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3 years ago

GIVING BRAINLIEST!!!!

Physics

2 answers:

Svetach [21]3 years ago

8 0

1 is 1 pecan tree and tree 4 answer 2. Is natural selection or seduction I’m not shur

Likurg_2 [28]3 years ago

6 0

Answer:

1. Pecan tree 1 and pecan tree 4

2. I don't know

Explanation:

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An object is hanging from a rope. When it is held in air the tension in the rope is 8.86 N, and when it is submerged in water th

SSSSS [86.1K]

Answer:

8684.2 kg/m³

Explanation:

Tension in the rope as a result of the weight = 8.86 N

Tension in the rope when submerge in water = 7.84

upthrust = 8.86 - 7.84 =1.02 N = mass of water displaced × acceleration due to gravity

Mass of water displaced = 1.02 / 9.81 = 0.104 kg

density of water = mass of water / volume of water

make volume subject of the formula

volume of water displaced = mass / density ( 1000) = 0.104 / 1000 = 0.000104 m³

volume of the object = volume of water displaced

density of the object = mass of the object / volume of the object = (8.86 / 9.81) / 0.000104 = 0.9032 / 0.000104 = 8684.2 kg/m³

8 0

3 years ago

You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with

Misha Larkins [42]

Answer:

Sam's and Abigail speeds before colliding were a. 12.34 m/s and b. 2.86 m/s, respectively. Their total kinetic energy was diminished by c. 1484.42 J, approximately

Explanation:

By conservation of momentum, we have

$\Delta \vec{p} = 0\\\\m\vec{v}_i = m\vec{v}_f\\m_S v_S\hat{i} + m_A v_A\hat{j} = m_S \times 7 \cos{(40)} \hat{i} + m_S \times 7 \sin{(40)} \hat{j} + m_A \times 9.4 \cos{(18)} \hat{i} - m_A \times 9.4 \sin{(18)} \hat{j}.$

Writing for each direction at a time,

$m_S v_S = 7m_S \cos{(40)} +9.4 m_A \cos{(18)}\\v_S = 7 \cos{(40)} + 9.4 \frac{m_A}{m_S} \cos{(18)} = 7 \cos{(40)} + 9.4\times\frac{57}{73} \cos{(18)} \approx \mathbf{12.34}\\m_A v_A = 7m_S\sin{(40)} - 9.4m_A\sin{(18)}\\v_A = 7\frac{m_S}{m_A}\sin{(40)} - 9.4\sin{(18)} = 7\times\frac{73}{57}\sin{(40)} - 9.4\sin{(18)} \approx \mathbf{2.86}.$

Their kinetic energy changed by

$K_f - K_i = \left( \frac{1}{2}m_s v_{fs}^2 + \frac{1}{2}m_a v_{fa}^2 \right) - \left( \frac{1}{2}m_s v_{is}^2 + \frac{1}{2}m_a v_{ia}^2 \right) = \left( \frac{1}{2}\times 73 \times 7^2 + \frac{1}{2}\times 57 \times 9.4^2 \right) - \left( \frac{1}{2}\times 73 \times 12.34^2 + \frac{1}{2}\times 57 \times 2.86^2 \right) \approx \mathbf{-1484.418 J}.$

3 0

3 years ago

A 1.150 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.900 N to

shutvik [7]

k = 5.29

a = 0.78m/s²

KE = 0.0765J

<u>Explanation:</u>

Given-

Mass of air tracker, m = 1.15kg

Force, F = 0.9N

distance, x = 0.17m

(a) Effective spring constant, k = ?

Force = kx

0.9 = k X0.17

k = 5.29

(b) Maximum acceleration, m = ?

We know,

Force = ma

0.9N = 1.15 X a

a = 0.78 m/s²

c) kinetic energy, KE of the glider at x = 0.00 m.

The work done as the glider was moved = Average force * distance

This work is converted into kinetic energy when the block is released. The maximum kinetic energy occurs when the glider has moved 0.17m back to position x = 0

As the glider is moved 0.17m, the average force = ½ * (0 + 0.9)

Work = Kinetic energy

KE = 0.450 * 0.17

KE = 0.0765J

4 0

3 years ago

How does an air mattress protect a stunt person landing on the ground after a stunt?

Fudgin [204]

Answer:

a) It reduces the kinetic energy loss of the stunt person.

6 0

3 years ago

A motorboat traveling on a straight course slows

never [62]

Answer:

2.572 m/s²

Explanation:

Convert the given initial velocity and final velocity rates to m/s:

65 km/h → 18.0556 m/s
35 km/h → 9.72222 m/s

The motorboat's displacement is 45 m during this time.

We are trying to find the acceleration of the boat.

We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.

v² = v₀² + 2aΔx

Substitute the known values into the equation.

(9.72222)² = (18.0556)² + 2a(45)
94.52156173 = 326.0046914 + 90a
-231.4831296 = 90a
a = -2.572

The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².

3 0

3 years ago