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3 years ago

A bullet with a mass 2.25g is fired up into the air with a velocity of 187.5 m/s. What is the maximum height of the bullet

Physics

1 answer:

Ksivusya [100]3 years ago

3 0

Answer:

1793.7m

Explanation:

From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.

Now the kinetic energy; is

K.E = 1/2 × m × v2

Where m is mass

v is velocity

Hence.

K.E = 1/2 × 2.25 × (187.5)^2

Now this should be same with the potential energy which is given as;

P.E = m× g× h

Where m is mass of object

g is acceleration of free fall due to gravity = 9.8m/S2

h is maximum height substain by the object.

Hence P.E = 2.25 × 9.8 × h

From the foregoing analysis of energy conversation it implies;

1/2 × 2.25 × (187.5)^2 =2.25 × 9.8 × h

=> 1/2 × (187.5)^2 = 9.8 × h

=>1/2 × (187.5)^2 / 9.8 = h

=> 1793.69m = h

h= 1793.69m

h =1793.7m to 1 decimal place

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We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top

notka56 [123]

Answer:

A,)FD= 114.1N

B)Torque=798.5Nm

Explanation:

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

A)The equation of Drag force equation can be expressed below,

FD =[ CD × A × ρ × (v^2/ 2)]

Where CD= Drag coefficient for cone-shape = 0.5

ρ = Density

Area of of the tree canopy = 9.0 m^2

density of air of = 1.2 kg/m^3

V= wind velocity= 6.5 m/s,

If we substitute those values to the equation, we have;

FD =[ CD × A × ρ × (v^2/ 2)]

F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]

FD= 114.1N

B) the torque can be calculated using below formula below

Torque= (Force × distance)

= 114.1 × 7

= 798.5Nm

8 0

3 years ago

Help with this physics task pls

cupoosta [38]

Answer:

Answers can be seen below

Explanation:

First we must explain the essential when we clear equations, and that is that if the term we need to clear is accompanied by other terms that are being added up, then those terms go to the other side of the equation to subtract if those terms are subtracting, then they go to the other side to add, if those terms are found multiplying then they go to the other side of the equation to divide and if those other terms are found dividing then they go to the other side of the equation to multiply.

(Primero debemos explicar lo esencial cuando despejamos ecuaciones, y es que si el término que necesitamos despejar va acompañado de otros términos que se están sumando, entonces esos términos van al otro lado de la ecuación para restar si esos términos están restando, luego van al otro lado para sumar, si esos términos se encuentran multiplicando luego van al otro lado de la ecuación a dividir, y si esos términos se encuentran dividiendo, pasan al otro lado de la ecuación a multiplicar.)

1 )

$t=\frac{v}{a}$ ; $d=s*(t-t_{0} )$

$k=\frac{2*U}{x^{2} }$ ; $T_{2}=\frac{P_{2}*V_{2}*T_{1} }{P_{1}*V_{1} } \\$

$L=\frac{F}{\pi*r*P}$ ; $d=\frac{w}{F*cos(o)}$

$t^{2}=\frac{2*x}{g} ; V_{2}=\frac{A_{1}*V_{1} }{A_{2} } \\$

$h=\frac{V}{\pi *r^{2} } ; r=\frac{t}{F*sin(o)}$

$h=\frac{m}{(1/2)*\pi *r^{2} } ; h_{2}=\frac{F_{2}*(1/2)*b_{1} *h_{1} }{F_{1}*(1/2)*b_{2}*h_{2} }$

$b=\frac{mg-ma}{v}; m=\frac{F+kx}{g*cos(o)}$

$a=\frac{v-v_{o} }{t} ; u=\frac{m_{1}+m_{2} }{M}$

$v_{o}=\frac{x-\frac{1}{2}*a*t^{2} }{t} ; F=\frac{W+uNd}{d*cos(o)}$

10)

$h=\frac{E-\frac{1}{2}*m*v^{2} }{mg} ; v_{2} ^{2} = \frac{Dk-\frac{1}{2} m*v_{1}^{2} }{\frac{1}{2}m }$

11)

