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3 years ago

7

A ball thrown straight upwards with an initial velocity of 15 m/s. Calculate how long the ball will remain in the air and the ma

ximum height the ball reaches?

1 answer:

GenaCL600 [577]3 years ago

3 0

Answer:

The ball remain in air for 3.06 seconds.

Maximum height reached = 22.94 m

Explanation:

We have equation of motion v = u + at

At maximum height, final velocity, v =0 m/s

Initial velocity = 15 m/s

acceleration = -9.81 m/s²

Substituting

0 = 15 -9.81 t

t = 1.53 s

Time of flight = 2 x 1.53 = 3.06 s

The ball remain in air for 3.06 seconds.

We also have equation of motion v² = u² + 2as

At maximum height, final velocity, v =0 m/s

Initial velocity = 15 m/s

acceleration = -9.81 m/s²

Substituting

0² = 15² - 2 x 9.81 x s

s = 22.94 m

Maximum height reached = 22.94 m

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A hiker is at the bottom of a canyon facing the canyon wall closest to her. She is 790.5 meters from the wall and the sound of h

tester [92]

Answer:

4.80 seconds

Explanation:

The velocity of sound is obtained from;

V= 2d/t

Where;

V= velocity of sound = 329.2 ms-1

d= distance from the wall = 790.5 m

t= time = the unknown

t= 2d/V

t= 2 × 790.5/ 329.2

t= 4.80 seconds

7 0

3 years ago

Two ropes are attached to either side of a 100.0 kg wagon as shown below. The rope on the right is pulled at an angle 40.0° rela

NikAS [45]

The acceleration of the wagon is found by applying Newton's Second Law of motion.

1. The responses for question 1 are;

x-component of the tension in the rope on the right is approximately <u>91.93 N</u>

y-component of the tension in the rope on the right is approximately <u>71.135 N</u>

x-component of the tension in the rope on the left is -80.0 N

y-component of the tension in the rope on the left is 0

2. The net force in the x-direction is approximately <u>11.93 N</u>

3. The net acceleration of the wagon in the horizontal direction is approximately <u>0.1193 m/s²</u>.

Reasons:

The given parameters are;

Mass of the wagon, m = 100.0 kg

Angle of inclination to the horizontal of the rope to the right, θ = 40.0°

Tension in the rope on the right = 120.0 N

Direction in which the rope on the left is pulled = To the west

Tension in the rope on the left = 80.0 N

1. The <em>x</em> and <em>y</em> component of the tension in the rope on the right are;

x-component = 120.0 N × cos(40.0°) ≈ <u>91.93 N</u>

y-component = 120.0 N × sin(40.0°) ≈ <u>77.135 N</u>

The <em>x</em> and <em>y</em> component of the tension in the rope on the left are;

x-component = 80.0 N × cos(180°) = <u>-80.0 N</u>

y-component = 80.0 N × sin(180°) = <u>0.0 N</u>

2. The net force in the horizontal direction, Fₓ, is found as follows;

Fₓ = The x-component of the rope on the left + The x-component of the rope on the right

Which gives;

Fₓ = 91.93 N - 80.0 N = <u>11.93 N</u>

3. The net acceleration of the block is given as follows;

According to Newton's Second Law of motion, we have;

Force in the horizontal direction, Fₓ = Mass of wagon, m × Acceleration of the wagon in the horizontal direction, aₓ

Fₓ = m × aₓ

Therefore;

$\displaystyle a_x = \frac{F_x}{m} \approx \frac{11.93 \, N}{100.0 \, kg} = \mathbf{0.1193 \ m/s^2}$

The acceleration of the wagon in the horizontal direction, aₓ ≈ <u>0.1193 m/s²</u>.

Learn more here:

brainly.com/question/20357188

8 0

3 years ago

A swift blow with the hand can break a pine board. As the hand hits the board, the kinetic energy of the hand is transformed int

Stells [14]

Answer:

A. The hand must move with a velocity of 6.98 m/s to break the board.

B. Average force on the hand = 1025 N

Explanation:

A.To determine the speed the hand must move with to break the board, the force workdone in breaking the board is found first.

Workdone = force × distance

Minimum force required = 870 N;

Distance moved by board/Deflection in order to break = 1.4 cm = 0.014 M

Workd done = 870 N × 0.014 m = 12.18 Nm or 12.18 J

This work done = Kinetic energy of the hand

Kinetic energy = mv²/2 ; where m is mass and v is velocity

Mass of hand = 0.50 Kg, velocity = ?, K.E. = 12.18 J

v² = 2 KE/m

v = √2KE/m

v = √(2 × 12.18/0.50)

v = 6.98 m/s

Therefore, the hand must move with a velocity of 6.98 m/s to break the board.

B. Average force on the hand

This can be determined using the equation of motion, v² = u² + 2as to find acceleration, since force = mass × acceleration

From the equation of motion, a = v² - u²/2s

At rest, v = 0, u = 6.98, s = 1.2 cm = 0.012 m

a = 0² - 6.98²/ 2 × 0.012

a = -2030 m/s²

Force = 2030 m/s² × 0. 50 kg = 1015 N

Therefore, Average force on the hand = 1025 N

3 0

3 years ago

For a beam of light, the direction of polarization is defined as

evablogger [386]

<span>The direction of the electric field's vibration</span>

5 0

3 years ago

Why is it important to be able to trace the pole connection on a meter back to the same type of pole at the electrical source?

lozanna [386]

Answer:

Explanation:

A grounded wire is sometimes strung along the tops of the towers to provide lightning protection.

In areas where the neutral is grounded or earthed, it is essential to endure that the neutral and the live or hot wires are not confused for each other.

When this happens, the fuses on the transformer will not operate unless the fault is very close to the transformer. The fuses in the consumer's intake box, will not operate.

6 0

3 years ago