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3 years ago

6

An object is held at rest and allowed to drop How far does the object fall in 3.3 seconds?

1 answer:

IgorC [24]3 years ago

6 0

After x seconds, an object will fall $\frac{1}{2}at^2$ where a is acceleration due to gravity and t is time

so when t=3.3
the distance it will fall is $(\frac{1}{2})(9.8\frac{m}{s^2})(3.3m)^2$ =53.361m
it will fal 53.361 meters

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Canadian geese migrate essentially along a north-south direction for well over a thousand kilometers in some cases, traveling at

Wewaii [24]

Answer:

θ=19.877⁰

Explanation:

Given data

Velocity Va=34.0 km/h

Velocity Va=100 km/h

To find

Angle θ

Solution

We want the bird to fly with velocity Vb=100 km/h with an angle θ relative to the ground so that the bird fly due south relative to the ground.From figure which is attached we got

Sinθ=(Va/Vb)

Sinθ=(34.0/100)

θ=Sin⁻¹(34.0/100)

θ=19.877⁰

5 0

3 years ago

Ls -2 a solution of 4x +3= -5?.<br>

torisob [31]

Answer:

yes

Explanation:

Let's solve your equation step-by-step.

4x+3=−5

Step 1: Subtract 3 from both sides.

4x+3−3=−5−3

4x=−8

Step 2: Divide both sides by 4.

4x / 4 = −8 / 4

x=−2

Hope it helps,

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5 0

3 years ago

X = 20 <br> v = 3.5 m/s <br> t = ?

N76 [4]

Answer:

16.5 I think

Explanation:

3 0

3 years ago

A particle leaves the origin with a speed of 3.6 106 m/s at 34 degrees to the positive x axis. It moves in a uniform electric fi

Andrej [43]

Answer:

E = -4556.18 N/m

Explanation:

Given data

u = 3.6×10^6 m/sec

angle = 34°

distance x = 1.5 cm = 1.5×10^-2 m (This data has been assumed not given in

Question)

from the projectile motion the horizontal distance traveled by electron is

x = u×cosA×t

⇒t = x/(u×cos A)

We also know that force in an electric field is given as

F = qE

q= charge , E= strength of electric field

By newton 2nd law of motion

ma = qE

⇒a = qE/m

Also, y = u×sinA×t - 0.5×a×t^2

⇒y = u×sinA×t - 0.5×(qE/m)×t^2

if y = 0 then

⇒t = 2mu×sinA/(qE) = x/(u×cosA)

Also, E = 2mu^2×sinA×cosA/(x×q)

Now plugging the values we get

E = 2×9.1×10^{-31}×3.6^2×10^{12}×(sin34°)×(cos34°)/(1.5×10^{-2}×(-1.6)×10^{-19})

E = -4556.18 N/m

4 0

3 years ago

A block is balanced on top of a frictionless sphere of radius R. When the block is given a slight nudge it starts to slide down

Nataly_w [17]

Answer: $\frac{R}{3}$

Explanation:

Given

Sphere of Radius R

Suppose mass of block is m

At any instant \theta Normal reaction(N) and weight(mg) is acting such that

$mg\sin \theta -N=\frac{mv^2}{R}$ , where v is velocity of block at any angle \theta

When block is just about to leave then N=0

therefore

$mg\sin \theta =\frac{mv^2}{R}$

$v^2=gR\sin \theta$ -------------------1

Also by conserving Energy we get

Potential Energy=kinetic Energy of block

$mgh=\frac{mv^2}{2}$

here h=vertical distance traveled by block

From diagram

$h=R-R\sin \theta$

$h=R(1-\sin \theta )$

$mgR(1-\sin \theta )=\frac{mv^2}{2}$

$2gR(1-\sin \theta )=v^2$ -----------------2

From 1 and 2

$2(1-\sin \theta )=\sin \theta$

$3\sin \theta =2$

$\sin \theta =\frac{2}{3}$

Thus from this value of h is

$h=R(1-\sin \theta )$

$h=R(1-\frac{2}{3})$

$h=\frac{R}{3}$

3 0

3 years ago