Ask question

Ask question

All categories

3 years ago

5

Read each scenario and then answer the question. Scenario A: 120 J of work is done in 6 s. Scenario B: 160 J of work is done in

8 s. Scenario C: 200 J of work is done in 10 s. Which scenario uses the most power?.

2 answers:

Alexeev081 [22]3 years ago

4 0

The power in every scenario is the same. Power is the measurement of the energy transfer over time.

<h3>What is power?</h3>

Power is the measurement of the energy transfer over time. It can be calculated by the formula,

$P = \dfrac Wt$

Where,

$P$ - power

$W$ - work

$t$ - time

In scenario A,

$P = \dfrac {120 }6 = 20 \rm \ W$

In scenario B,

$P = \dfrac {160 }8 = 20 \rm \ W$

In scenario B,

$P = \dfrac {200 }{10} = 20 \rm \ W$

Therefore, the power in every scenario is the same.

Learn more about Power:

brainly.com/question/1618040

Andrej [43]3 years ago

3 0

Answer:

Read each scenario and then answer the question.

Scenario A: 120 J of work is done in 6 s.

Scenario B: 160 J of work is done in 8 s.

Scenario C: 200 J of work is done in 10 s.

You might be interested in

A 56 kg astronaut stands on a bathroom scale inside a rotating circular space station. The radius of the space station is 250 m.

Zielflug [23.3K]

Answer:

The speed of space station floor is 49.49 m/s.

Explanation:

Given that,

Mass of astronaut = 56 kg

Radius = 250 m

We need to calculate the speed of space station floor

Using centripetal force and newton's second law

$F=mg$

$\dfrac{mv^2}{r}=mg$

$\dfrac{v^2}{r}=g$

$v=\sqrt{rg}$

Where, v = speed of space station floor

r = radius

g = acceleration due to gravity

Put the value into the formula

$v=\sqrt{250\times9.8}$

$v=49.49\ m/s$

Hence, The speed of space station floor is 49.49 m/s.

6 0

3 years ago

Four objects are situated along the y axis as follows: a 1.99-kg object is at 2.99 m, a 2.96-kg object is at 2.57 m, a 2.43-kg o

Dominik [7]

Answer:

The center of mass for the object is $y_c = 1.063 \ m$ from the origin

Explanation:

From the question we are told that

The mass of the first object is $m_1 = 1.99 \ kg$

The position of first object with respect to origin $y_1 = 2.99 \ m$

The mass of the second object is $m_2 = 2.96 \ kg$

The position of second object with respect to origin $y_2 = 2.57 \ m$

The mass of the third object is $m_3 = 2.43 \ kg$

The position of third object with respect to origin $y_3 = 0 \ m$

The mass of the fourth object is $m_3 = 3.96 \ kg$

The position of fourth object with respect to origin $y_3 = -0.502 \ m$

Generally the center of mass of the object along the x-axis is zero because all the mass lie on the y axis

Generally the location of the center mass of the object is mathematically represented as

$y_c = \frac{m_1 * y_1 + m_2 * y_2 + m_3 * y_3 + m_4 * y_4}{m_1 + m_2 + m_3 + m_4}$

=> $y_c = \frac{1.99 * 2.99 + 2.96 * 2.57 + 2.43 * 0 + 3.96 * (-0.502)}{1.99+ 2.96 + 2.43 + 3.96}$

=> $y_c = 1.063 \ m$

3 0

3 years ago

What is the mass of an object that is accelerated at 25 m/s2 by a force of 135 N?

yKpoI14uk [10]

Answer:

<h3>The answer is 5.4 kg</h3>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{135}{25} = \frac{27}{5} \\$

We have the final answer as

<h3>5.4 kg</h3>

Hope this helps you

4 0

3 years ago

A chef places an open sack of flour on a kitchen scale. The scale reading of

Novay_Z [31]

Answer:

Explanation:

Given

Initial reading on scale =40 N

So, we can conclude that weight of the sack is 40 N

After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)

This net downward force is the resultant of earth graviational pull and the applied upward force.

So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.

This situation can be diagramatically represented by figure given below

4 0

4 years ago

Two parallel-plate capacitors have the same plate area, but the plate gap in capacitor 1 is twice as big as capacitor 2. If capa

-BARSIC- [3]

Answer:

Capacitance of the second capacitor = 2C

Explanation:

$\texttt{Capacitance, C}=\frac{\varepsilon_0A}{d}$

Where A is the area, d is the gap between plates and ε₀ is the dielectric constant.

Let C₁ be the capacitance of first capacitor with area A₁ and gap between plates d₁.

We have

$\texttt{Capacitance, C}_1=\frac{\varepsilon_0A_1}{d_1}=C$

Similarly for capacitor 2

$\texttt{Capacitance, C}_2=\frac{\varepsilon_0A_2}{d_2}=\frac{\varepsilon_0A_1}{\frac{d_1}{2}}=2\times \frac{\varepsilon_0A_1}{d_1}=2C$

Capacitance of the second capacitor = 2C

6 0

3 years ago