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Snowcat [4.5K]
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? -4.3 km -2.5 km 2.5 km 4.3 km
Physics
2 answers:
mixer [17]3 years ago
4 0

Given :

A jogger runs 5.0 km on a straight trail at an angle of 60° south of west.

To Find :

The southern component of the run rounded to the nearest tenth of a kilometre.

Solution :

We know, southern component is given by :

d = Dsin \ \theta\\\\d = 5\times sin \ 60^{\circ}\\\\d = -4.33\  km

Here, negative sign indicates south direction.

Therefore, the southern component of the run rounded to the nearest tenth of a kilometre is 4.3 km.

Hence, this is the required solution.

jasenka [17]3 years ago
3 0

Answer:

d

Explanation:

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mixas84 [53]

Answer:

v = 8.65 m/s

Explanation:

Given that,

Distance covered by the doge, d = 45 m

Time taken, t = 5.2 s

We need to find its average speed. The total distance covered divided by the total time taken is called the average speed of an object. So,

v=\dfrac{45\ m}{5.2\ s}\\\\=8.65\ m/s

So, the average speed is 8.65 m/s.

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3 years ago
If only an external force can change the velocity of a body, how can the internal force of the brakes bring a car to rest? 1. It
RoseWind [281]

Answer:

4. It is the force of the road on the tires (an external force) that stops the car.

Explanation:

If there is no friction between the road and the tires, the car won't stop.

You can see this, for example, when there is ice on the road. You can still apply the brakes (internal force), but since there is no friction (external force) the car won't stop.

The force of the brakes on the wheels is not what makes the car stop, it is the friction of the road against still tires that makes it stop.

3 0
3 years ago
Write an expression for the magnitude of charge moved, Q, in terms of N and the fundamental charge e
NeTakaya

We have that for the Question "Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be said its equation is

Q=\frac{E}{Nr^2}

       

From the question we are told

Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e

<h3>An Expression for the <em>magnitude </em>of charge moved</h3>

Generally the equation for the  <em>magnitude </em>of charge moved, Q   is mathematically given as

Q=\frac{E}{Nr^2}

Therefore

An expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be

 Q=\frac{E}{Nr^2}

 

For more information on this visit

brainly.com/question/16517842

3 0
2 years ago
What is the voltage across each resistor?
kogti [31]

Answer:

see solution below

Explanation:

The given resistors are connected in series.

Equivalent resistance in series = 30 + 55 + 15

Equivalent resistance in series Rt = 100 ohms

Since the potential difference in the circuit = 36V

Get the current in the circuit first

I = V/Rt

I = 36/100

I = 0.36A

Get the voltage across 30ohms resistor;

V30 = 0.36 * 30

V30  = 10.8volts

Hence the voltage across the 30ohms resistor is 10.8volts

Get the voltage across 55ohms resistor;

V55 = 0.36 * 55

V55  = 19.8volts

Hence the voltage across the 55ohms resistor is 19.8volts

Get the voltage across 15ohms resistor;

V15 = 0.36 * 15

V15  = 5.4volts

Hence the voltage across the 15ohms resistor is 5.4volts

4 0
3 years ago
S Four point charges each having charge Q are located at the corners of a square having sides of length a. Find expressions for(
Ket [755]

The total electric potential at the center of the square due to the four charges is V  = √2Q/πÈa.

<h3>What do you mean by electric potential? </h3>

The amount of work needed to move a unit charge from a reference point to a specific point against an electric field. It's SI unit is volt.       

V = kq/r

Where V represents electric potential, K is coulomb constant, q  is Charge and r is distance between any  two around charge to the point charge.

Electric potential at O due to four charges is given by,

V = 4KQ/ r

where, r = √2a/2 = a/√2

V = 4k × Q√2/a

V  = √2Q/πÈa

The total electric potential at the center of the square due to the four charges is V  = √2Q/πÈa.

To learn more about electric potential refer to:

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