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3 years ago

Steam is more dangerous than the boiling water because it has more a.Temperature

Physics

1 answer:

balandron [24]3 years ago

4 0

Answer:

c) heat

Explanation:

steam has more heat energy than boiling water due to its latent heat of vaporisation.

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A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle

Sunny_sXe [5.5K]

Answer:

a) 5.09 seconds

b) 107.07 meters

Explanation:

a) As we know

$t_2- t_1 = \sqrt{\frac{2 X}{a} }$

Substituting the given values we get

$t_2 - t_1 = \sqrt{\frac{2 * 52}{4} } \\t_2 - t_1 = 5.09$

It takes 5 .09 s for the motorcycle to accelerate until it catches up with the car

$X_{t`2} = v_i \sqrt{\frac{2X}{a} } + 0.5 a\sqrt{\frac{2X}{a} }\\X_{t`2} = (v_i + 0.5 a) \sqrt{\frac{2X}{a} }\\X_{t`2} = ( 19 + 2) \sqrt{\frac{2* 52}{4} }\\X_{t`2} = 21 * 5.09\\X_{t`2} = 107.07$

4 0

3 years ago

the freezing point of water is the same as ice's __________ point. a. boiling b. melting c. condensing d. sublimation

nika2105 [10]

Melting. Water freezes at 0 degrees celsius or 32 degrees fahrenheit and therefore water freezes at that temperature or ice melts back to its liquid state at those temperatures.

3 0

4 years ago

If the temperature outside increases from 30 degrees to 60 degrees, how has the kinetic energy of the air particles changed?

musickatia [10]

The kinetic energy increase because as temperature warms up particles move faster

7 0

3 years ago

Read 2 more answers

A plane takes off in san francisco at noon and flies toward the southeast. an hour later, it is 400 kilometers east and 300 kilo

eduard

This first step involves a right triangle. If the plane is 400 km east and 300 km south of the origin, and it flew in a straight line, then you can construct a right triangle with side lengths 300, 400, and c. You may recognize that these are multiples of the Pythagorean triple 3, 4, 5, so the side length c is 500 km. Otherwise, you would write
$c^2 = 300^2+400^2 \\ c = 500$ .

This second step is, if I am correctly interpreting "degrees south of east," to find the angle formed by the horizontal line representing the east and the path of the plane. I made a diagram that does just that (see attached). You can use a trig function of one of the angles to solve. I chose
$tan(C)= \frac{3}{4} \\ arctan( \frac{3}{4})=36.870$ . Thus, I believe it is 37° south of east.

5 0

3 years ago

A projectile is fired over level ground with an initial velocity that has a vertical component of 20 m/s and a horizontal compon

Anettt [7]

First of all, let's write the equation of motions on both horizontal (x) and vertical (y) axis. It's a uniform motion on the x-axis, with constant speed $v_x=30 m/s$ , and an accelerated motion on the y-axis, with initial speed $v_y=20 m/s$ and acceleration $g=9.81 m/s^2$ :
$S_x(t)=v_xt$
$S_y(t)=v_y t- \frac{1}{2} gt^2$
where the negative sign in front of g means the acceleration points towards negative direction of y-axis (downward).

To find the distance from the landing point, we should find first the time at which the projectile hits the ground. This can be found by requiring
$S_y(t)=0$
Therefore:
$v_y t - \frac{1}{2}gt^2=0$
which has two solutions:
$t=0$ is the time of the beginning of the motion,
$t= \frac{2 v_y}{g} = \frac{2\cdot 20 m/s}{9.81 m/s^2}=4.08 s$ is the time at which the projectile hits the ground.

Now, we can find the distance covered on the horizontal axis during this time, and this is the distance from launching to landing point:
$S_x(4.08 s)=v_x t=(30 m/s)(4.08 s)=122.4 m$

4 0

3 years ago