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Vlad1618 [11]
4 years ago
12

Steelhead trout migrate up stream to spawn.Occasionally they need to leap up small waterfalls to continue their journey. Fortuna

tely, steelhead are remarkable jumpers, capable of leaving the water at a speed of 8.0 m/s.
a. What is the maximum height a steelhead trout can jump?
b.Leaving the water at 8.0 m/s, the trout lands on top of the water fall 1.8 m high. How long was it in the air?
Physics
2 answers:
frozen [14]4 years ago
8 0

A) 3.3m

B)  1.36 sec

Explanation:

Cause I know its right

Ainat [17]4 years ago
5 0

Part a)

since steelhead jump under the influence of gravity

So we will use kinematic equations

v_f^2 - v_i^2 = 2ad

here at the maximum position we know that final speed will be zero

v_f = 0

a = -9.8 m/s^2

v_i = 8m/s

now we will have

0 - 8^2 = 2(-9.8)d

d = 3.26 m

so the maximum height that it can reach will be 3.26 m

Part B)

now if it reaches to height 1.8 m with initial speed 8 m/s

so here we will have

y = v_i t + \frac{1}{2}at^2

1.8 = 8 t - \frac{1}{2}(9.8)t^2

4.9 t^2 - 8t + 1.8 = 0

t = 0.27 s

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A ball is rolling along a road at the top of a hill, at a velocity of 10.0 km/hr. The ball rolls down a 50.0 m high hill. Neglec
Andru [333]

Answer:

v_f = 31.44 m/s

Explanation:

initial height of the ball is given as

H = 50.0 m

initial speed of the ball is given as

v = 10.0 km/h

now we know that

v = 10 \times \frac{1000}{3600}

v = 2.78 m/s

now by energy conservation we can say

initial kinetic energy + initial potential energy = final kinetic energy + final potential energy

\frac{1}{2}mv_i^2 + mgh_1 = \frac{1}{2}mv_f^2

\frac{1}{2}(2.78)^2 + (9.81)(50.0) = \frac{1}{2}v_f^2

v_f = 31.44 m/s

8 0
3 years ago
A 500 g block of metal absorbs 5016 J of heat when its temperature changes from 20.0°C to 30.0°C. Calculate the specific heat of
Step2247 [10]

Explanation:

the answer is

s= 1.0032 J/ g .C

7 0
3 years ago
A typical mirror in a persons house would be an example of a/an____mirror
Aloiza [94]
It would be a plane mirror, since the average household mirror is flat.
4 0
4 years ago
Read 2 more answers
A car is driving at 85 km/h and the driver spots a stop sign ahead. What coefficient of friction is needed to stop the car at th
almond37 [142]

Answer:

μ = 0.0315

Explanation:

Since the car moves on a horizontal surface, if we sum forces equal to zero on the Y-axis, we can determine the value of the normal force exerted by the ground on the vehicle. This force is equal to the weight of the cart (product of its mass by gravity)

N = m*g (1)

The friction force is equal to the product of the normal force by the coefficient of friction.

F = μ*N (2)

This way replacing 1 in 2, we have:

F = μ*m*g (2)

Using the theorem of work and energy, which tells us that the sum of the potential and kinetic energies and the work done on a body is equal to the final kinetic energy of the body. We can determine an equation that relates the frictional force to the initial speed of the carriage, so we will determine the coefficient of friction.

\frac{1}{2} *m*v_{i}^{2}-(F*d)=  \frac{1}{2} *m*v_{f}^{2}

where:

vf = final velocity = 0

vi = initial velocity = 85 [km/h] = 23.61 [m/s]

d = displacement = 900 [m]

F = friction force [N]

The final velocity is zero since when the vehicle has traveled 900 meters its velocity is zero.

Now replacing:

(1/2)*m*(23.61)^2 = μ*m*g*d

0.5*(23.61)^2 = μ*9,81*900

μ = 0.0315

5 0
3 years ago
A 250-kg crate is on a rough ramp, inclined at 30° above the horizontal. The coefficient of kinetic friction between the crate a
astraxan [27]

The force that is pushing the crate up the ramp is competing with at least another two forces, those being G force= defines the attraction to the earth and Ff, the friction between the crate and the rough material of the ramp

Any Force can be defined by the weigh of a particular body and the acceleration

the kinetic friction indicative would be defined by the report between the active force and the friction force, considering G

So 5000-22%×5000=250× a

a=3900/250 m/s

6 0
4 years ago
Read 2 more answers
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