A shell fired into the air explodes into two equal fragments. Which of the following is conserved.

In forced convection, the fluid is forced to move by external means such as a fan, pump, or the wind. You must provide an answer

Grace [21]

Answer:

True

Explanation:

Convection simply means the transfer of heat resulting from the movement of the molecules of fluid, which can be liquid or gas.

Convection can either be natural or forced. natural convection occurs by a natural means ,such as the difference in density of warm air and cool air. This occurs when the cooler air falls and the warmer air rises.

However in forced convection, the fluid is agitated by an external force eg a fan or pump.

4 0

3 years ago

Speeding truck slams on the brakes and accelerates at -7.40m/s^2 Before coming to a stop. It leaves skid marks on the pavement t

cluponka [151]

Answer:

25.2563 m/s

Explanation

This is the equation needed

$$v^2 = u^2 + 2ad$ where v is the final velocity, u is the initial velocity, a is the acceleration, and d the distance$

So Just plug in!

6 0

3 years ago

A gas at 5.00 atm pressure was stored in a tank during the winter at 5.0 °C. During the summer, the temperature in the storage a

saw5 [17]

Answer:

a) 5.63 atm

Explanation:

We can use combined gas law

<em>The combined gas law</em> combines the three gas laws:

Boyle's Law, (P₁V₁ =P₂V₂)
Charles' Law (V₁/T₁ =V₂/T₂)
Gay-Lussac's Law. (P₁/T₁ =P₂/T₂)

It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant.

P₁V₁/T₁ =P₂V₂/T₂

where P = Pressure, T = Absolute temperature, V = Volume occupied

The volume of the system remains constant,

So, P₁/T₁ =P₂/T₂

a) $\frac{5}{278} =\frac{P_2}{313} \\\\P_2=\frac{5*313}{278}\\ P_2 = 5.63 atm$

7 0

4 years ago

Assume both snowballs are thrown with the same initial speed 27.2 m/s. The first snowball is thrown at an angle of 75◦ above the

Leona [35]

Answer:

15 deg

Explanation:

Assume both snowballs are thrown with the same initial speed 27.2 m/s. The first snowball is thrown at an angle of 75◦ above the horizontal. At what angle should you throw the second snowball to make it hit the same point as the first? The acceleration of gravity is 9.8 m/s 2 . Answer in units of ◦ .

Given:

For first ball, θ1 = 75◦

initial velocity for both the balls, u = 27.2 m/s

for second ball, θ2 = ?

since distance covered by both the balls is same.

Therefore,..

R1=(u^{2} sin2\alpha _{1}) /g[/tex]

the range for the first ball

the range for the second ball

R2=(u^{2} sin2\alpha _{2}) /g[/tex]

(u^{2} sin2\alpha _{2}) /g[/tex]=(u^{2} sin2\alpha _{1}) /g[/tex]

sin2\alpha _{2})=sin2\alpha _{1})

$2\alpha _{2}$ =sin^-1(sin2\alpha _{1})

$\alpha _{2}$ =1/2sin^-1(sin2\alpha _{1})

$\alpha _{2}$ =

15 deg

8 0

4 years ago

high school physics, no need detail explain, just give the answer, but you have to make sure thank you

andrey2020 [161]

Answer:

approximately 30 degrees

Explanation:

If it takes the cannonball 2 seconds to reach the maximum height, we can use the analysis of the vertical component of the velocity and the fact that the acceleration of gravity is the one acting opposite to this initial vertical component $(v_y)$ of the velocity. We know as well that at the top of the trajectory, the vertical component of the velocity is zero, and then the movement starts going down in it trajectory. So, the final velocity for the first part of the ascending movement is zero, giving us the following equation for the velocity under an accelerated movement (with acceleration of gravity "g" acting):

$v_f=v_i-a\,t\\v_f=v_y-g\,t\\0=v_y-9.8\,*\,2\\v_y=9.8\,*\,2=19.6 \frac{m}{s}$

By knowing the vertical component of the initial velocity (19.6 m/s), and the actual magnitude of the total initial velocity (40 m/s), we can calculate what angle was the initial velocity vector forming above the horizontal. We use for such the fact that the sine of the angle relates the opposite side of a right angle triangle with the hypotenuse, and solve for the angle using the arcsin function:

$sin(\theta)=\frac{opp}{hyp} \\sin(\theta)=\frac{19.6}{40}\\\theta=arcsin(\frac{19.6}{40})\\\theta=29.34^o$

which tells us that the closer answer shown is $30^o$

7 0

4 years ago