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3 years ago

Formula:

Physics

1 answer:

s2008m [1.1K]3 years ago

8 0

Answer:

55N

Explanation:

Using Newton's second law of motion:

F=ma

Force=mass × acceleration

F=25×2.2

F=55N

So 55 Newtons are needed

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Jason has 13720 J of gravitational potential energy standing at the top of a cliff over the lake. If he jumps off the cliff and

Anna [14]

The conservation of energy and Newton's second law allows us to find the results about Jason's falling motion are;

The energy when reaching the water is K = 13720 J

The average force of the water to stop it is: F = 2744 N

<h3>Energy conservation.</h3><h3> </h3>

The conservation of energy is one of the most important principles of physics, stable that if there is no friction force, mechanical energy is conserved at all points.

Mechanical energy is the sum of kinetic energy plus potential energy.

Let's look for the energy at two points

Starting point. Get higher.

Em₀ = U = 13720 J

Final point. Lower down.

$Em_f$ = K

Friction in the air is negligible, so energy is conserved.

$Em_o= Em_f$

K = 13720J

<h3>Kinematics and Newton's law.</h3><h3> </h3>

They indicate that it stops 5m under the water, if we assume that the water acts with a constant force, we can use kinematics and Newton's second law to find this force.

The kinematics expression to find the acceleration is

v² =v₀² – 2ay

When it stops the speed is zero.

$a = \frac{v_o^2}{2y}$

Newton's second law is:

F = ma

F = m ( $\frac{v_o^2}{2y}$ )

The expression for the kinetic energy is:

K = ½ m v₀²

$v_o^2 = \frac{2K}{m}$

Let's substitute.

F = $m (\frac{2K}{m}) \frac{1}{2y}$

$F= \frac{K}{y}$

Let's calculate.

F= $\frac{13720}{5}$

F = 2744N

In conclusion using conservation of energy and Newton's second law we can find the results about Jason's falling motion are;

The energy when reaching the water is K = 13720 J

The average force of the water to stop it is: F = 2744 N

Learn more about energy here: brainly.com/question/14274074

6 0

2 years ago

A bicyclist is in a 50-km race. She says she had an average velocity of 35.

zlopas [31]

The correct answer to this is (A. Units Only).

It shows that there is a velocity of 35, but the units are missing.

8 0

4 years ago

The position of a particle moving on x-axis is given by x(t)=t^2 + 2, it’s average velocity in the final interval from t=1 to t=

san4es73 [151]

Answer:

The average velocity is 2 m/s.

Explanation:

The velocity of the particle is the time derivative of its position $x(t):$

$v =\frac{dx(t)}{dt} = \frac{d}{dt}[t^2+2]$

$v =2t$

Now the average from $t=1$ and $t=2$ is

$v_{avg} = \dfrac{v(2)-v(1)}{2-1} = \dfrac{2(2)-2(1)}{1}$

$\boxed{v_{avg} = 2 m/s} \text{ ( If the units are m/s)}$

Thus, the average velocity is 2 m/s.

8 0

4 years ago

An Alaskan rescue plane traveling 39 m/s drops a package of emergency rations from a height of 198 m to a stranded party of 3 ex

MAVERICK [17]

Answer:

y = 0m

y0 = 166m

v0y = 0 m/s

g = 9.8 m/s^2

t = ?

Solve for t:

y = y0 + v0y*t - (0.5)gt^2

0 = 166 - (0.5)(9.8)t^2

t = 5.82 s

Now, using time, we can solve for the range using the equation:

x = vx(t)

x = (40)(5.82)

x = 232.8 m

The impact horizontal component of velocity will be 40 m/s as velocity in terms of x is always constant. To find the impact vertical component of velocity, we use the equation:

v = v0y - gt

v = 0 - (9.8)(5.82)

v = -57.04 m/s

4 0

3 years ago

Read 2 more answers

If a projectile is fired straight up at a speed of 10 m/s, the time it takes to reach the top of its path is about

UNO [17]

Answer:

A. The time it takes the projectile to reach the top of its path is about 1 second.

Explanation:

Hi there!

The equation of the velocity of a projectile fired straight up is the following:

v = v0 + g · t

Where:

v = velocity of the projectile.

v0 = initial velocity.

g = acceleration due to gravity (≅ -9.8 m/s² considering the upward direction as positive)

t = time.

When the projectile reaches the top of its path, its velocity is zero, then, using the equation of velocity, we can solve it for the time:

v = v0 + g · t

0 = 10 m/s - 9.8 m/s² · t

t = -10 m/s / -9.8 m/s²

t = 1.0 s

The time it takes the projectile to reach the top of its path is about 1 second.

7 0

3 years ago