Landon attends an early childhood program that is located at a community center which also runs an adult care program.

An arrow, starting from rest, leaves the bow with a speed of25

xz_007 [3.2K]

Answer:

The speed will be $v_f= 43.30 \frac{m}{s}$

Explanation:

We can use the following kinematics equation

$v_f^2-v_i^2=2 \ a \ d$

where $v_f$ is the final speed, $v_i$ its the initial speed, a is the acceleration, and d the distance.

The force will be tripled, the force is:

$\vec{F} = m \vec{a}$

in 1D

$F = m a$

Now, for the original problem, we have

$(25 \frac{m}{s})^2-(0 \frac{m}{s})^2=2 \ a' \ d'$

$(25 \frac{m}{s}))^2=2 \ a' \ d'$

$625 \frac{m^2}{s^2}=2 \ a' \ d'$

For the second problem, we have

$(v_f})^2-(v_i)^2=2 \ a'' \ d''$

starting from the rest, we have the same initial velocity.

$(v_f})^2-(0\frac{m}{s})^2=2 \ a'' \ d''$

$(v_f})^2=2 \ a'' \ d''$

As the force is tripled, we have:

$F'' = 3 F'$

$m'' \ a'' = 3 m' \ a'$

But the mass its the same, so

$m' \ a'' = 3 m' \ a'$

$a'' = 3 \ a'$

So the acceleration its also tripled.

$(v_f})^2=2 \ (3 * a') \ d''$

$(v_f})^2=3 ( 2 \ ( a' \ d'' )$

As the distance traveled by the arrow must also be the same, we have:

$(v_f})^2=3 ( 2 \ ( a' \ d' )$

$(v_f})^2=3 (625 \frac{m^2}{s^2})$

$v_f= \ sqrt{3 (625 \frac{m^2}{s^2})}$

$v_f= \ sqrt{3} 25 \frac{m}{s}$

$v_f= 43.30 \frac{m}{s}$

And this will be the speed from the arrow leaving the bow.

5 0

3 years ago

A 81 kg block is released at a 3.8 m height. the track is frictionless. the block travels down the track, hits a massless spring

Alenkinab [10]

Since the track is friction less, blockâ€™s kinetic energy at the bottom of the track is equal to its potential energy at the top of the track.
PE = 81 * 9.8 * 3.8 = 3016.44 J
Work = 1/2 * 1888 * d^2
PE = Kinetic energy at the base.
1/2 * 1888 * d^2 = 3016.44
d = 1.78 approx 1.8
F = Ke = 1888 * 1.8 = 3398.4N

8 0

3 years ago

Total mechanical energy (The sum of kinetic and potential energy

victus00 [196]

Answer:

Emechanical=mgh+ $\frac{1}{2}$ mν²

Explanation:

The equation for the total mechanical energy is:

Emechanical=Epotential+Ekinetic

In which,

Epotential=mgh; m: mass of the body, g: gravity; h: height

Ekinetic= $\frac{1}{2}$ mν²; m: mass of the body, ν: velocity of the body

So,

Emechanical=mgh+ $\frac{1}{2}$ mν²

5 0

3 years ago

A boy runs at a speed of 3.3 m/s straight off the end of a diving board that is 3 meters above the water. How long is he airborn

svp [43]

Answer:

45

Explanation:

7 0

3 years ago

A satellite with mass 6000 kg is orbiting the planet at 2500 km above the planet's

9966 [12]

By the law of universal gravitation, the gravitational force F between the satellite (mass m) and planet (mass M) is

F = G M m / R ²

where

• G = 6.67 × 10⁻¹¹ m³/(kg•s²) is the universal gravitation constant

• R = 2500 km + 5000 km = 7500 km is the distance between the satellite and the center of the planet

Solve for M :

M = F R ² / (G m)

M = ((3 × 10⁴ N) (75 × 10⁵ m)²) / (G (6 × 10³ kg))

M ≈ 2.8 × 10¹⁴ kg

6 0

2 years ago