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vitfil [10]
2 years ago
7

A chinook salmon needs to jump a waterfall that is 1.5 m high. If the fish starts from a distance of 1.00 m from the base of the

ledge over which the waterfall flows, find the x- and y- components of the initial velocity the salmon would need to reach the ledge at the top of its trajectory. Can the fish make this jump? ( A Chinook salmon can jump out of the water with a speed of 6.26m/s)
Physics
1 answer:
____ [38]2 years ago
8 0
 <span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal 

it can cross and reach top of trajectory if its top height h = 1.5m 
and horizontal distance d = (1/2) Range 
--------------------------------------... 
let t be top height time 
at top height, vertical component of its velocity =0 
vy = 0 = u sin p - gt 
t = u sin p/g 
h = [u sin p]*t - 0.5 g[t[^2 
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g 
u^2 sin^2 p/2g = 1.5 
u^2 sin^2 p = 1.5*2*9.8 = 29.4 
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component 
===================== 
t = HALF the time of flight 
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g] 
1 = u^2 sin p cos p/g 
u sin p * u cos p = 9.8 
5.42 * u cos p = 9.8 
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component 
check>> 
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s 
u < less than fish's potential jump speed 6.26 m/s 

so it will able to cross</span>
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8 0
2 years ago
a ballistic pendulum is used to measure the speed of high-speed projectiles. A 6 g bullet A is fired into a 1 kg wood block B su
Galina-37 [17]

Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

(c) T = 2.59 N

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

U=(M+m)gh       (1)

U: potential energy

M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

g: gravitational constant = 9.8m/s^2

h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m

The potential energy is:

U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

Mv_1+mv_2=(M+m)v    (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

M(0)+mv_2=(M+m)v    

v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}

The speed of the bullet before the impact with the wood block is 399.04 m/s

(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:

I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N    

The force on the cord after the impact is 2.59N

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3 years ago
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