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Rus_ich [418]
3 years ago
6

qual o nome da posição que o planeta tem maior velocidade? este ponto é mais próximo ou mais afastado do sol?

Physics
1 answer:
klio [65]3 years ago
7 0

Em inglês, esse ponto é 'perihelion' ... 'periélio'. É o mais próximo que a Terra ou o planeta chega do sol.

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An experiment begins with crystalline salt at the bottom of an otherwise empty glass. First, students add water to the glass. Th
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Answer:

Explanation:

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Therefore entropy of the system after the salt dissolves is highest among them all and the least entropy is associated when mixing begins.

Entropy is the degree of randomness and it increases when salt dissociates to Na^{+}\ and\ Cl^{-} as these ions are freer to move inside an aqueous solution compared to Previous state.

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3 years ago
When driving at night switch to low-beams whenever you come within ___ ft of an oncoming vehicle.
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B. 500ft.  If you are driving at night with high beams on you must dim them within 500 feet of an oncoming vehicle.
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3 years ago
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A jogger runs 6 km north ,5 km east then another 4km north her average speed 8 km how long will it take her to complete her run
LekaFEV [45]

The time taken to complete her run is 1.9 hr.

<u>Explanation:</u>

Speed is a scalar quantity and it is defined as the ratio of distance covered to the time taken to cover that distance. As distance is also a scalar quantity, so the directions given in the problem can be ignored. Thus, the distance covered by the jogger is the sum of kilometers given in problem.

Distance covered = 6+5+4 = 15 km

And the speed is given as 8 km/hr.

So the time taken will be ratio of distance to speed.

\text { time }=\frac{\text {distance}}{\text {speed}}=\frac{15}{8}=1.9 \text { hour }

So the jogger will take 1.9 hr to complete her run.

8 0
3 years ago
A single-turn current loop carrying a 4.00 A current, is in the shape of a right-angle triangle with sides of 50.0 cm, 120 cm, a
pantera1 [17]

Given that,

Current = 4 A

Sides of triangle = 50.0 cm, 120 cm and 130 cm

Magnetic field = 75.0 mT

Distance = 130 cm

We need to calculate the angle α

Using cosine law

120^2=130^2+50^2-2\times130\times50\cos\alpha

\cos\alpha=\dfrac{120^2-130^2-50^2}{2\times130\times50}

\alpha=\cos^{-1}(0.3846)

\alpha=67.38^{\circ}

We need to calculate the angle β

Using cosine law

50^2=130^2+120^2-2\times130\times120\cos\beta

\cos\beta=\dfrac{50^2-130^2-120^2}{2\times130\times120}

\beta=\cos^{-1}(0.923)

\beta=22.63^{\circ}

We need to calculate the force on 130 cm side

Using formula of force

F_{130}=ILB\sin\theta

F_{130}=4\times130\times10^{-2}\times75\times10^{-3}\sin0

F_{130}=0

We need to calculate the force on 120 cm side

Using formula of force

F_{120}=ILB\sin\beta

F_{120}=4\times120\times10^{-2}\times75\times10^{-3}\sin22.63

F_{120}=0.1385\ N

The direction of force is out of page.

We need to calculate the force on 50 cm side

Using formula of force

F_{50}=ILB\sin\alpha

F_{50}=4\times50\times10^{-2}\times75\times10^{-3}\sin67.38

F_{50}=0.1385\ N

The direction of force is into page.

Hence, The magnitude of the magnetic force on each of the three sides of the loop are 0 N, 0.1385 N and 0.1385 N.

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