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1 month ago

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You do 174 J of work while pulling your sister back on a swing, whose chain is 5.10 m long, until the swing makes an angle of 32

.0° with the vertical. What is your sister's mass?

1 answer:

inna [3.1K]1 month ago

6 0

Potential energy U = 174J = m*g*Δh

m is mass
g = 9.81 m/s²
Δh = 5.1 - 5.1cos32° = 0.77

Now solve for the mass (m).

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When a submarine dives to a depth of 5.0 × 102 m, how much pressure, in pa, must its hull be able to withstand? how many times l

ValentinkaMS [3465]

Depth = 5.0 × 10^2 m
Density of seawater = 1.025 x 10^3
Pd = Po + pgh
Standard atmospheric pressure is Patm = 1.01325 x 10^5 Pa
Since the pressure inside the hull is normal, we can disregard Po.
Thus, Pd = pgh = 1.025 x 10^3 x 9.8 x 5.0 x 10^2 = 50.225 x 10^5
Now calculating Pd / Patm gives us 50.225 x 10^5 / 1.01325 x 10^5 = 49.56
This indicates the pressure is 49.56 times greater.

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1 month ago

Read 2 more answers

The wavelength of light is 5000 angstrom. Express it in nm and m.

serg [3582]

Answer:

1 angstrom equals 0.1 nm.

To convert 5000 angstroms: 5000 angstrom = 5000 / 1 × 0.1 nm.

= 500 nm

$1 \: angstrom = 1 \times {10}^{ - 10} m$

To express 5000 angstroms in meters: 5000 angstrom = 5000 × 1 × 10^-10.

= 5 × 10^-7 m

Hope this explanation is useful for you.

7 0

2 months ago

Two racecars are driving at constant speeds around a circular track. Both cars are the same distance away from the center of the

Yuliya22 [3333]

Response:

The acceleration of car 2 is four times that of car 1.

Rationale:

Centripetal acceleration occurs when an object travels in a circular route. It can be expressed as:

$a=\dfrac{v^2}{r}$

In this scenario, two race cars are moving at consistent speeds around a circular course. Both automobiles are located at an equal distance from the center, but car 2 is operating at twice the speed of car 1.

Thus,

$\dfrac{a_1}{a_2}=\dfrac{v_1^2}{v_2^2}$

1 and 2 represent the first and second cars, respectively.

$v_2=2v_1$

Consequently,

$\dfrac{a_1}{a_2}=\dfrac{v_1^2}{(2v_1)^2}\\\\\dfrac{a_1}{a_2}=\dfrac{v_1^2}{4v_1^2}\\\\\dfrac{a_1}{a_2}=\dfrac{1}{4}\\\\a_2=4\times a_1$

Therefore, car 2's acceleration is four times that of car 1.

4 0

1 month ago

A diver explores a shallow reef off the coast of Belize. She initially swims d1 = 74.8 m north, makes a turn to the east and con

inna [3103]

Response:R=1607556m

θ=180degrees

Clarification:

d1=74.8m

d2=160.7km=160.7km*1000

d2=160700m

d3=80m

d4=198.1m

Utilizing an analytical approach:

Rx=-(160700+75*cos(41.8))= -160755.9m

Ry= -(74.8+75sin(41.8))-198.1=73m

Magnitude, R:

R=√Rx+Ry

R=√160755.9^2+20^2=160755.916

R=160756m

Direction,θ:

θ=arctan(Rx/Ry)

θ=arctan(-73/160755.9)

θ=-7.9256*10^-6

It is worth noting that since θ is in the second quadrant, 180 is added

θ=180-7.9256*10^6=180degrees

8 0

14 days ago

Two projectiles are launched with the same initial speed from the same location, one at a 30° angle and the other at a 60° angle

Ostrovityanka [3204]

Answer:

Explanation:

Let us denote the launch angle as $\theta _1=30^{\circ}$ .

$\theta _2=60^{\circ}$ , $\theta _1$ , and $\theta _2$ are complementary angles which means their ranges are identical.

$H_{max}=\frac{u^2(\sin \theta )^2}{2g}$

$H_{30}=\frac{u^2(\sin 30)^2}{2g}$

$H_{30}=\frac{u^2}{8g}$

Time of flight is indicated as $=\frac{2u\sin \theta }{g}$ .

$t_{30}=\frac{u}{g}$

For $\theta =60^{\circ}$

$H_{60}=\frac{u^2(\sin 60)^2}{2g}$

$H_{60}=\frac{3u^2}{8g}$

$t_{60}=\frac{2u\sin 60}{g}=\frac{\sqrt{3}u}{g}$

$H_{60}=3\times H_{30}$

$t_{60}=\sqrt{3}\times t_{30}$

7 0

1 month ago