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OleMash [197]
2 years ago
6

A jogger runs 5.0 km on a straight trail at an angle of 60° south of west. What is the southern component of the run rounded to

the nearest tenth of a kilometer?
Physics
2 answers:
AnnyKZ [126]2 years ago
8 0
Without rounding off, the answer is -4.33. Since, the southern component of the run of the jogger should be rounded to the nearest tenth of a kilometer, then the answer is -4.3. The answer is negative because it is going south. 
spin [16.1K]2 years ago
3 0

The southern component of the jogger's path is given by:

S_y = S \sin \theta

where

S=5.0 km is the total distance of the jogger's run

\theta=60^{\circ} is the angle with respect to the horizontal direction


Substituting numbers into the equation, we find

S_y=(5.0 km)\sin (60^{\circ})=4.33 km

and the direction is south.

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ivanzaharov [21]
D. Teach the public energy conservation
7 0
2 years ago
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A cylinder is sliced in half along its diagonal. Determine the location of the center of mass and the inertia properties relativ
Black_prince [1.1K]

Answer:

See the explanation below

Explanation:

To better understand this problem, a cylinder sketch is attached before and after the cut, we see that after the cut, the shape of this resembles that of a right triangle.

We can find, the centroid in the xy plane, knowing that the centroid for a triangle is located a third of its base.

In the z axis there is no displacement of the centroid.

3 0
2 years ago
The expressions for e/m and the relative error of e/m due to all of the parameters measured:
bija089 [108]

Answer:

Term 1 = (0.616 × 10⁻⁵)

Term 2 = (7.24 × 10⁻⁵)

Term 3 = (174 × 10⁻⁵)

Term 4 = (317 × 10⁻⁵)

(σ ₑ/ₘ) / (e/m) = (499 × 10⁻⁵) to the appropriate significant figures.

Explanation:

(σ ₑ/ₘ) / (e/m) = (σᵥ /V)² + (2 σᵢ/ɪ)² + (2 σʀ /R)² + (2 σᵣ /r)²

mean measurements

Voltage, V = (403 ± 1) V,

σᵥ = 1 V, V = 403 V

Current, I = (2.35 ± 0.01) A

σᵢ = 0.01 A, I = 2.35 A

Coils radius, R = (14.4 ± 0.3) cm

σʀ = 0.3 cm, R = 14.4 cm

Curvature of the electron trajectory, r = (7.1 ± 0.2) cm.

σᵣ = 0.2 cm, r = 7.1 cm

Term 1 = (σᵥ /V)² = (1/403)² = 0.0000061573 = (0.616 × 10⁻⁵)

Term 2 = (2 σᵢ/ɪ)² = (2×0.01/2.35)² = 0.000072431 = (7.24 × 10⁻⁵)

Term 3 = (2 σʀ /R)² = (2×0.3/14.4)² = 0.0017361111 = (174 × 10⁻⁵)

Term 4 = (2 σᵣ /r)² = (2×0.2/7.1)² = 0.0031739734 = (317 × 10⁻⁵)

The relative value of the e/m ratio is a sum of all the calculated terms.

(σ ₑ/ₘ) / (e/m)

= (0.616 + 7.24 + 174 + 317) × 10⁻⁵

= (498.856 × 10⁻⁵)

= (499 × 10⁻⁵) to the appropriate significant figures.

Hope this Helps!!!

6 0
2 years ago
A circular loop of diameter 10 cm, carrying a current of 0.20 A, is placed inside a magnetic field B⃗ =0.30 Tk^. The normal to t
arlik [135]

Answer:

The magnitude of the torque on the loop due to the magnetic field is 4.7\times10^{-4}\ N-m.

Explanation:

Given that,

Diameter = 10 cm

Current = 0.20 A

Magnetic field = 0.30 T

Unit vectorn=-0.60\hat{i}-0.080\hat{j}

We need to calculate the torque on the loop

Using formula of torque

\tau=NIAB\sin\theta

Where, N = number of turns

A = area

I = current

B = magnetic field

Put the value into the formula

\tau=1\times0.20\times\pi\times(5\times10^{-2})^2\times0.30\times\sin90^{\circ}

\tau=4.7\times10^{-4}\ N-m

Hence, The magnitude of the torque on the loop due to the magnetic field is 4.7\times10^{-4}\ N-m.

5 0
2 years ago
6. Starting from 1.5 miles away, a car drives toward a speed checkpoint and then passes it. The car travels at a constant rate o
loris [4]
For the answer to the question above, 

<span>To be 0.1 miles away from the check point ,
 the car has to travel 1.4 miles OR 1.6 miles. </span>


53 miles = 60 minutes 

1.4 miles = 1.4 / 53 X 60 = 1.5849056 minutes OR 95.1 seconds 

<span>1.6 miles = 1.6 /53 X 60 = 1.8113207 minutes OR 108.7 seconds 
</span>So the answer is <span>95.1s and 108.7s
I hope my answer helped you</span>
7 0
2 years ago
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