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3 years ago

Both cars start from the same point. Which describes the motion shown?

Physics

2 answers:

denis-greek [22]3 years ago

6 0

jane walks 4 blocks south while jeff walks 2 blocks east

evablogger [386]3 years ago

4 0

Is there a graph we can look at?

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List two of the number of strategies you have learned

jolli1 [7]

Answer:

what?

Explanation:

does that means also i learned Ratios and multiplication i keared them like 5 years ago?

7 0

2 years ago

Compare the results of applying the acceleration equation in the following 2 cases : (1) an object that goes from 0 to 10 m/s in

erastovalidia [21]

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8 0

3 years ago

The pilot of an aircraft wishes to fly due west in a 33.9 km/h wind blowing toward the south. The speed of the aircraft in the a

diamong [38]

Answer: $\theta =9.96^{\circ}$ North of west

Explanation:

Given

Plane wishes to fly in west

but wind with speed 33.9 km/h towards south obstructing its path

so plane must fly at an angle of \theta w.r.t west such that it final velocity is towards west

Plane absolute speed=195 km/h

To fly towards west velocity in Y direction should be zero

thus $195sin\theta =33.9$

$\theta =9.96^{\circ}$

so Plane should head towards $9.96^{\circ}$ North of west in order to fly in west.

So plane

actual velocity is

$v=-195cos9.96\hat{i}+195sin9.96\hat{j}$

5 0

3 years ago

A proton is released from rest at the positive plate of a parallelplatecapacitor. It crosses the capacitor and reaches the negat

Triss [41]

Answer:

2.1406 × $10^6$ m/sec

Explanation:

we know that energy is always conserved

so from the law of energy conservation

$qV=\frac{1}{2}mv^2$

here V is the potential difference

we know that mass of proton = 1.67× $10^{-27}$ kg

we have given speed =50000m/sec

so potential difference $V=\frac{\frac{1}{2}\times 1.67\times 10^{-27}50000^2}{1.6\times 10^{-19}}=13.045$

now mass of electron =9.11× $10^{-31}$

so for electron

$\frac{1}{2}\times 9.11\times 10^{-31}v^2=1.6\times 10^{-19}\times 13.045=2.1406\times 10^6 m/sec$

so the velocity of electron will be 2.1406× $10^6$ m/sec

4 0

3 years ago

A small drop of water is suspended motionless in air by a uniform electric field that is directed upward and has a magnitude of

Gennadij [26K]

To solve this problem we will apply the concepts related to the electric field such as the smelting of the Force and the load (In this case the force is equivalent to the weight). Later we will apply the ratio of the total charge as a function of the multiplication of the number of electrons and their individual charge.

$E = \frac{mg}{q}$

Here,

m = mass

g = Acceleration due to gravity

Rearranging to find the charge,

$q = \frac{mg}{E}$

Replacing,

$q = \frac{(3.37*10^{-9})(9.8)}{11000}$

$q = 3.002*10^{-12}C$

Since the field is acting upwards the charge on the drop should be negative to balance it in air. The equation to find the number of electrons then is

$q = ne$