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3 years ago

5

A crate is sliding down an inclined ramp at a constant speed of 0.55 m/s. The vector sum of all the forces acting on this crate

must point down the ramp

1 answer:

Lemur [1.5K]3 years ago

8 0

Answer:

$F_{net} = 0$

Explanation:

here it is given that the crate is sliding down with constant speed

so we will have

$v = constant$

now we can say

$a = \frac{dv}{dt}$

so here we have

a = 0

now from Newton's II law we can say that sum of all forces on the system is always product of mass and its acceleration

so here we will have

$F_{net} = ma$

$F_{net} = 0$

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consider two stars, star a and star b. star a has a temperature of 4900 k , and star b has a temperature of 9900 k . how many ti

ValentinkaMS [17]

The Energy flux from Star B is 16 times of the energy flux from Star A.

We have Two stars - A and B with 4900 k and 9900 k surface temperatures.

We have to determine how many times larger is the energy flux from Star B compared to the energy flux from Star A.

<h3>State Stephen's Law?</h3>

Stephens law states that if E is the energy radiated away from the star in the form of electromagnetic radiation, T is the surface temperature of the star, and σ is a constant known as the Stephan-Boltzmann constant then-

$$\frac{Energy}{Area} = \sigma\times T^{4}$

Now -

Energy emitted per unit surface area of Star is called Energy flux. Let us denote it by E. Then -

$$E= \sigma\times T^{4}$

Now -

For Star A →

$T_{A}$ = 4900 K

For Star B →

$T_{B}$ = 9900 K

Therefore -

$$\frac{T_{B} }{T_{A} } =\frac{9900}{4900}$

$\frac{T_{B} }{T_{A} }=$ 2.02 = 2 (Approx.)

Now -

Assume that the energy flux of Star A is E(A) and that of Star B is E(B). Then -

$$\frac{E(B)}{E(A)} = \frac{\sigma\times T(B)^{4} }{\sigma \times T(A)^{2} }$

E(B) = E(A) x $(\frac{T(B)}{T(A)} )^{4}$

E(B) = E(A) x $2^{4}$

E(B) = 16 E(A)

Hence, the Energy flux from Star B is 16 times of the energy flux from Star A.

To learn more about Stars, visit the link below-

brainly.com/question/13451162

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2 years ago

Careers in this career cluster also include related professional and technical support activities such as production planning an

qwelly [4]

True, all of those careers go in the same category

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3 years ago

until a train is a safe distance from the station, it must travel at 5 m/s. once the train is on open track, it can speed up to

MatroZZZ [7]

Answer:

5 meters per second squared

Explanation:

We calculate the acceleration using the formula:

a = (vf - vi) / t

where "vf" is the final velocity, "vi" the initial velocity, and "t" the time it took to change from the initial velocity to the final one.

In our case:

a = (45 - 5) / 8 = 40 / 8 = 5 m/s^2

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4 years ago

What would the coefficient of friction be if the frictional force was equal to the normal force? (Answer will be an number)

e-lub [12.9K]

That coefficient is 1.00 .

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3 years ago

An air conditioner runs 15 minutes each hour on a hot summer day. It is on a 240 volt circuit and uses 21 amps. Rate is $.10/kWh

Helen [10]

Answer:

Approximately $\$ 3.02$ .

Explanation:

Note that the electric rate in this question is in the unit dollar-per- ${\rm kWh}$ , where $1\; {\rm kWh}$ is the energy to run an appliance of power $1\; {\rm kW}$ for an hour.

Number of minutes for which the air conditioner is running in that day: $15 \times 24 = 360\; \text{minute}$ . Apply unit conversion and ensure that this time is measured in hours (same as the unit of the electric rate.)

$\begin{aligned} \text{time} &= 360\; \text{minute} \times \frac{1\; \text{hour}}{60\; \text{minute}} = 6\; \text{hour} \end{aligned}$ .

The power of this air conditioner is:

$\begin{aligned} \text{power} &= \text{voltage} \times \text{current} \\ &= 240\; {\rm V} \times 21\; {\rm A} \\ &= 5040\; {\rm W} \\ &= 5.04\; {\rm kW} \end{aligned}$ .

Thus, the energy that this air conditioner would consume would be:

$\begin{aligned}\text{energy} &= \text{power} \times \text{time} \\ &= 5.04\; {\rm kW} \times 6\; \text{hour} \\ &= 30.24\; {\rm kWh} \end{aligned}$ .

At a rate of $0.1$ dollar-per- ${\rm kWh}$ , the cost of that much energy would be approximately $3.02$ dollars (rounded to the nearest cent.)

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2 years ago