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3 years ago

8

The temperature light will come on in your vehicle when

1 answer:

nikitadnepr [17]3 years ago

5 0

If your engine is not at the right temprature, usually when it is too hot.

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If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?

sdas [7]

Data:
$f_{2} = 42 Hz$
n (Wave node)
V (Wave belly)
L (Wave length)
<span>The number of bells is equal to the number of the harmonic emitted by the string.
</span>
$f_{n} = \frac{nV}{2L}$

Wire 2 → 2º Harmonic → n = 2

$f_{n} = \frac{nV}{2L}$
$f_{2} = \frac{2V}{2L} 
$
$2V = f_{2} *2L$
$V = \frac{ f_{2}*2L }{2}$
$V = \frac{42*2L}{2}$
$V = \frac{84L}{2}$
$V = 42L$

Wire 1 → 1º Harmonic or Fundamental rope → n = 1

$f_{n} = \frac{nV}{2L}$
$f_{1} = \frac{1V}{2L}$
$f_{1} = \frac{V}{2L}$

If, We have:
V = 42L
Soon:
$f_{1} = \frac{V}{2L}$
$f_{1} = \frac{42L}{2L}$
$\boxed{f_{1} = 21 Hz}$

Answer:

<span>The fundamental frequency of the string:
</span>21 Hz

7 0

3 years ago

Read 2 more answers

A cart is uniformly decelerating from rest. The net force acting on the cart is

mina [271]

Answer:

i think the answer is constant

7 0

3 years ago

if the instantaneous current in the circuit is giveen by I=3 sin theta amperes, the rms value of the current will be

Kisachek [45]

Answer:

$I_{rms}=2.12\ A$

Explanation:

Given that,

The instantaneous current in the circuit is giveen by :

$I=3\sin\theta\ A$

We need to find the rms value of the current.

The general equation of current is given by :

$I=I_o\sin\theta$

It means, $I_o=3\ A$

We know that,

$I_{rms}=\dfrac{I_o}{\sqrt2}\\\\=\dfrac{3}{\sqrt2}\\\\=2.12\ A$

So, the rms value of current is 2.12 A.

4 0

3 years ago

For the different values given for the radius of curvature RRR and speed vvv, rank the magnitude of the force of the roller-coas

nirvana33 [79]

Explanation:

The force of the roller-coaster track on the cart at the bottom is given by :

$F=\dfrac{mv^2}{R}$ , m is mass of roller coaster

Case 1.

R = 60 m v = 16 m/s

$F=\dfrac{(16)^2m}{60}=4.26m\ N$

Case 2.

R = 15 m v = 8 m/s

$F=\dfrac{(8)^2m}{15}=4.26m\ N$

Case 3.

R = 30 m v = 4 m/s

$F=\dfrac{(4)^2m}{30}=0.54m\ N$

Case 4.

R = 45 m v = 4 m/s

$F=\dfrac{(4)^2m}{45}=0.36m\ N$

Case 5.

R = 30 m v = 16 m/s

$F=\dfrac{(16)^2m}{30}=8.54m\ N$

Case 6.

R = 15 m v =12 m/s

$F=\dfrac{(12)^2m}{15}=9.6m\ N$

Ranking from largest to smallest is given by :

F>E>A=B>C>D

5 0

3 years ago

what is the force of friction between an 80kg box and the ground on Earth if the coefficient is 0.2?

Reil [10]

Answer:

160N

Explanation: When 80kg mass is one group . It's reaction force acting on a ground.

Weight of the object = 80*10

= 800 N

Here we are given cofficient of static friction its 0.2. It should be smaller than 1

Friction force = Reaction * Friction Cofficient

Reaction = 800N ( Considering Vertical Equilibrium )

F = 800* 0.2

F = 160N

3 0

3 years ago

Read 2 more answers