Ask question

Ask question

All categories

3 years ago

10

An object originally at rest, is accelerated uniformly along a straight line to a speed of 8m/s in 2s. What is the acceleration

of the object?

1 answer:

Anna11 [10]3 years ago

3 0

Answer:

4m/s²

Explanation:

Initial velocity (u) = 0 m/s

Final velocity (v) = 8 m/s

Time taken (t) = 2 sec

Acceleration (a) = ?

We know

$a = \frac{v - u}{t} \\ = \frac{8 - 0}{2} \\ = \frac{8}{2} \\ = 4 \: m |s ^{2}$

Hope it will help :)

You might be interested in

Most people can hear sound of pitch raning from to hertz

MrMuchimi

Since our earlobes repeatedly bounce from our inner ear twords the eardrums, this is compeltely true.

4 0

3 years ago

What does the atomic number of an atom tell us?

Trava [24]

Answer:

the number of protons in the atom.

Explanation:

5 0

3 years ago

Read 2 more answers

The kinetic energy of a proton and that of an a-particle are 4 eV and 1 eV,

madam [21]

Answer:

<u><em>(b) 1:1</em></u>

Explanation:

<u></u>

<u>1. Formulae:</u>

E = hf

E = h.v/λ

λ = h/(mv)

E = (1/2)mv²

Where:

E = kinetic energy of the particle
λ = de-Broglie wavelength
m = mass of the particle
v = speed of the particle
h = Planck constant

<u><em>2. Reasoning</em></u>

An alha particle contains 2 neutrons and 2 protons, thus its mass number is 4.

A proton has mass number 1.

Thus, the relative masses of an alpha particle and a proton are:

$\dfrac{m_\alpha}{m_p}=4$

For the kinetic energies you find:

$\dfrac{E_\alpha}{E_p}=\dfrac{m_\alpha \times v_\alpha^2}{m_p\times v_p^2}$

$\dfrac{1eV}{4eV}=\dfrac{4\times v_\alpha^2}{1\times v_p^2}\\\\\\\dfrac{v_p^2}{v_\alpha^2}=16\\\\\\\dfrac{v_p}{v_\alpha}=4$

Thus:

$\dfrac{m_\alpha}{m_p}=4=\dfrac{v_p}{v__\alpha}$

$m_\alpha v_\alpha=m_pv_p$

From de-Broglie equation, λ = h/(mv)

$\dfrac{\lambda_p}{\lambda_\alpha}=\dfrac{m_\lambda v_\lambda}{m_pv_p}=\dfrac{1}{1}=1:1$

5 0

3 years ago

When you jump from an elevated position you usually bend your knees upon reaching the ground. By doing this, you make the time o

dsp73

Answer:

c. about 1/10 as great.

Explanation:

While jumping form a certain height when we bend our knees upon reaching the ground such that the time taken to come to complete rest is increased by 10 times then the impact force gets reduced to one-tenth of the initial value when we would not do so.

This is in accordance with the Newton's second law of motion which states that the rate of change in velocity is directly proportional to the force applied on the body.

Mathematically:

$F\propto\frac{d}{dt} (p)$

$\Rightarrow F=\frac{d}{dt} (m.v)$

since mass is constant

$F=m\frac{d}{dt}v$

when $dt=10t$

then,

$F'=m.\frac{v}{10\times t}$

$F'=\frac{1}{10} \times \frac{m.v}{t}$

$F'=\frac{F}{10}$ the body will experience the tenth part of the maximum force.

where:

$\frac{d}{dt} =$ represents the rate of change in dependent quantity with respect to time

$p=$ momentum

$m=$ mass of the person jumping

$v=$ velocity of the body while hitting the ground.

7 0

3 years ago

A cubical Gaussian surface surrounds two positive charges, each has a charge q 1 1 = + 3.90 × 10 − 12 3.90×10−12 C, and three ne

Masteriza [31]

Answer:

The electric flux is zero because charge is zero.

Explanation:

Given that,

Positive charge $q_{1}=3.90\times10^{-12}\ C$

Negative charge $q_{2}=-2.60\times10^{-12}\ C$

We need to calculate the total charged

Using formula of charge

$Q_{enc}=2q_{1}+3q_{2}$

Put the value into the formula

$Q_{enc}=2\times3.90\times10^{-12}+3\times(-2.60\times10^{-12})$

$Q_{enc}=0$

We need to calculate the electric flux

Using formula of electric flux

$\phi=\dfrac{Q_{enc}}{\epsilon_{0}}$

Put the value into the formula

$\phi=\dfrac{0}{8.85\times10^{-12}}$

Hence, The electric flux is zero because charge is zero.

7 0

3 years ago