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3 years ago

9

How are typhoons formed? Need help!

2 answers:

yan [13]3 years ago

8 0

Answer:

typhoons are formed by the people cant judge or observe

Explanation:

mark me brainliest

Mila [183]3 years ago

5 0

Answer:

typhoon forms when winds blow into areas of the ocean where the water is warm. These winds collect moisture and rise, while colder air moves in below. This creates pressure, which causes the winds to move very quickly. The winds rotate, or spin, around a center called an eye.

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If you apply a 4 N force to a 2 kg object, how fast will it accelerate? (Your

zubka84 [21]

Answer:2n

Explanation:

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3 years ago

How do you get rid of depression?

german

A sicitrist might help you or go to a doctor and get some marijawana

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3 years ago

Read 2 more answers

A negative slope on the velocity vs. time graph indicates that the object is not accelerating

liubo4ka [24]

AnswerNo

However, the acceleration is the opposite of the movement:

Explanation:

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3 years ago

An emf of 22.0 mV is induced in a 519-turn coil when the current is changing at the rate of 10.0 A/s. What is the magnetic flux

zhenek [66]

Answer:

$\phi=1.56\times 10^{-5}\ Wb$

Explanation:

Given that,

Emf, V = 22 mV

Number of turns in the coil us 519

Rate of change of current is 10 A/s.

We need to find the magnetic flux through each turn of the coil at an instant when the current is 3.70 A.

Let's find the inductance first. So,

$L=\dfrac{\epsilon}{(dI/dt)}\\\\L=\dfrac{0.022}{10}\\\\L=0.0022\ H$

We have,

$L=\dfrac{N\phi}{I}$ , $\phi$ is magnetic flux

$\phi=\dfrac{LI}{N}\\\\\phi=\dfrac{0.0022\times3.7}{519}\\\\\phi=1.56\times 10^{-5}\ Wb$

So, the magnetic flux is $1.56\times 10^{-5}\ Wb$ .

8 0

3 years ago

(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly

nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car, a:

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2$

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

$F=(1100 kg)(-2.23 m/s^2)=-2451 N$

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) $-1.53\cdot 10^5 N$

We can use again the equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car. This time we have

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2$

So now we can find the force exerted on the car by using again Newton's second law:

$F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N$

As we can see, the force is much stronger than the force exerted in part a).

8 0

3 years ago