6. A 2000 kg spacecraft is located 9.2x10 m from the center of the earth. The mass of the earth

Match each letter to the description

Makovka662 [10]

A woman walks in a straight line with the sun to her right at six o'clock in the morning.

The sun rises East of her, so the woman is walking toward the North pole.

A man walks in a straight line with the sun to his right at six o'clock in the evening.

The sun sets West of him, so the man is walking toward the South pole.

The woman and the man are both walking along lines of constant longitude.

5 0

3 years ago

(4.56 x 10^-13)-(1.17 x 10^-13)

avanturin [10]

3.39 x 10^-13

Please mark brainliest!

3 0

3 years ago

16) A man ran a 5 mile race. The race looped around a city park and back

finlep [7]

Answer:

The man's total displacement is equal to 0.

Explanation:

Given that,

A man ran a 5 mile race. The race looped around a city park and back to the starting line.

We need to find the total displacement of the man.

We know that,

Displacement = shortest path covered

Also,

Displacement = final position - initial position

As it reaches back to its starting line, it means, the displacement is equal to 0.

Hence, the man's total displacement is equal to 0.

3 0

3 years ago

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

3 years ago

One of the dangers of tornados and hurricanes is the rapid drop in air pressurethat is associated with such storms. Assume that

Oksi-84 [34.3K]

Answer:

The net outward force is exerted on a square window of the house is $4.5930\times10^{4}\ N$ .

Explanation:

Given that,

Pressure = 1.02 atm

External air pressure = 0.910 atm

Distance = 2.03 m

We need to calculate the net outward force

Using formula of force

$F= PA$

Where, P = pressure

F = force

A = area

Put the value into the formula

$F=(1.02-0.910)\times101325\times(2.03)^2$

$F=4.5930\times10^{4}\ N$

Hence, The net outward force is exerted on a square window of the house is $4.5930\times10^{4}\ N$ .

6 0

3 years ago