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3 years ago

If a students clothing catches fire the student should

Physics

1 answer:

Roman55 [17]3 years ago

4 0

Answer:

stop, drop and roll.

Explanation:

This is because rolling on the ground can help put out the fire by depriving it of oxygen.

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A devout halloweener not only dressed as an astronaut, but travelled to the moon for the full experience. The astronaut jumps on

nikdorinn [45]

Answer:

2.78 m

Explanation:

At the peak, the velocity is 0.

Given:

a = -1.6 m/s²

v₀ = 2.98 m/s

v = 0 m/s

x₀ = 0 m

Find:

v² = v₀² + 2a(x - x₀)

(0 m/s)² = (2.98 m/s)² + 2(-1.6 m/s²) (x - 0 m)

x = 2.775 m

Rounded to 3 sig-figs, the astronaut halloweener reaches a maximum height of 2.78 meters.

5 0

3 years ago

A football kicker kicks a ball with a mass of .42 kg. The average acceleration of the football was 14.8 m/s squared. How much fo

mrs_skeptik [129]

To find the force we use the formula,

F = ma , where m is mass and a acceleration

Using the formula,

F = ma

F = 0.42 x 14.8

F = 6.216 N / 6.22 N

Hope you liked the answer !

4 0

3 years ago

A ship maneuvers to within 2.46×10³ m of an island’s 1.80 × 10³ m high mountain peak and fires a projectile at an enemy ship 6.1

Nesterboy [21]

Answer:

The distance close to the peak is 597.4 m.

Explanation:

Given that,

Distance of the first ship from the mountain $d=2.46\times10^{3}\ m$

Height of island $h=1.80\times10^{3}\ m$

Distance of the enemy ship from the mountain $d'=6.10\times10^{2}\ m$

Initial velocity $v=2.55\times10^{2}\ m/s$

Angle = 74.9°

We need to calculate the horizontal component of initial velocity

Using formula of horizontal component

$v_{x}=v\cos\theta$

Put the value into the formula

$v_{x}=2.55\times10^{2}\cos74.9$

$v_{x}=66.42\ m/s$

We need to calculate the vertical component of initial velocity

Using formula of vertical component

$v_{y}=v\sin\theta$

Put the value into the formula

$v_{y}=2.55\times10^{2}\sin74.9$

$v_{y}=246.19\ m/s$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{v_{x}}$

$t=\dfrac{2.46\times10^{3}}{66.42}$

$t=37.03\ sec$

We need to calculate the height of the shell on reaching the mountain

Using equation of motion

$H= v_{y}t-\dfrac{1}{2}gt^2$

Put the value in the equation

$H=246.19\times37.03-\dfrac{1}{2}\times9.8\times(37.03)^2$

$H=2397.4\ m$

We need to calculate the distance close to the peak

Using formula of distance

$H'=H-h$

Put the value into the formula

$H'=2397.4-1800$

$H'=597.4\ m$

Hence, The distance close to the peak is 597.4 m.

6 0

3 years ago

2. A test reveals that 150 J of work is required to lift an object 3 m at a

Nuetrik [128]

Answer:

50N

Explanation:

W=Fd

150=F(3)

50N=F

7 0

3 years ago

How do I ship a pringle in the mail without it breaking? This is a physics project.

vekshin1

Assuming you want it to be as small and lightweight as possible :

Cut a solid box roughly twice as big as the pringle. Put the pringle inside the box, and fill the remaining space with cotton, that will cushion the impacts. Be sure to apply the mention <em>FRAGILE</em> to the box, so that they'll take care of it properly.

5 0

4 years ago