$N=\frac{mg*sin(o)-F}{u} ; x^{2}=\frac{W+\frac{1}{2}k*x_{1}^{2} }{\frac{1}{2}*k }$

12)

$x=x_{o} +\frac{v^{2-v_{o}^{2} } }{2a} ; m=\frac{P*A-F_{1}-F_{2} }{g}$

13)

$x_{o} = x-\frac{F}{k} ; u=\frac{cos(o)-\frac{a}{g} }{sin(o)}$

14)

$t=\frac{d}{v} +t_{o}$ ; $t_{o} = t-(\frac{v-v_{o} }{a} )$

15)

$F_{2}=\frac{W-F_{1} *d}{d}+F_{3} ; v_{2}^{2}=v_{1}^{2}+\frac{2*Dk}{m}$

16)

$y_{1}=y-\frac{u}{mg} ; x^{2} = \frac{2W}{k}+x_{o} ^{2}$

7 0

3 years ago

A bus travels 280 km south along a straight path with an average velocity of 88 km/h to the south. The bus stops for 24 min, the

IgorC [24]

Answer:

76.78 km/h To calculate the average velocity for the total trip, you need to first determine the total distance traveled and the total time taken. First, let's calculate the total distance traveled. The trip consists of 2 legs. The 1st leg is 280 km and the 2nd leg is 210 km. So the total distance is 280 km + 210 km = 490 km. Now you need to calculate the total time taken. For this problem, there are 3 intervals that need to be accounted for. The travel time for the 1st leg, the duration of the rest stop in the middle, and the travel time for the 2nd leg. The travel time for both legs is calculated by dividing the distance traveled by the average speed. So for the first leg we have 280 km / (88 km / h) = 3.181818 h The 2nd leg is 210 km / (75 km/h) = 2.8 h The rest stop in hours is 24 min / (60 min/h) = 0.4 h The total time is 3.181818 h + 2.8 h + 0.4 h = 6.381818 h The average velocity is the distance divided by the time, giving: 490 km / (6.381818 h) = 76.78 km/h

Explanation:

Hope this helps!!

3 0

3 years ago

Read 2 more answers

An electric company charges $9.55 per month plus $0.14 per? kilowatt-hour (kwh) for the first 250 kwh used per month and $0.03 p

storchak [24]

$34.75 per month

It is a trick question at the end because it says that anything over 250 kwh is $0.03. Although, you are only calculating for 180 kwh and the monthly charge.

8 0

3 years ago

A 120-V motor has mechanical power output of 2.50hp. It is 90.0% efficient in converting power that it takes in by electrical tr

EastWind [94]

The current in this motor is equal to 17.27 Ampere.
The energy delivered to this motor in 3.00 hours is equal to 22.38 Megajoules.
At $0.110/kWh, the cost to run the motor for 3.00 hours is equal to $0.684.

<h3>How to determine the current (in A) delivered to the motor?</h3>

Assuming this electric motor is a single-phase motor and it operates by using DC current, its mechanical power output would be given by:

W = ηIV

Making current (I) the subject of formula, we have:

I = ηV/W

Substituting the given parameters into the formula, we have;

I = (2.50 × 0.746 × 1000)/(0.9 × 120)

I = 1,865/108

Current, I = 17.27 Ampere.

For the energy delivered to this motor, we have:

First of all, we would determine the power delivered to this motor as follows:

Power, P = IV

Power, P = 17.27 × 120

Power, P = 2,072.4 Watt.

Therefore, the energy delivered to this motor in 3.00 hours is given by:

Energy = power × time

Energy = 2,072.4 × 3.00 × 3,600 × 1/1000000

Energy = 22.38 Megajoules.

<h3>How to determine the cost?</h3>

At $0.110/kWh, the cost to run the motor for 3.00 hours is given by:

Cost = 0.110 × 22.38 × 0.278

Cost = $0.684.

Read more on energy here: brainly.com/question/15567897

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Complete Question:

A 120-V motor has mechanical power output of 2.50 hp. It is 90.0% efficient in converting power that it takes in by electrical transmission into mechanical power.

(a) Find the current in the motor.

(b) Find the energy delivered to the motor by electrical transmission in 3.00 h of operation.

6 0

1 year ago