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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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Rank the following situations according to the magnitude of the impulse of the net force, from largest value to smallest value.
wolverine [178]

Answer:

V

I and II

III and IV

Explanation:

The impulse is equal to the change in momentum of the object involved, so we can calculate the change in momentum in each situation and compare them all.

Taking always east as positive direction, and labelling

u the initial velocity

v the final velocity

m = 1000 kg the mass (which is always equal)

We find:

(i)

u = 25 m/s

v = 0

|I|=m(v-u)=(1000)(0-25)=25,000 Ns

(II)

u = 25 m/s

v = 0

|I|=m(v-u)=(1000)(0-25)=25,000 Ns

(III)

In this case,

F = 2000 N is the force

\Delta t = 10 s is the time

So the magnitude of the impulse is

|I| =F\Delta t = (2000N)(10)=20,000 Ns

(IV)

F = 2000 N is the force

\Delta t = 10 s is the time

So the magnitude of the impulse is

|I| =F\Delta t = (2000N)(10)=20,000 Ns

(V)

u = 25 m/s

v = -25 m/s

|I|=m(v-u)=(1000)(-25-25)=50,000 Ns

So the ranking from largest to smallest is:

V

I and II

III and IV

5 0
2 years ago
A system dissipates 12 J of heat into the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the
timurjin [86]

Answer:

option C

Explanation:

given,

energy dissipated by the system to the surrounding = 12 J

Work done on the system = 28 J

change in internal energy of the system

Δ U = Q - W

system losses energy = - 12 J

work done = -28 J

Δ U = Q - W

Δ U = -12 -(-28)

Δ U = 16 J

hence, the correct answer is option C

6 0
2 years ago
A particular material has an index of refraction of 1.25. What percent of the speed of light in a vacuum is the speed of light i
beks73 [17]

Answer:

80% (Eighty percent)

Explanation:

The material has a refractive index (n) of 1.25

Speed of light in a vacuum (c) is 2.99792458 x 10⁸  m/s

We can find the speed of light in the material (v) using the relationship

n = c/v, similarly

v = c/n

therefore v = 2.99792458 x 10⁸  m/s ÷ (1.25) = 239 833 966 m/s

v = 239 833 966 m/s

Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as

(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%

Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)

3 0
2 years ago
Read 2 more answers
Jack's model includes one object consisting of two molecules and one object consisting of six molecules. He put 16 energy cubes
Novay_Z [31]

Answer:

the molecules have different energy and the system is not in equilibrium

Explanation:

The model developed by Jack has the same energy for each of the two objects, but as each object is made up of a different number of molecules, in the system with more molecules, object 2, each one has approximately 2.4 less energy and the molecules of Object 1 have an E / 2 energy.

 So when you book them together the molecules have different energy and the system is not in equilibrium

5 0
2 years ago
Suppose the value of one division of vernier scale is 0.5mm and the value of one main scale
gregori [183]

Answer:

-0.01 mm

Explanation:

We are given that

The value of one division of vernier scale =0.5 mm

The value of one main scale division=0.49 mm

We have to find the value of least count of the instrument in mm.

We know that

Leas count of vernier caliper=1 main scale division-1 vernier scale division

Least count of vernier caliper=0.49-0.50=-0.01 mm

Hence, the least count of the instrument=-0.01 mm

Answer: -0.01 mm

8 0
2 years ago
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