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SOVA2 [1]
3 years ago
5

At a circus, a human cannonball is shot from a cannon at 15m/s at an angle of 40° above horizontal. She

Physics
1 answer:
Citrus2011 [14]3 years ago
7 0

Answer:

a

The height is   H  = 6.74 \  m

b

The horizontal distance is  D  = 23.74 \  m

Explanation:

From the question we are told that

  The speed is  v  =  15 \  m/s

  The angle is  \theta  =  40^o

   The height of the cannon from the ground is  h  =  2 m

  The distance of the net from the ground is k  =  1 m

 

Generally the maximum height she reaches is mathematically represented as  

     H  =  \frac{v^2 sin^2 \theta }{2 *  g }  +  h

=>    H  =  \frac{(15)^2 [sin (40)]^2 }{2 * 9.8}  +  2

=>    H  = 6.74 \  m

Generally from kinematic equation  

    s = ut + \frac{1}{2} at^2

Here s is the displacement which is mathematically represented as

         s  =  [-(h-k)]  

    =>  s =  -(2-1)

    =>  s  = -1 m

There reason why s =  -1 m is because upward motion canceled the downward motion remaining only the distance of the net from the ground which was covered during the first half but not covered during the second half

     a =  -g = -9.8

    u  =  v sin (\theta)

So

    -1 = (vsin 40 )t + \frac{1}{2} * (-9.8) t^2

=>  -4.9t^2 + 9.6418t + 1 = 0

using  quadratic formula to solve the equation we have

    t  =  2.07 \  s

Generally distance covered along the horizontal is  

   D  =  v cos (40) *  2.07

=>   D  =  15 cos (40) *  2.07

=>   D  = 23.74 \  m

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A sculpture is suspended in equilibrium by two cables, one from a wall and the other
aleksandrvk [35]

Answer:

T_1=6655.295917 \approx 6655.3N

Explanation:

From the question we are told that

Angle of cable 2 \theta=37.0\textdegree

Weight of sculpture W=5000 N

Generally the Tension from cable 2 T_2 is mathematically given by

   T_2sin37\textdegree=5000N

   T_2=5000N/sin37\textdegree

   T_2=8308.2N

Generally the Tension from Cable 1 T_1 is mathematically given by

   T_1=T_2 cos37\textdegree

   T_1=8308.2* cos 37\textdegree

   T_1=6655.295917 \approx 6655.3N

7 0
3 years ago
Which question would most likely fill in the blank
Citrus2011 [14]
Definitely not the last 2. My bet is on the first option. If it is wrong don't hit me please...
8 0
3 years ago
A small balloon is released at a point 150 feet away from an observer, who is on level ground. If the balloon goes straight up a
Elza [17]

Answer:

\dfrac{dz}{dt}=0.65\ ft/s

Explanation:

Given that

x= 150 ft

\dfrac{dy}{dt}= 7\ ft/s

y= 14 ft

From the diagram

z^2=x^2+y^2

When ,x= 150 ft and y= 14 ft

z^2=150^2+14^2

z=\sqrt{150^2+15^2}

z=150.74 ft

z^2=x^2+y^2

By differentiating with respect to time t

2z\dfrac{dz}{dt}= 2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt}

z\dfrac{dz}{dt}= x\dfrac{dx}{dt}+y\dfrac{dy}{dt}

Here x is constant that is why

\dfrac{dx}{dt}=0

z\dfrac{dz}{dt}= y\dfrac{dy}{dt}

Now by putting the values in the above equation we get

150.74\times \dfrac{dz}{dt}=14\times 7

\dfrac{dz}{dt}=\dfrac{14\times 7}{150.74}\ ft/s

\dfrac{dz}{dt}=0.65\ ft/s

Therefore the distance between balloon and observer increasing with 0.65 ft/s.

5 0
3 years ago
A box has a mass of 35kg.he pulls the rope horizontally with a force of 175 N. Find the horizontal acceleration of the box as it
Cloud [144]

Answer:

3.43 m/s^2

Explanation:

Force is equal to mass times acceleration. (F=ma). You can use inverse operations to get the formula for acceleration, which is acceleration is equal to force divided by mass. (a=F/m). Since there are two forces here, the force friction (55 N), and the force applied (175 N), we must solve for the net force. To solve for the net force, you take the applied force (175 N) and subtract the frictional force from it (55 N). Thus, the net force is 120 N. With this done, we can now solve for our acceleration.

Using the equation for acceleration, we take the force and divide it by mass.

120/35

Answer: 3.43* m/s^2**

*Note: This is rounded to the nearest hundredth, the full answer is: 3.42857143

**Note: In case you're confused, this is meters per second squared.

8 0
3 years ago
A person hits a tennis ball with a mass of 0.058 kg against a wall.
horrorfan [7]

Explanation:

Mass of the ball, m = 0.058 kg

Initial speed of the ball, u = 11 m/s

Final speed of the ball, v = -11 m/s (negative as it rebounds)

Time, t = 2.1 s

(a) Let F is the average force exerted on the wall. It is given by :

F=\dfrac{m(v-u)}{t}

F=\dfrac{0.058\times (11-(-11))}{2.1}

F = 0.607 N

(b) Area of wall, A=3\ m^2

Let P is the average pressure on that area. It is given by :

P=\dfrac{F}{A}

P=\dfrac{0.607\ N}{3\ m^2}

P = 0.202 Pa

Hence, this is the required solution.

8 0
3 years ago
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