 \: \: joules$

Therefore, the kinetic Energy of the car is 56 joules

4 0

1 year ago

Read 2 more answers