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3 years ago

How is pressure related to force and surface area

1 answer:

7 0

Pressure is Force per Unit Area.Pressure is the force on an object that is spread over a surface area. The equation for pressure is the forcedivided by the area where the force is applied.

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Introduced species often thrive and multiply in an environment very different from their original one. Why are they often able t

Anestetic [448]

They begin to adapt into their new location. They then end up having adaptations to help them survive.

8 0

2 years ago

Why do you choose MCB in place of a fuse？

astraxan [27]

It protects the electrical appliance and the person from electrical shocks/faults.

Hope it Helped!

3 0

2 years ago

An object of mass 3.00 kg, moving with an initial velocity of 5.05 m/s, collides with and sticks to an object of mass 2.76 kg wi

Lynna [10]

Answer:

0.752 m/s

Explanation:

m1 = 3.00kg

u1 = 5.05m/s

m2 = 2.76kg

u2 = -3.66m/s

According to the law of conservation of momentum,

m1u1 + m2u2 = (m1+m2)v

3(5.05) + 2.76(-3.66) = (5.05+2.76)v

15.15 - 9.2736 = 7.81v

5.8764 = 7.81v

v = 5.8764/7.81

v = 0.752m/s

6 0

2 years ago

Consider two ideal gases, A & B, at the same temperature. The rms speed of the molecules of gas A is twice that of gas B. Ho

uysha [10]

Answer:

option (d)

Explanation:

The relation between the rms velocity and the molecular mass is given by

v proportional to \frac{1}{\sqrt{M}} keeping the temperature constant

So for two gases

$\frac{v_{A}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}$

$\frac{2v_{B}}{v_{B}}=\sqrt{\frac{M_{B}}{M_{A}}}$

${\frac{M_{B}}{M_{A}}} = 4$

${\frac{M_{B}}{4}} = M_{A}$

7 0

3 years ago

During takeoff, an airplane climbs with a speed of 195 m/s at an angle of 15° above the horizontal. The speed and angle constitu

matrenka [14]

Answer:

The horizontal component of the velocity is 188 m/s

The vertical component of the velocity is 50 m/s.

Explanation:

Hi there!

Please, see the figure for a graphic description of the problem. Notice that the x-component of the vector velocity (vx), the y-component (vy) and the vector velocity form a right triangle. Then, we can use trigonometry to obtain the magnitude of vx and vy:

We can find vx using the following trigonometric rule of a right triangle:

cos α = adjacent / hypotenuse

cos 15° = vx / 195 m/s

195 m/s · cos 15° = vx

vx = 188 m/s

The horizontal component of the velocity is 188 m/s

To calculate the y-component we will use the following trigonometric rule:

sin α = opposite / hypotenuse

sin 15° = vy / 195 m/s

195 m/s · sin 15° = vy

vy = 50 m/s

The vertical component of the velocity is 50 m/s.

4 0

3 years ago