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1 year ago

8

If you see fire or smoke coming from your vehicle's hood, you should pull over and get away from the vehicle as soon as possible

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1 answer:

hammer [34]1 year ago

8 0

The given statement "If you see fire or smoke coming from your vehicle's hood, you should pull over and get away from the vehicle as soon as possible." is True.

What to do when there is smoking engine in your car?

You should take rapid action if the smoke seems to be coming from a fire under your bonnet.

shut off your vehicle

Don't try to prop the bonnet open; instead, pull the bonnet release lever.

Remove everyone from the vehicle and maintain your distance.

Ask for the fire service by dialing 999.

If it is safe to do so, warn oncoming traffic.

Learn more about the safety precaution with the help of the given link:

brainly.com/question/2347255

#SPJ4

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A soccer player takes a corner kick, lofting a stationary ball 33.0° above the horizon at 15.0 m/s. If the soccer ball has a mas

Alexxandr [17]

Explanation:

It is given that,

Mass of the soccer ball, m = 0.425 kg

Speed of the ball, u = 15 m/s

Angle with horizontal, $\theta=33^{\circ}$

Time for which the player's foot is in contact with it, $\Delta t = 5.1\times 10^{-2}\ s$

Part A,

The x component of the soccer ball's change in momentum is given by :

$\Delta p_x=mv\ cos\theta$

$\Delta p_x=0.425\times 15\ cos(33)$

$p_x=5.34\ kg-m/s$

The y component of the soccer ball's change in momentum is given by :

$\Delta p_y=mv\ sin\theta$

$\Delta p_y=0.425\times 15\ sin(33)$

$p_y=3.47\ kg-m/s$

Hence, this is the required solution.

3 0

2 years ago

An aerobatic airplane pilot experiences weightlessness as she passes over the top of a loop-the-loop maneuver. The acceleration

Debora [2.8K]

Answer:

The radius of the loop is 20.66 km

Explanation:

let the radius of the loop be r

mass of airplane is m

At the top, the pilot experiences two radial forces, which are

1) Gravitational force is mg

2) Centrifugal forces mv²/r out of the center

When the pilot experiences no weight,

then, mg = mv²/r

r = v² / g

= 450² / 9.8

= 20.66 x 10³3

= 20.66 km

3 0

2 years ago

Why cable color codes are important ? what if we use three phase cables without color

Degger [83]

Answer:

Cable color codes are very important for example, If something is not wired properly, it can result in injuries, deaths, fires, and many other problems. This is why there are well-established wire color codes to ensure those working with an around this type of equipment can ensure everything is wired safely and effectively. If we used three phase cables without color code, this can lead too incorrect things that might be a hazard too get not fix and learning new color codes would be hard therefore we should stick too the same color codes for cables and it saves the hassle for needing too check every cable cord color then the original one signed too it.

4 0

2 years ago

Bob is pushing a box across the floor at a constant speed of 1.7 m/s, applying a horizontal force whose magnitude is 70 N. Alice

OlgaM077 [116]

Answer:

a) 70 N, b) b. Each initially applied a force bigger than static friction to get the box moving and accelerating, then when the desired final speed was achieved they reduced the force to make the net force zero.

Explanation:

a) A constant speed means that magnitude of friction force is equal to the magnitude of the external force. The friction force is directly proportional to the normal force, which is equal to the weight of the box. Therefore, the magnitude of the force is 70 N.

b) Alice used initially a greater force to accelerate the box up to needed speed and later reduced the external force to keep speed constant. The right choice is option b.

3 0

2 years ago

A projectile is thrown with velocity v at an angle θ with horizontal. When the projectile is at a height equal to half of the ma

raketka [301]

Let, the maximum height covered by projectile be $\sf{H_m}$

$\purple{ \longrightarrow \bf{h_m = \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} }}$

Projectile is thrown with a velocity = v
Angle of projection = θ

Velocity of projectile at a height half of the maximum height covered be $\sf{v_0}$

$\qquad$ ______________________________

Then –

$\qquad$ $\pink{ \longrightarrow \bf{ \dfrac{h_m}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }}$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2g} \times \dfrac{1}{2} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{4g} = \dfrac{ {v_0}^{2} \: {sin}^{2} \theta }{2g} }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} \: {sin}^{2} \theta }{2} = {v_0}^{2} \: {sin}^{2} \theta }$

$\qquad$ $\longrightarrow \sf{ \dfrac{ {v}^{2} }{2} = {v_0}^{2} }$

$\qquad$ $\longrightarrow \bf{v_0 = \sqrt{ \dfrac{ {v}^{2} }{2} } = \dfrac{v}{ \sqrt{2} } }$

Now, the vertical component of velocity of projectile at the height half of $\sf{h_m}$ will be –

$\qquad$ $\longrightarrow \bf{v_{(y)}=v_0 \: sin \theta }$

$\qquad$ $\longrightarrow \bf{v_{(y)} = \dfrac{v}{ \sqrt{2} } \: sin \theta = \dfrac{v \: sin \: \theta}{ \sqrt{2} } }$

Therefore, the vertical component of velocity of projectile at this height will be–

☀️ $\qquad$ $\pink {\bf{ \dfrac{v \: sin \: \theta}{ \sqrt{2} }} }$

6 0

2 years ago

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