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BigorU [14]
3 years ago
8

When a mortar shell is fired with an initial

Physics
1 answer:
Shtirlitz [24]3 years ago
5 0
The equation y = Vo cos(alfa) t - 16t^2 implies that the mortar landed at the same level that it was fire and that fire angle is also 75°.

With that said, let us work on the parametric equations

y = 0 = Vo sin(alfa) t - 16t^2, which by factoring =>

     t (Vo sin(alfa) - 16t) = 0 => t = Vo sin(alfa) / 16    .....(1)

x = 2500 = Vo cos(alfa) t => t = 2500 / [Vo cos(alfa)]      ..... (2)

Now make (1) equal to (2)

Vo sin(alfa) / 16 = 2500 / [Vo cos(alfa)] =>

Vo^2 sin(alfa)cos(alfa) = 2500ft*16ft/s^2 =>

Vo^2 = 2500*16 / [sin(75°)cos(75)] = 2500*16/0.25 = 160,000 ft^2/s^2

Vo = √(160,000) ft/s = 400 ft/s

Answer: option 5. 400 ft/s


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7 0
3 years ago
A 5,257 kg rocket blasts off to the moon with an acceleration of 76 m/s ^2 what is the net force on the rocket
frutty [35]

Newton's subsequent law expresses that power is corresponding to what exactly is needed for an object of consistent mass to change its speed. This is equivalent to that item's mass increased by its speed increase.

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7 0
3 years ago
two porters are available to carry a long timber wood.out of them one is weak.how to do you make less load to the weak one? writ
marshall27 [118]

Answer:

Please find the answer in the explanation.

Explanation:

Given that two porters are available to carry a long timber wood.out of them one is weak. how do you make less load to the weak one?

We can make the weak one to carry less load through two different ways or means.

First, if we can locate the centre of gravity and centre of mass of the long timbre wood, the week one can carry the other end away from the center of gravity and centre of mass.

Second, the strong porter can carry the long timbre wood almost to the fulcrum and allow the weak one to support at the other end. By doing this, the weak one will only carry light portion of the load.

4 0
3 years ago
Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm
Burka [1]

Answer:

The atmospheric pressure is 0.97622\times10^{5}\ Pa.

Explanation:

Given that,

Atmospheric pressure P_{atm}= 1.013\times10^{5}\ Pa    

drop height h'= 27.1 mm

Density of mercury \rho= 13.59 g/cm^3

We need to calculate the height

Using formula of pressure

p = \rho g h

Put the value into the formula

1.013\times10^{5}=13.59\times10^{3}\times9.8\times h

h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}

h=0.76\ m

We need to calculate the new height

h''=h - h'

h''=0.76-27.1\times10^{-3}

h''=0.76-0.027

h''=0.733\ m

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

P=\rho g h

Put the value into the formula

P= 13.59\times10^{3}\times9.8\times0.733

P=0.97622\times10^{5}\ Pa

Hence, The atmospheric pressure is 0.97622\times10^{5}\ Pa.

7 0
3 years ago
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