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katrin2010 [14]

2 years ago

12

Think of a famous person and gather information about him/her

1 answer:

dybincka [34]2 years ago

4 0

Abraham Lincoln

Explanation:

hope it helps you a little

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Suppose a police officer is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph. At the same tim

blagie [28]

Answer:

75.36 mph

Explanation:

The distance between the other car and the intersection is,

$x=x_{0}+V t \\ x=\frac{1}{2}+V t$

The distance between the police car and the intersection is,

$y=y_{0}+V t$

$y=\frac{1}{2}-40 t$

(Negative sign indicates that he is moving towards the intersection)

Therefore the distance between them is given by,

$z^{2}=x^{2}+y^{2}(\text { Using Phythogorous theorem })$

$z^{2}=\left(\frac{1}{2}+V t\right)^{2}+\left(\frac{1}{2}-40 t\right)^{2} \ldots \ldots \ldots(1)$

The rate of change is,

$2 z \frac{d z}{d t}=2\left(\frac{1}{2}+V t\right) V+2\left(\frac{1}{2}-40 t\right)(-40)$

$2 z \frac{d z}{d t}=V+2 V^{2} t-40+3200 t \ldots \ldots \ldots$

Now finding $z$ when $t=0,$ from (1) we have

$z^{2}=\left(\frac{1}{2}+V(0)\right)^{2}+\left(\frac{1}{2}-40(0)\right)^{2}$

$z^{2}=\frac{1}{4}+\frac{1}{4}=\frac{1}{2} \\ z=\sqrt{\frac{1}{2}} \approx 0.7071$

The officer's radar gun indicates 25 mph pointed at the other car then, $\frac{d z}{d t}=25$ when $t=0,$ from

From (2) we get

$2(0.7071)(25)=V+2 V^{2}(0)-40+3200(0)$

$2(0.7071)(25)=V+2 V^{2}(0)-40$

$35.36=V-40$

$V=35.36+40=75.36$

Hence the speed of the car is $75.36 mph$

7 0

3 years ago

Which shows the correct order of forces, from weakest to strongest?

Norma-Jean [14]

^^^^^^^^^^^^^^^^^^^^^^^^^^^

7 0

2 years ago

A small bolt with a mass of 41.0 g sits on top of a piston. The piston is undergoing simple harmonic motion in the vertical dire

EleoNora [17]

Answer:

Amplitude will be equal to 0.091 m

Explanation:

Given mass of the slits = 41 gram = 0.041 kg

Frequency f = 1.65 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 1.65=10.362rad/sec$

Angular frequency is equal to $\omega =\sqrt{\frac{k}{m}}$

$10.362 =\sqrt{\frac{k}{0.041}}$

Squaring both side

$107.371 ={\frac{k}{0.041}}$

k = 4.40 N/m

For vertical osculation

$mg=kA$

$0.041\times 9.8=4.40\times A$

A = 0.091 m

So amplitude will be equal to 0.0391 m

8 0

3 years ago

A 3.0 \Omega Ω resistor is connected in parallel to a 6.0 \Omega Ω resistor, and the combination is connected in series with a 4

GarryVolchara [31]

Answer:

The power dissipated in the 3 Ω resistor is P= 5.3watts.

Explanation:

After combine the 3 and 6 Ω resistor in parallel, we have an 2 Ω and a 4 Ω resistor in series.

The resultating resistor is of Req=6Ω.

I= V/Req

I= 2A

the parallel resistors have a potential drop of Vparallel=4 volts.

I(3Ω) = Vparallel/R(3Ω)

I(3Ω)= 1.33A

P= I(3Ω)² * R(3Ω)

P= 5.3 Watts

7 0

3 years ago

In your own words, define the following terms. Radiation, Convection, and conduction.

Dafna11 [192]

Radiation: transferred energy
Convection: Transfer of heat
Conduction: transfer of electric charge

3 0

3 years ago