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3 years ago

The smallest particle of an element that retains the properties of that element

Physics

1 answer:

Vika [28.1K]3 years ago

5 0

Answer:

Explanation:

Atom, smaller than that and you get into subatomic particles which don't really have the properties of elements they make up.

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A. A ball is thrown directly up with an initial speed of 4.00 m/s at y = 0. What is the maximum height h that it achieves, and w

sveta [45]

Answer:

A) t = 0.40816 s , y = 0.916 m

Explanation:

A) For this problem we use the kinematic relations

v = v₀ - g t

the highest point zero velocities (v = 0)

t = (v₀-v) / g

t = (4 - 0) / 9.8

t = 0.40816 s

to calculate the height let's use

v² = v₀² - 2 g y

y = vo2 / 2g

y = 4 2 / (2 9.8)

y = 0.916 m

To find the number of photos, we can use a direct proportions rule, if you take 30 photos in a second in 0.40816 s how many photos does it take

# _photos1 = 0.40916 (30/1)

# _photos1 = 12

yes i take 120 fps

#_fotod = 0.40916 (120/1)

#photos = 5.87 10³

B) The ball is released from a latura h how long it takes to reach the floor

v² = v₀² + 2 g y

where the initial velocity is zero and the velocity with which the expert leaves is equal to the velocity with which v = 4 m / s leaves

v² = 2gy

v = √ (2 9.8 0.916)

v = √ (2.1397 101)

v = 4.6257 m / s

c) we ask us for the time for latura

y = L / 2

y = 0.916 / 2

y = 0.458 m

now we can use the formula

y = v₀ t - ½ g t²

0.458 = 4.00 t - ½ 9.8 t²

4.9 t² - 4t + 0.458 = 0

t² -0.8163 t +0.09346 = 0

we solve second degree execution

t = [0.8163 ±√ (0.8163² - 4 0.09346)] / 2

t = [0.8163 ± 0.540] / 2

t₁ = 0.678 m

t₂ = 0.2763 m

the shortest time is for when the ball goes up and the longest when it goes down

D) the graph of vs Vs is expected to be a closed line

and the graph of position versus time a parabola

0 0

4 years ago

A 2.0 kg particle moves in a circle of radius 3.1 m. As you look down on the plane of its orbit, the particle is initially movin

Ghella [55]

Answer

given,

L(t) = 10 - 3.5 t

mass of particle = 2 Kg

radius of the circle = 3.1 m

a) torque

τ = $\dfrac{dL}{dt}$

τ = $\dfrac{d}{dt}(10 - 3.5 t)$

τ = -3.5 N.m

Particle rotates clockwise as i look down the plane. Hence, its angular velocity is downward.

L decreases the angular acceleration upward. so, net torque is upward.

b) Moment of inertia of the particle

I = m R^2

I = 2 x 3.1²

I = 19.22 kg.m²

L = I ω

ω = $\dfrac{L}{I}$

ω = $\dfrac{10 - 3.5 t}{19.22}$

ω = $0.520 - 0.182 t$

A = 0.52 rad/s B = -0.182 rad/s²

5 0

3 years ago

Read 2 more answers

A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2.

Bumek [7]

Answer:

the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

Explanation:

Given that;

speed of car V = 120 km/h = 33.3333 m/s

Reaction time of an alert driver = 0.8 sec

Reaction time of an alert driver = 3 sec

extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec

now, extra distance that car will travel in case of sleepy driver will be'

S_d = V × 2.2 sec

S_d = 33.3333 m/s × 2.2 sec

S_d = 73.3333 m

hence, number of car of additional car length n will be;

n = S_n / car length

n = 73.3333 m / 5m

n = 14.666 ≈ 15

Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15

8 0

3 years ago

If a stone dropped into a well reaches the water's surface after 3.0 seconds, how far did the stone drop before hitting the wate

pogonyaev

Let h = distance (m) to the water surface.

Initial velocity, u = 0 (because the stone was dropped).
Use the formula
h = ut + (1/2)gt^2
where g = 9.8 m/s^2 (acc. due to graity)
t = time (s)

h = (1/2)*(9.8)*(3^2) = 44.1 m

8 0

3 years ago

You’ve made the finals of the science Olympics. As one of your tasks you’re given 1.0 g of copper and asked to make a cylindrica

Pani-rosa [81]

Answer:

Length = 2.92 m

Diameter = 0.11 mm

Explanation:

We have $m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A}$ , where:

$l$ is the length

$m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3$

We divide the first equation by the second equation to get:

$\frac{m}{R} = \frac{d A^2}{\rho}$

$A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8} \ m^2$