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3 years ago

Can someone check my answers and tell me if their correct?

Physics

1 answer:

Otrada [13]3 years ago

6 0

Seven

The magnitude is pointing towards the origin and is at - 20 degrees. The combination makes 160 with the x axis: C answer

Eight

They keep doing this. They use distance where they should use displacement but they use distance to try and fool you. It's a mighty poor practice.

The distance between the start and end points is the displacement. That "distance" is 180*sqrt(25) = 900 . The actual distance should be 180*4 + 180*3 = 720 + 540 = 1260. That's what a car's odometer or a bicycle odometer would read. the difference is 360.

I really do object to the wording, but what can I do?

Nine

Nine is the same thing as 8.

Displacement = sqrt(400^2 + 80^2)= sqrt(166400) = 408

The actual distance is 400 + 80 = 480

The difference is the answer = 480 - 408 = 72 <<<< Answer

Ten

This is just the displacement magnitude.

dis = sqrt(30^2 + 80^2)

dis = sqrt(900 + 6400)

dis = sqrt(7300)

dis = 85.44 <<<< Answer D

Twelve

Vi = 2.15*Sin(30) = 1.075 m/s

vf = 0

a = - 9.81

t = ?

Formula

a = (vf - vi)/t

Solve

-9.81 = (0 - 1.075)/t

- 9.81 * t = -1.075

t = 0.11 seconds

Thirteen

I'm leaving this last one to you. You need the initial height xo to answer it properly. Judging by the other questions, this one is right.

Edit

That is a surprise! Really quickly

d = 3.2 m

a = - 9.82

vf = 0

vi = ?

vf^2 = vi^2 - 2*a*d

0 = vi^2 - 2*9.81*3.2

vi = sqrt(19.62*3.2)

vi = 8.0 m/s But that is the vertical component of the speed

v = vi/sin(25)

v = 8.0/sin(25) = 11

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A silver wire has a cross sectional area a = 2.0 mm2. a total of 9.4 × 1018 electrons pass through the wire in 3.0 s. the conduc

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This problem uses the relationships among current I, current density J, and drift speed vd. We are given the total of electrons that pass through the wire in t = 3s and the area A, so we use the following equation to to find vd, from J and the known electron density n, so:

$v_{d} = \frac{J}{n\left | q \right |}$

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 $I = \frac{\Delta Q}{\Delta t} = \frac{9.4x1018electrons}{3s} = 3189.73(A)$

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2 years ago

A bottle rocket takes off with a = 34.5 m/s2. It

sergiy2304 [10]

Answer:3.4 seconds

Explanation:

Initial velocity(u)=0

acceleration=34.5m/s^2

Height(h)=200m

Time =t

h=u x t - (gxt^2)/2

200=0xt+(34.5xt^2)/2

200=34.5t^2/2

Cross multiply

200x2=34.5t^2

400=34.5t^2

Divide both sides by 34.5

400/34.5=34.5t^2/34.5

11.59=t^2

t^2=11.59

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3 years ago

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Misha Larkins [42]

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3 years ago

Read 2 more answers

THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THANKS!!! THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THAN

timurjin [86]

Answer:

$f_{o}$ = 391.67 Hz

Explanation:

The sound of lowest frequency which is produced by a vibrating sting is called its fundamental frequency ( $f_{o}$ ).

The For a vibrating string, the fundamental frequency ( $f_{o}$ ) can be determined by:

$f_{o}$ = $\frac{v}{2L}$

Where v is the speed of waves of the string, and L is the length of the string.

L = 42.0 cm = 0.42 m

v = 329 m/s

$f_{o}$ = $\frac{329}{2*0.42}$

= $\frac{329}{0.84}$

$f_{o}$ = 391.6667 Hz

The fundamental frequency of the string is 391.67 Hz.

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2 years ago