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3 years ago

Turn the ignition switch to start and release the key immediately or you could destroy the______________.

Physics

1 answer:

vredina [299]3 years ago

5 0

Starter

Explanation:

Turn the ignition switch to start and release the key immediately or you could destroy the starter.

The car starter is used to cause ignition in the internal combustion engine in order to fire the piston and cause mechanical motion. The starter is used to start the cyclic process of the internal combustion engine.

Once the engine starts by igniting the starter, it is best to release it.
The starter ensures that the spark plug is engaged and the motor is brought into work.
If the ignition is still engaged, the process continues repeatedly and it can damage the starter of the car.

learn more:

Automobile brainly.com/question/2599962

#learnwithBrainly

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A 1900 kg car rounds a curve of 55 m banked at an angle of 11 degrees? . If the car is traveling at 98 km/h, How much friction f

Nat2105 [25]

Answer:

22000 N

Explanation:

Convert velocity to SI units:

98 km/h × (1000 m / km) × (1 h / 3600 s) = 27.2 m/s

Draw a free body diagram. There are three forces acting on the car. Normal force perpendicular to the bank, gravity downwards, and friction parallel to the bank.

I'm going to assume the friction force is pointed down the bank. If I get a negative answer, that'll just mean it's actually pointed up the bank.

Sum of the forces in the radial direction (+x):

∑F = ma

N sin θ + F cos θ = m v² / r

Sum of the forces in the y direction:

∑F = ma

N cos θ - F sin θ - W = 0

To solve the system of equations for F, first solve for N and substitute.

N = (W + F sin θ) / cos θ

Substituting:

((W + F sin θ) / cos θ) sin θ + F cos θ = m v² / r

(W + F sin θ) tan θ + F cos θ = m v² / r

W tan θ + F sin θ tan θ + F cos θ = m v² / r

W tan θ + F (sin θ tan θ + cos θ) = m v² / r

W tan θ + F sec θ = m v² / r

F sec θ = m v² / r - W tan θ

F = m v² cos θ / r - W sin θ

F = m (v² cos θ / r - g sin θ)

Given that m = 1900 kg, θ = 11°, v = 27.2 m/s, and r = 55 m:

F = 1900 ((27.2)² cos 11 / 55 - 9.8 sin 11)

F = 21577 N

Rounding to two sig-figs, you need at least 22000 N of friction force.

4 0

2 years ago

10. How far does a transverse pulse travel in 1.23 ms on a string with a density of 5.47 × 10−3 kg/m under tension of 47.8 ?????

KATRIN_1 [288]

Answer: Tension = 47.8N, Δx = 11.5× $10^{-6}$ m.

Tension = 95.6N, Δx = 15.4× $10^{-5}$ m

Explanation: A speed of wave on a string under a tension force can be calculated as:

$|v| = \sqrt{\frac{F_{T}}{\mu} }$

$F_{T}$ is tension force (N)

μ is linear density (kg/m)

Determining velocity:

$|v| = \sqrt{\frac{47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{0.00874 }$

$|v| =$ 0.0935 m/s

The displacement a pulse traveled in 1.23ms:

$\Delta x = |v|.t$

$\Delta x = 9.35.10^{-2}*1.23.10^{-3}$

Δx = 11.5× $10^{-6}$

With tension of 47.8N, a pulse will travel Δx = 11.5× $10^{-6}$ m.

Doubling Tension:

$|v| = \sqrt{\frac{2*47.8}{5.47.10^{-3}} }$

$|v| = \sqrt{2.0.00874 }$

$|v| = \sqrt{0.01568}$

|v| = 0.1252 m/s

Displacement for same time:

$\Delta x = |v|.t$

$\Delta x = 12.52.10^{-2}*1.23.10^{-3}$

$\Delta x =$ 15.4× $10^{-5}$

With doubled tension, it travels $\Delta x =$ 15.4× $10^{-5}$ m

4 0

3 years ago

After being struck by a bowling ball, a 1.8 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.8 kg

kaheart [24]

Answer:

a) v₂ = 4.2 m/s

b) v₂ = 5 m/s

Explanation:

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0.8 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0.8\ m/s)+(1.8\ kg)(v_2)\\v_2 = 5\ m/s - 0.8\ m/s$

v₂ = 4.2 m/s

We will use the law of conservation of momentum here:

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

where,

m₁ = m₂ = mass of bowling pin = 1.8 kg

u₁ = speed of first pin before collsion = 5 m/s

u₂ = speed of second pin before collsion = 0 m/s

v₁ = speed of first pin after collsion = 0 m/s

v₂ = speed of second after before collsion = ?

Therefore,

$(1.8\ kg)(5\ m/s)+(1.8\ kg)(0\ m/s)=(1.8\ kg)(0\ m/s)+(1.8\ kg)(v_2)$

v₂ = 5 m/s

5 0

3 years ago

A department store sells an ""astronomical telescope"" with an objective lens of 30 cm focal length and an eyepiece lens of foca

Yuliya22 [10]

Answer:

The magnifying power of this telescope is (-60).

Explanation:

Given that,

The focal length of the objective lens of an astronomical telescope, $f_o=30\ cm$

The focal length of the eyepiece lens of an astronomical telescope, $f_e=5\ mm=0.5\ cm$

To find,

The magnifying power of this telescope.

Solution,

The ratio of focal length of the objective lens to the focal length of the eyepiece lens is called magnifying of the lens. It is given by :

$m=\dfrac{-f_o}{f_e}$

$m=\dfrac{-30}{0.5}$

m = -60

So, the magnifying power of this telescope is 60. Therefore, this is the required solution.

7 0

3 years ago

What happens to particles of the medium in a mechanical wave?

Karo-lina-s [1.5K]

They flow i guess but dont quote me on that

3 0

2 years ago

Read 2 more answers