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3 years ago

5

Sara walks part way around a swimming pool. she walks 50 yards north, then 20 yards east, then 50 yards south. the magnitude of

her total displacement during this walk is __________ yards.

2 answers:

Semmy [17]3 years ago

6 0

|---20----|
| |
| 50 |50
|---D--->|
Start End

Total displacement(D) 20 yards (East).

FinnZ [79.3K]3 years ago

5 0

her total displacement is: 20 (yards)

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The best scientific reason for a scientist to accept a specific theory is?

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The theory is supported by all the available observations and data.

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9. Pick the best example of Newton's Second Law in action.

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3 years ago

T-Joe (65 kg) is running at 3 m/s. T-Brud (50 kg) is running at 4 m/s. What would be T-Joe's momentum?

Debora [2.8K]

Answer:

$P_J=195N$

Explanation:

From the question we are told that

$Mass\ of T-joe\ M_J=65\\Velocity\ of T-joe\ V_J=3m/s\\Mass of\ T-Brud\ M_B=50kg\\Velocity\ of T-Brud\ V_B=3m/s\\$

Generally the equation for momentum is mathematically given by

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Therefore

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3 years ago

cylindrical container is to be constructed to be open at the top with a volume of 27π cubic meters using the least amount of mat

Llana [10]

Answer:

radius comes out to be 3 m

height of the cylinder comes out to be 3m

Explanation:

given

volume of cylinder = 27π m³

π r² h = 27π

r² h = 27.............(1)

surface area of cylinder open at the top

S = 2πrh + π r²

$S = 2\pi \dfrac{27}{r} + \pi r^2$

$\frac{\mathrm{d} s}{\mathrm{d} r}=\frac{\mathrm{d}}{\mathrm{d} r} (2\pi \dfrac{27}{r} + \pi r^2)$

$\frac{\mathrm{d} s}{\mathrm{d} r}=54\pi \dfrac{-1}{r^2}+2\pi r$

$\frac{\mathrm{d} s}{\mathrm{d} r}=0$

for least amount of material requirement.

$\dfrac{54\pi }{r^2} = 2\pi r\\r=3m$

hence radius comes out to be 3 m

for height put the value in the equation 1

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3 0

3 years ago

Read 2 more answers

A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t

Semenov [28]

Answer:

a. $I=2.77x10^{-8} kg*m^2$

b. $K=4.37 x10^{-6} N*m$

Explanation:

The inertia can be find using

a.

$I = m*r^2$

$m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg$

$r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m$

$I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2$

$I=2.77x10^{-8} kg*m^2$

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

$T=2\pi *\sqrt{\frac{I}{K}}$

Solve to K'

$K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}$

$K=4.37 x10^{-6} N*m$

3 0

3 years ago