The table below shows the difference between the average temperatures in April and October in four locations.

Which statement best describes a characteristic of gases?

salantis [7]

Assumes the shape and volume of its container
<span>particles can move past one another</span>

8 0

3 years ago

A car approaches you at a constant speed, sounding its horn, and you hear a frequency of 76 Hz. After the car goes by, you hear

Talja [164]

Answer:

70.07 Hz

Explanation:

Since the sound is moving away from the observer then

$f_o = f_s\frac {(v+vs)}{v}$ and $f_o = f_s\frac {(v-vs)}{v}$ when moving towards observer

With $f_o$ of 76 then taking speed in air as 343 m/s we have

$76 = f_s\times\frac {(343-vs)}{343}$

$f_s=\frac {343\times 76}{343-v_s}$

Similarly, with $f_o$ of 65 we have

$65 = f_s\times\frac {(343+vs)}{343}\\f_s=\frac {343\times 65}{343+v_s}$

Now

$f_s=\frac {343\times 65}{343+v_s}=\frac {343\times 76}{343-v_s}$

v_s=27.76 m/s

Substituting the above into any of the first two equations then we obtain

$f_s=70.07 Hz$

4 0

3 years ago

Kim took 5 hours to complete a journey with an average speed

monitta

Answer:110

Explanation:

6 0

3 years ago

A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long

Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

$W=-\mu mg\times l$

Put the value into the formula

$W=-0.30\times62\times9.8\times3.50$

$W=-637.98\ J$

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

$K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}$

$\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}$

Final potential energy is zero

So, $\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2$

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl$

Put the value into the formula

$\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50$

$v_{2}=\sqrt{2\times55.915}$

$v_{2}=10.57\ m/s$

The speed of skier is 10.57 m/s.

Hence, (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0

3 years ago

If someone throws a 3 gram fry accelerating at 5 meter/second2, what is the fry’s force?

Alex

Answer:

0.015 m/s2

Explanation:

Using Newtons 2nd law.

F = ma where F = Force applied, m = mass of the object and a = acceleration acquired.

So substitute the values in SI units.

m = $\frac{3}{1000}$ kg

Therefore F = 0.003×5 = 0.015 m/s2

3 0

2 years ago