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2 years ago

¿Qué distancia recorre un móvil que lleva una aceleración de 5m/s durante 10seg?

Physics

1 answer:

coldgirl [10]2 years ago

3 0

Answer:

250 m

Explanation:

The car in this problem is moving of uniform accelerated motion, so we can use the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance covered

u is the initial velocity

t is the time

a is the acceleration

Assuming the car starts from rest,

u = 0

Also we know that

a = 5 m/s^2 (acceleration of the car)

t = 10 s

Substituting, we find the distance covered:

$s=0+\frac{1}{2}(5)(10)^2=250 m$

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50 points if brainliest !!!!!!!!!!

Natasha_Volkova [10]

Answer:

$weight = mg \\ = 70 \times 10 \\ = 700 \: newtons$

$force = 700 \: newtons$

$force = mass \times acceleration \\ 700 = 70 \times a \\ a = 10 \: {ms}^{ - 2}$

4 0

3 years ago

What is the only possible value of ml for an electron in an s orbital?

Archy [21]

Answer:

zero

Explanation:

$m_l$ is the magnetic quantum number.

The only possible value for the magnetic quantum number for an electron in an s orbital is 0.

The first three quantun numbers are:

n: principal quantum number. It may have positive integer values: 1, 2, 3, 4,5, 6, 7, ...

$l$ : Azimuthal or angular momentum quantum number. It may have integer values from 0 to n - 1.

This quantum number is related to the type (or shape) of the orbital:

For s orbitals $l=0$

For p orbitals $l=1$

For d orbitals $l=2$

For f orbitals $l=3$

In this case, it is an s orbital, so we have $l=0$ .

$m_l$ , the third quantum number can have integer values ${from-l}$ to ${+l}$

Since, for the s orbitals $l=0$ , the only possible value for ${m_l}$ is zero.

4 0

3 years ago

A batter hits a pop fly, and the baseball (with a mass of 148 g) reaches an altitude of 265 ft. If we assume that the ball was 3

den301095 [7]

Answer:

The increase in potential energy of the ball is 115.82 J

Explanation:

Conceptual analysis

Potential Energy (U) is the energy of a body located at a certain height (h) above the ground and is calculated as follows:

U = m × g × h

U: Potential Energy in Joules (J)

m: mass in kg

g: acceleration due to gravity in m/s²

h: height in m

Equivalences

1 kg = 1000 g

1 ft = 0.3048 m

1 N = 1 (kg×m)/s²

1 J = N × m

Known data

$h_2 = 265ft * \frac{0.3048m}{ft} = 80.77m$

$h_1 = 3ft * \frac{0.3048m}{ft} = 0.914m$

$m = 148g*\frac{1kg}{1000g} = 0.148kg$

$g = 9.8 \frac{m}{s^2}$

Problem development

ΔU: Potential energy change

ΔU = U₂ - U₁

U₂ - U₁ = mₓgₓh₂ - mₓgₓh₁

U₂ - U₁ = mₓg(h₂ - h₁)

$U_2 - U_1 = 0.148kg * 9.8 \frac{m}{s^2}*(80.77m - 0.914m) = 115.82 N * m = 115.82J$

The increase in potential energy of the ball is 115.82 J

5 0

3 years ago

A clarinetist, setting out for a performance, grabs his 3.070 kg clarinet case (including the clarinet) from the top of the pian

Cerrena [4.2K]

Answer:

the vertical acceleration of the case is 1.46 m/s

Explanation:

Given;

mass of the clarinet case, m = 3.07 kg

upward force applied by the man, F = 25.60 N

Apply Newton's second law of motion;

the upward force on the clarinet case = its weight acting downwards + downward force due to its downward accelaration

F = mg + m(-a)

the acceleration is negative due to downward motion from the top of the piano.

F = mg - ma

ma = mg - F

$a = \frac{mg - F}{m} \\\\a = \frac{(3.07 \times 9.8) \ - \ 25.6}{3.07} \\\\a = 1.46 \ m/s^2$

Therefore, the vertical acceleration of the case is 1.46 m/s²

4 0

3 years ago

An electron is released from rest at the negative plate of a parallel plate capacitor and accelerates to the positive plate (see

mash [69]

Answer:

(7.90 × 10⁻¹⁵) J

Explanation:

The electric force exerted on the elecrron by rhe electric field is given by

F = qE

where |q| = charge on the particle = (1.602 × 10⁻¹⁹) C

E = magnitude of the electric field = (2.9 × 10⁶) V/m or N/C

F = 1.602 × 10⁻¹⁹ × 2.9 × 10⁶ = (4.646 × 10⁻¹³) N

From Newton's first law of motion relation, we can obtain the acceleration this force confers on the electron

F = ma

m = mass of the electron = (9.11 × 10⁻³¹) kg

a = acceleration of the electron caused by the electric force = ?

(4.646 × 10⁻¹³) = (9.11 × 10⁻³¹) × a

a = (4.646 × 10⁻¹³)/(9.11 × 10⁻³¹)

a = (5.10 × 10¹⁷) m/s²

Now, using the equations of motion, we can obtain the velocity with which the electron reaches the positive plate

u = initial velocity of the electron = 0 m/s (since the electron was initially at rest)

v = final velocity of the electron = ?

a = acceleration of the electron = (5.10 × 10¹⁷) m/s²

y = distance covered by the electron = 1.7 cm = 0.017 m

v² = u² + 2ay

v² = 0² + 2(5.10 × 10¹⁷)(0.017)

v² = (1.734 × 10¹⁶)

v = 131,677,182.5 m/s = (1.32 × 10⁸) m/s

Kinetic energy with which the electron hits the positive plate = (1/2)(m)(v²) = (1/2)(9.11 × 10⁻³¹)(1.32 × 10⁸)² = (7.90 × 10⁻¹⁵) J

Hope this Helps!!!

3 0

2 years ago