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faltersainse [42]
3 years ago
15

A Chinook salmon can jump out of water with a speed of 7.20 m/s . How far horizontally can a Chinook salmon travel through the a

ir if it leaves the water with an initial angle of =29.0° with respect to the horizontal? (Neglect any effects due to air resistance.)
d=
Physics
1 answer:
Alborosie3 years ago
5 0

Explanation:

The vertical component of the salmon's velocity is 7.2 m/s x sin 29 = 3.49 m/s

If g = 9.81 m/s^2, the salmon takes

(3.49 m/s) / (9.81 m/s^2) = 0.356 s to reach the highest point of its trajectory.

It takes another 0.356 s to fall back into the water again.

So the salmon is out of the water for a total of 0.712 s.

In this time the salmon travels horizontally with a velocity of 7.2 m/s x cos 29 = 6.30 m/s

We can now calculate the horizontal distance tavelled by multiplying the horizontal velocity by the time spent out of water;

0.712 s x 6.30 m/s = 4.48 m

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Shale, sandstone, and limestone are the most commoc types of sedimentary rocks. They are formed by the most common mineral that is found on or near the surface of the Earth
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It's important to do cardiovascular exercise on a regular basis because it _____.
n200080 [17]

Answer:

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4 0
3 years ago
A 4,000 kg truck is moving at +10 m/s hits a 1500 kg parked car which moves off at +10 m/s. What is the velocity of the truck
Igoryamba

Answer:

10m/s

Explanation:

Using the law of conservation of momentums

M1u1+m2u2 = (m1+m2)v

Substitute.

4000(10)+1500(10) = (4000+1500)v

40,000+15,000 = 5,500v

55000 = 5500v

v = 55000/5500

v= 10m/s

Hence the velocity of the truck after Collision is 10m/s

6 0
3 years ago
A 56 kg sprinter, starting from rest, runs 49 m in 7.0 s at constant acceleration.what is the sprinter's power output at 2.0 s,
alexgriva [62]
The sprinter is in uniform accelerated motion, and its initial velocity is zero, so the relationship betwen space (S) and time (t) is
S= \frac{1}{2} a t^2
where a is the acceleration. Using the data of the problem, we can find a:
a= \frac{2S}{t^2} = \frac{2 \cdot 49 m}{(7.0 s)^2} =2.0 m/s^2
So now we can solve the 3 parts of the problem.

a) power output at t=2.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(2.0 s)=4.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(4.0 m/s)^2=448 J

and so the power output is
P= \frac{E}{t} = \frac{448 J}{2.0 s} =224 W

b) power output at t=4.0s 
The velocity at t=4.0 s is
v(t)=at=(2.0 m/s^2)(4.0 s)=8.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(8.0 m/s)^2=1792 J

and so the power output is
P= \frac{E}{t} = \frac{1792 J}{4.0 s} =448 W

c) Power output at t=6.0 s
The velocity at t=2.0 s is
v(t)=at=(2.0 m/s^2)(6.0 s)=12.0 m/s

the kinetic energy of the sprinter is
K= \frac{1}{2} mv^2= \frac{1}{2}(56 kg)(6.0 m/s)^2=4032 J

and so the power output is
P= \frac{E}{t} = \frac{4032 J}{6.0 s} =672 W
8 0
3 years ago
What holds the positively charged metal ions together within a metallic bond?
enot [183]

Answer:

The electron cloud

Explanation:

Metallic bonds result from interaction of positively charged metal ions with free valence electrons which now forms an electron cloud around the metal ions. Electrostatic interaction between the metal ions and the electron cloud holds the metal ions together in the metallic bond.

4 0
3 years ago
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