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kherson [118]
3 years ago
12

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:
harkovskaia [24]3 years ago
8 0

Answer:

The work for the previous answer is correct, however, since the final velocity is in the southward direction, the answer is -4.3.

Explanation:

Edgenuity will tell you this is the correct answer as well.

liberstina [14]3 years ago
6 0
60 ° is the angle between W- direction and the run direction.

You need the angle betwee S-direction and the run direction. This is 90° - 60° = 30 °.

By geometry you can trace a right triangle, where the S-component is the adyacent side and the run  is the hypotenuse.

Then cos 30° = adyacent side / hypotenuse = S-component / run

Then S-component = run * cos 30° = 5.0 km * 0.866 = 4.3 km

Answer: 4.3 km
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A car is traveling at 39.7 mi/h on a horizontal highway. The acceleration of gravity is 9.8 m/s 2 . If the coefficient of fricti
777dan777 [17]

Answer:

The minimum distance in which the car will stop is

x=167.38m

Explanation:

39.7\frac{mi}{h}*\frac{1km}{0.621371mi}*\frac{1000m}{1km}*\frac{1h}{3600s}=17.747\frac{m}{s}

∑F=m*a

∑F=u*m*g

The force of friction is the same value but in different direction of the force moving the car so it can stop so

F=m*a\\a=\frac{F}{m}\\a=\frac{u*m*g}{m}\\a=u*g\\a=0.096*-9.8\frac{m}{s^{2} }

a=-0.9408 \frac{m}{s^{2}}

v_{f}^{2}=v_{o}^{2}+2*a*(x_{f}-x_{o})\\v_{f}=0 \\x_{o}=0\\0=v_{o}^{2}+2*a*x_{f}\\x_{f}=\frac{v_{o}^{2}}{2*a} \\x_{f}=\frac{(-17.747\frac{m}{s})^{2}}{2*(-0.9408)} \\x_{f}=167.38m

4 0
3 years ago
A student is examining a bacterium under the microscope. The E. coli bacterial cell has a mass of m = 0.900 fg (where a femtogra
just olya [345]

Answer:

9.73 x 10⁻¹⁰ m

Explanation:

According to Heisenberg uncertainty principle

Uncertainty in position x uncertainty in momentum ≥ h / 4π

Δ X x Δp ≥ h / 4π

Δp = mΔV

ΔV = Uncertainty in velocity

= 2 x 10⁻⁶ x 3 / 100

= 6 x 10⁻⁸

mass m = 0.9 x 10⁻¹⁵ x 10⁻³ kg

m = 9 x 10⁻¹⁹

Δp = mΔV

= 9 x 10⁻¹⁹ x 6 x 10⁻⁸

= 54 x 10⁻²⁷

Δ X x Δp ≥ h / 4π

Δ X x  54 x 10⁻²⁷ ≥ h / 4π

Δ X = h / 4π x 1 /  54 x 10⁻²⁷

= \frac{6.6\times10^{-34}}{4\times3.14\times54\times10^{-27}}

= 9.73 x 10⁻¹⁰ m

7 0
3 years ago
Applied force is the force of support exerted by an object that holds up another object.
kenny6666 [7]
False, applied force is when a person or an object pushes on another object 
6 0
3 years ago
Read 2 more answers
A charge Q is uniformly spread over one surface of a very large nonconducting square elastic sheet having sides of length d. At
yaroslaw [1]

The electric field of a very large (essentially infinitely large) plane of charge is given by:

E = σ/(2ε₀)

E is the electric field, σ is the surface charge density, and ε₀ is the electric constant.

To determine σ:

σ = Q/A

Where Q is the total charge of the sheet and A is the sheet's area. The sheet is a square with a side length d, so A = d²:

σ = Q/d²

Make this substitution in the equation for E:

E = Q/(2ε₀d²)

We see that E is inversely proportional to the square of d:

E ∝ 1/d²

The electric field at P has some magnitude E. Now we double the side length of the sheet while keeping the same amount of charge Q distributed over the sheet. By the relationship of E with d, the electric field at P must now have a quarter of its original magnitude:

E_{new} = E/4

4 0
3 years ago
True or false. The greater the distance that the plane moves from an object, the lower the force that will be applied when the a
vovikov84 [41]

Answer:

I think it's false I am not that sure

3 0
3 years ago
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