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3 years ago

Help!!! If anyone could do one of these, i'm confused on how I should write the equation down.

Physics

1 answer:

balu736 [363]3 years ago

4 0

F=ma=m(change in velocity/change in time)

Number 1

F=ma

F=55kg(1.1ms^-1/1.6s)=37.8N

Number 2

F=ma

F=0.440kg(10ms^-1/0.02s)=220N

Number 3

F=ma

F=1400kg(15ms^-1/0.73s)=2.88*10^3N or 28,767N

Any questions please feel free to ask.

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A fatigue test was conducted in which the mean stress was 50 MPa (7250 psi) and the stress amplitude was 225 MPa (32,625 psi).

Tems11 [23]

Answer:

275 MPa, -175 MPa

-0.63636

450 MPa

Explanation:

$\sigma_{max}$ = Maximum stress

$\sigma_{min}$ = Minimum stress

$\sigma_m$ = Mean stress = 50 MPa

$\sigma_a$ = Stress amplitude = 225 MPa

Mean stress is given by

$\sigma_m=\frac{\sigma_{max}+\sigma_{min}}{2}\\\Rightarrow \sigma_{max}+\sigma_{min}=2\sigma_m\\\Rightarrow \sigma_{max}+\sigma_{min}=2\times 50\\\Rightarrow \sigma_{max}+\sigma_{min}=100\ MPa\\\Rightarrow \sigma_{max}=100-\sigma_{min}$

Stress amplitude is given by

$\sigma_a=\frac{\sigma_{max}-\sigma_{min}}{2}\\\Rightarrow \sigma_{max}-\sigma_{min}=2\sigma_a\\\Rightarrow \sigma_{max}-\sigma_{min}=2\times 225\\\Rightarrow \sigma_{max}-\sigma_{min}=450\ MPa\\\Rightarrow 100-\sigma_{min}-\sigma_{min}=450\\\Rightarrow -2\sigma_{min}=350\\\Rightarrow \sigma_{min}=-175\ MPa$

$\sigma_{max}=100-\sigma_{min}\\\Rightarrow \sigma_{max}=100-(-175)\\\Rightarrow \sigma_{max}=275\ MPa$

Maximum stress level is 275 MPa

Minimum stress level is -175 MPa

Stress ratio is given by

$R=\frac{\sigma_{min}}{\sigma_{max}}\\\Rightarrow R=\frac{-175}{275}\\\Rightarrow R=-0.63636$

The stress ratio is -0.63636

Stress range is given by

$\sigma_{max}-\sigma_{min}=450\ MPa$

Magnitude of the stress range is 450 MPa

8 0

3 years ago

I’m still lost on this please help me on this I know it’s not D I got it wrong

RSB [31]

Before you even look at the questions, look over the graph, so you know what kind of information is there.

The x-axis is "time". OK. You know that as the graph moves from left to right, it shows what's happening as time goes on.

The y-axis is "speed" of something. OK. When the graph is high, the thing is moving fast. When the graph is low, the thing is moving slow. When the graph slopes up, the thing is gaining speed. When the graph slopes down, the thing is slowing down. When the graph is flat, the speed isn't changing, so the thing is moving at a constant speed.

NOW you can look at the questions.

OMG ! It's only ONE question: What's happening from 'c' to 'd' ? Well I don't know. Perhaps we can figure it out if we LOOK AT THE GRAPH !

-- Between c and d, the graph is flat. The speed is not changing. It's the same speed at d as it was back at c .

What speed is it ?

-- Look back at the y-axis. The speed at the height of c and d is 'zero' .

-- The 2nd and 4th choices are both correct. From c to d, <em>the speed is constant</em>. The constant speed is zero. <em>The car is not moving</em>.

5 0

3 years ago

A solar cell generates a potential difference of 0.25 V when a 550 Ω resistor is connected across it, and a potential difference

Andre45 [30]

a) $400 \Omega$

b) 0.43 V

c) 0.44 %

Explanation:

For a battery with internal resistance, the relationship between emf of the battery and the terminal voltage (the voltage provided) is

$V=E-Ir$ (1)

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

In this problem, we have two situations:

1) when $R_1=550 \Omega$ , $V_1=0.25 V$

Using Ohm's Law, the current is:

$I_1=\frac{V_1}{R_1}=\frac{0.25}{550}=4.5\cdot 10^{-4} A$

2) when $R_2=1000 \Omega$ , $V_2=0.31 V$

Using Ohm's Law, the current is:

$I_2=\frac{V_2}{R_2}=\frac{0.31}{1000}=3.1\cdot 10^{-4} A$

Now we can rewrite eq.(1) in two forms:

$V_1 = E-I_1 r$

$V_2=E-I_2 r$

And we can solve this system of equations to find r, the internal resistance. We do it by substracting eq.(2) from eq(1), we find:

$V_1-V_2=r(I_2-I_1)\\r=\frac{V_1-V_2}{I_2-I_1}=\frac{0.25-0.31}{3.1\cdot 10^{-4}-4.5\cdot 10^{-4}}=400 \Omega$

To find the electromotive force (emf) of the solar cell, we simply use the equation used in part a)

$V=E-Ir$

where

V is the terminal voltage

E is the emf of the battery

I is the current

r is the internal resistance

Using the first set of data,

$V=0.25 V$ is the voltage

$I=4.5\cdot 10^{-4}A$ is the current

$r=400\Omega$ is the internal resistance

Solving for E,

$E=V+Ir=0.25+(4.5\cdot 10^{-4})(400)=0.43 V$

In this part, we are told that the area of the cell is

$A=4.0 cm^2$

While the intensity of incoming radiation (the energy received per unit area) is

$Int.=5.5 mW/cm^2$

This means that the power of the incoming radiation is:

$P=Int.\cdot A=(5.5)(4.0)=22 mW = 0.022 W$

This is the power in input to the resistor.

The power in output to the resistor can be found by using

$P'=I^2R$

where:

$R=1000 \Omega$ is the resistance of the resistor

$I=3.1\cdot 10^{-4} A$ is the current on the resistor (found in part A)

Susbtituting,

$P'=(3.1\cdot 10^{-4})^2(1000)=9.61\cdot 10^{-5} W$

Therefore, the efficiency of the cell in converting light energy to thermal energy is:

$\epsilon = \frac{P'}{P}\cdot 100 = \frac{9.6\cdot 10^{-5}}{0.022}=0.0044\cdot 100 = 0.44\%$

7 0

3 years ago

I NEED HELP ASAP WILL MARK BRAINLIEST:

Evgen [1.6K]

Answer:

It is b for sure.

Explanation:

because they are examining urine now that's pure science.

7 0

3 years ago

Which of the following statements is NOT correct?

Flauer [41]

The answer is
D:if the

6 0

2 years ago