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3 years ago

While on an ice-fishing trip, you find yourself holding a 10 kg fish but stuck in the middle of

Physics

1 answer:

Anna007 [38]3 years ago

3 0

The forward force you exert on the fish and your backward action will allow you to reach the shore.

<h3>Newton's third law of motion</h3>

Newton's third law of motion states that for every action, there is an equal and opposite reaction.

Fa = -Fb

Let's assume the fish is held in the hook, this will give you the opportunity to throw the fish forward while still holding it.

When the the fish is thrown forward, you will move backwards with an equal force based on Newton's third law. Your backward momentum towards the shore will help to maintain equal linear momentum between you and the fish.

Thus, this forward force of the fish and your backward action will allow you to reach the shore.

Learn more Newton's third law of motion here: brainly.com/question/25998091

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A key falls from a bridge that is 32 m above the water. It falls directly into a model boat, moving with constant velocity, that

FinnZ [79.3K]

Answer:

Speed of the boat, v = 4.31 m/s

Explanation:

Given that,

Height of the bridge, h = 32 m

The model boat is 11 m from the point of impact when the key was released, d = 11 m

Firstly, we will find the time needed for the boat to get in this position using second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

Here, u = 0 and a = g

$t=\sqrt{\dfrac{2s}{g}}$

$t=\sqrt{\dfrac{2\times 32}{9.8}}$

t = 2.55 seconds

Let v is the speed of the boat. It can be calculated as :

$v=\dfrac{d}{t}$

$v=\dfrac{11\ m}{2.55\ s}$

v = 4.31 m/s

So, the speed of the boat is 4.31 m/s. Hence, this is the required solution.

3 0

3 years ago

A boy 12.0 m above the ground in a tree throws a ball for his dog, who is standing right below the tree and starts running the i

miss Akunina [59]

Answer:

Explanation:

Height covered = 12m

time to fall by 12 m

s = 1/2 gt²

12 = 1/2 g t²

t = 1.565 s

Horizontal distance of throw

= 8.5 x 1.565

= 13.3 m

This distance is to be covered by dog during the time ball falls ie 1.565 s

Speed of dog required = 13.3 / 1.565

= 8.5 m /s

b ) dog will catch the ball at a distance of 13.3 m .

4 0

3 years ago

A truck has shock absorbers with a spring constant of 24200 N/m. When it hits a bump, it oscillates at 0.429 Hz. What is the mas

siniylev [52]

Answer:

3331.5 kg

Explanation:

Given:

Spring constant of the spring (k) = 24200 N/m

Frequency of oscillation (f) = 0.429 Hz

Let the mass be 'm' kg.

Now, we know that, a spring-mass system undergoes Simple Harmonic Motion (SHM). The frequency of oscillation of SHM is given as:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

Rewrite the above equation in terms of 'm'. This gives,

$2\pi f=\sqrt{\frac{k}{m}}\\\\Squaring\ both\ sides,\ we\ get:\\\\(2\pi f)^2=\frac{k}{m}\\\\m=\frac{k}{4\pi^2 f^2}$

Now, plug in the given values and solve for 'm'. This gives,

$m=\frac{24200\ N/m}{4\pi^2\times (0.429\ Hz)^2 }\\\\m=\frac{24200\ N/m}{4\pi^2\times 0.184\ Hz^2}\\\\m\approx3331.5\ kg$

Therefore, the mass of the truck is 3331.5 kg.

3 0

3 years ago

Are scientific theories and laws developed in the acquisition of scientific knowledge

grigory [225]

Answer:

Scientific theories and laws develop from the acquisition of scientific knowledge.

Explanation:

4 0

3 years ago

As a science fair project, you want to launch an 800 g model rocket st raight up and hit a horizontally moving target as it pass

fredd [130]

Sum the forces in the y (upward) direction

$\sum F_y = ma$

$F_t +F_w = ma$

$15N - (0.800kg)(9.81m/s^2) = 0.800kg * a$

$a = 8.94 m/s^2$

Applying the kinematic equations of linear motion we have that the displacement as a function of the initial speed, acceleration and time is

$s = v_0t + \frac{1}{2} at^2$

$30 = 0 +\frac{1}{2} (8.94) t^2$

$t = 2.59 s$

Again through the kinematic equation of linear motion that describes velocity as the change of displacement in a given time, we have to

$v = \frac{d}{t} \rightarrow d = vt$

$d = (15m/s)(2.59s)$

$d = 38.85m$

Therefore the horizontal distance between the target and the rocket should be 38.83m

4 0

3 years ago