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AfilCa [17]
3 years ago
13

One student runs with a velocity of +10 m/s while a second student runs with a velocity of –10 m/s. Which student has the faster

velocity? Why?
Physics
1 answer:
larisa86 [58]3 years ago
7 0

Answer:

The one with the faster velocity is the one with a velocity of -10m/s

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Could someone summarize these into a paragraph? i'll give brainliest<br> DUE TODAY!!
slamgirl [31]

Answer:

I can't see the picture

Explanation:

Im sorry can you right whats on the picture down in the comments plz.

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The graph represents the reaction 2H2 + 02 32H20 as it reaches
Alex

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C and D

Explanation:

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When a carpenter shuts off his circular saw, the 10.0 inch diameter blade slows from 4250 rpm to 0.00 in 4.00 s. (a) What is the
MaRussiya [10]

Answer:

(a) \alpha=-111.26rad/s

(b) s=4450.6in

(c) 8.66in

Explanation:

First change the units of the velocity, using these equivalents 1rev=2\pi rad and 1 min =60s

4250rpm(\frac{2\pi rad}{1rev})(\frac{1 min}{60 s} )=445.06rad/s

The angular acceleration \alpha the time rate of change of the angular speed \omega according to:

\alpha=\frac{\Delta \omega}{\Delta t}

\Delta  \omega=\omega_i-\omega_f

Where \omega_i is the original velocity, in the case the velocity before starting the deceleration, and \omega_f is the final velocity, equal to zero because it has stopped.

\alpha=\frac{\Delta \omega}{\Delta t} =\frac{\omega_i-\omega_f}{4}\frac{0-445.06}{4} =\frac{-445.06}{4} =-111.26rad/s

b) To find the distance traveled in radians use the formula:

\theta = \omega_i t + \frac{1}{2} \alpha t^2

\theta = 445.06 (4) + \frac{1}{2}(-111.26) (4)^2=1780.24-890.12=890.12rad

To change this result to inches, solve the angular displacement \theta for the distance traveled s (r is the radius).

\theta=\frac{s}{r} \\s=\theta r

s=890.12(5)=4450.6in

c) The displacement is the difference between the original position and the final. But in every complete rotation of the rim, the point returns to its original position. so is needed to know how many rotations did the point in the 890.16 rad of distant traveled:

\frac{890.12}{2\pi}=141.6667

The real difference is in the 0.6667 (or 2/3) of the rotation. To find the distance  between these positions imagine a triangle formed with the center of the blade (point C), the initial position (point A) and the final position (point B). The angle \gamma=\frac{2\pi}{3}=\frac{360^o}{3}=120 is between the two sides known. Using the theorem of the cosine we can find the missing side of the the triangle(which  is also the net displacement):

c^2=a^2+b^2-2abcos(\gamma)

c^2=5^2+5^2-2(5)(5)cos(\frac{2\pi}{3} )\\c^2=25+25+25\\c^2=75\\c=5\sqrt{3}=8.66in

4 0
4 years ago
An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi
mel-nik [20]

Answer:

Explanation:

All the displacement will be converted into vector, considering east as x axis and north as y axis.

5.3 km north

D = 5.3 j

8.3 km at 50 degree north of east

D₁= 8.3 cos 50 i + 8.3 sin 50 j.

= 5.33 i + 6.36 j

Let D₂ be the displacement which when added to D₁ gives the required displacement D

D₁ + D₂ = D

5.33 i + 6.36 j + D₂ = 5.3 j

D₂ = 5.3 j - 5.33i - 6.36j

= - 5.33i - 1.06 j

magnitude of D₂

D₂²= 5.33² + 1.06²

D₂ = 5.43 km

Angle θ

Tanθ = 1.06 / 5.33

= 0.1988

θ =11.25 ° south of due west.

4 0
3 years ago
What is the weight on the surface of Earth of an object of mass 2.00 kilograms?
Anni [7]

Answer:

D

Explanation:

19.6 newtons

A 2.00-kilogram object weighs 19.6 newtons on Earth.

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