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Hatshy [7]
3 years ago
10

Frim the castle wall 20 m high shot an arrow. The initial speed of the bow is 45 m/s directed 40 degrees above horizontal. Find

the range and hI e final velocity
Physics
1 answer:
True [87]3 years ago
7 0

Answer:

Range of arrow = 225.09 meter

Final horizontal velocity = 34.47 m/s

Explanation:

We have equation of motion s=ut+\frac{1}{2} at^2, where u is the initial velocity, t is the time taken, a is the acceleration and s is the displacement.

Considering the vertical motion of arrow ( up direction as positive)

   We have u = 45 sin40 = 28.93 m/s, s = -20 m, a = acceleration due to gravity = -9.8m/s^2.

   -20=28.93*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-28.93t-20=0

   t = 6.53 seconds or t = -0.63 seconds

   So time = 6.53 seconds.

Considering the horizontal motion of arrow

   u = 45 cos 40 = 34.47 m/s, t = 6.53 s, a = 0m/s^2

   s=34.47*6.53+\frac{1}{2} *0*6.53^2\\ \\ s=225.09m

So range of arrow = 225.09 meter

Horizontal velocity will not change , final horizontal velocity = 34.47 m/s.

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A hoop of mass 2 kg, radius 0.5 m is rotating about its center with an angular speed of 3 rad's. A force of 10N is applied tange
Degger [83]

Answer:

The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.

Explanation:

Given that,

Mass = 2 kg

Radius = 0.5 m

Angular speed = 3 rad/s

Force = 10 N

(I). We need to calculate the rotational kinetic energy

Using formula of kinetic energy

K.E =\dfrac{1}{2}\timesI\omega^2

K.E=\dfrac{1}{2}\times mr^2\times\omega^2

K.E=\dfrac{1}{2}\times2\times(0.5)^2\times(3)^2

K.E=2.25\ J

(II). We need to calculate the instantaneous change rate of the kinetic energy

Using formula of kinetic energy

K.E=\dfrac{1}{2}mv^2

On differentiating

\dfrac{K.E}{dt}=\dfrac{1}{2}m\times2v\times\dfrac{dv}{dt}

\dfrac{K.T}{dt}=mva....(I)

Using newton's second law

F = ma

a= \dfrac{F}{m}

a=\dfrac{10}{2}

a=5 m/s^2

Put the value of a in equation (I)

\dfrac{K.E}{dt}=mva

\dfrac{K.E}{dt}=mr\omega a

\dfrac{K.E}{dt}=2\times0.5\times3\times5

\dfrac{K.E}{dt}=15\ J/s

Hence, The rotational kinetic energy of the hoop and the instantaneous change rate of the kinetic energy are 2.25 J and 15 J.

3 0
3 years ago
I need help in this please as soon as possible
topjm [15]

Answer:

u itls

Explanation:

sorry look it up

7 0
3 years ago
A car traveling at 40 ft/sec decelerates at a constant 5 feet per second per second. how many feet does the car travel before co
Tema [17]
<span>This problem is solved by the equation of motion: x = x0 + v0*t + 1/2*a*t^2, Here x0 = 0, v0 = 40ft/sec and a = -5 ft/s^2, we need to solve for t: v = v0 + a*t, solve how long does it take to stop: 0 = v0 + a*t --> a*t = -v0 --> t = -v0/a -- > 40/5 = 8 seconds to stop. In this time, the car travels x = 0 + 40*8 + 0.5*-5*8^2 ft ~ 160 ft. Answer: The car travels 160 ft.</span>
3 0
3 years ago
A 3.90 kg block is in equilibrium on an incline of 31.0◦. The acceleration of gravity is 9.81 m/s2 . What is Fn of the incline o
storchak [24]

Answer:

Explanation:

The sum of the pore along the plane is expressed according to Newton's law

Fn-Ff = ma

Fn is the moving force

Ff = nR = frictional force

m is the Mass

a is the acceleration

Substitute the given values

Fn - nR = ma

Fn - tan31°(mgcostheta) =3.9(9.8)

Fn - tan31(3.9(9.8)cos31) = 3.9(9.8)

Fn - tan31(38.22cos31)= 38.22

Fn - 32.76tan31 = 38.22

Fn-19.68 = 38.22

Fn = 38.22+19.68.

Fn = 57.90N

Hence Fn (moving force) of the inclined block is 57.90

4 0
3 years ago
The weight of the windsurfer is 700 newtons. Calculate the moment exerted by the windsurfer on the sailboat ​
borishaifa [10]
Hey, I think someone should help u cause I’m stuck too

Hope this helps !
6 0
2 years ago
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