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vova2212 [387]
3 years ago
11

a projectile is fired from the ground with an initial horizontal velocity of 8 meters per second. its initial vertical velocityi

s 6 meters per second. whats is the "total/resultant" velocity of the projectile at its peak?

Physics
1 answer:
Stels [109]3 years ago
6 0

a projectile is fired from the ground with an initial horizontal velocity of 8 meters per second. its initial vertical velocityis 6 meters per second. whats is the "total/resultant" velocity of the projectile at its peak?


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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} )

2)

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

3)

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

4)

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

5)

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

6)

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}   }

7)

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

8)

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

9)

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
3.
Ivan

Answer:

car B will be 30 Km ahead of car A.

Explanation:

We'll begin by calculating the distance travelled by each car. This is illustrated below:

For car A:

Speed = 40 km/h

Time = 3 hours

Distance =?

Speed = distance / time

40 = distance / 3

Cross multiply

Distance = 40 × 3

Distance = 120 Km

For car B:

Speed = 50 km/h

Time = 3 hours

Distance =?

Speed = distance / time

50 = distance / 3

Cross multiply

Distance = 50 × 3

Distance = 150 Km

Finally, we shall determine the distance between car B an car A. This can be obtained as follow:

Distance travelled by car B (D₆) = 150 Km

Distance travelled by car A (Dₐ) = 120 Km

Distance apart =?

Distance apart = D₆ – Dₐ

Distance apart = 150 – 120

Distance apart = 30 Km

Therefore, car B will be 30 Km ahead of car A.

7 0
3 years ago
Find the focal length of a lens of power-2.0D . What type of lens is this ?
olga_2 [115]

Answer: f = - 0.50 m, negative (diverging) lens

Explanation: D-value (diopter ) means 1/f, unit is 1/m. Thus -2.0D

Means f = ( 1/ -2.0 ) m = -0.50 m

3 0
2 years ago
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