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3 years ago

A 10N force pulls to the right and friction opposes 2N. If the object is 20kg,find the acceleraton.

Physics

2 answers:

zmey [24]3 years ago

8 0

Force = mass * acceleration

10 N - 2 N = 20 kg * acceleration

8 N = 20 kg * acceleration

8 / 20 = acceleration

2/5 m/s^2 = acceleration

tia_tia [17]3 years ago

7 0

Answer:

The acceleration is 0.4 m/s²

Explanation:

Given that,

Force = 10 N

Opposing force = 2 N

Mass of object = 20 kg

We need to calculate the acceleration

Using formula of newton's second law

$F = ma$

Where, F = force

m = mass

a = acceleration

Put the value into the formula

$10N-2N = 20\times a$

$8 N=20\times a$

$a =\dfrac{8}{20}$

$a=0.4\ m/s^2$

Hence, The acceleration is 0.4 m/s²

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Fiona and her twin sister April are enjoying the bumper cars at an amusement park. Fiona drives her car toward her sisters and t

vlabodo [156]

The mass of April and her bumper car is 80 kg

Explanation:

This problem can be solved by using the law of conservation of momentum.

In fact, the total momentum of the system must be conserved before and after the collision, so we have:

$p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

where:

$m_1 = 80 kg$ is the mass of Fiona+her bumper

$u_1 = 3 m/s$ is the initial velocity of Fiona (we take her direction as positive direction)

$v_1 = -1 m/s$ is the final velocity of Fiona (she rebounds back)

$m_2$ is the mass of April+her bumper

$u_2 = - 1 m/s$ is the initial velocity of April (in the opposite direction)

$v_2 = 3 m/s$ is the final velocity of April

Re-arranging the equation, we can find the mass of April+bumper:

$m_2 = \frac{m_1 u_1 - m_1 v_1}{v_2-u_2}=\frac{(80)(3)-(80)(-1)}{3-(-1)}=80 kg$

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5 0

3 years ago

) Electric charge is uniformly distributed inside a nonconducting sphere of radius 0.30 m. the electric field at a point P, whic

aleksandr82 [10.1K]

Answer:

Explanation:

Point P is situated outside sphere . For a non conducting sphere , electric field outside sphere is given by the relation

E = k Q / d²

15000 = 9 x 10⁹ x Q / .5²

Q = 416.67 X 10⁻⁹

Maximum field will be when d = r ( radius )

maximum E = kQ / r²

= 9 x 10⁹ x 416.67 X 10⁻⁹ / .3²

= 41667 N / C

8 0

3 years ago

an astronaut is said to be weightless when he/she travels in a satellite.does it mean that the earth does not attract him/her?

IrinaK [193]

Answer:

No, gravitational force is a force having infinite range. It only reduces to an extent that astronauts cannot feel it.

Explanation:

4 0

3 years ago

Which quantity can be used to describe the complete motion of a projectile that is launched at an angle?

andreev551 [17]

That would be the velocity ... speed and direction ... of
the launch out of the cannon.

3 0

3 years ago

Coherent light with wavelength 594 nm passes through two very narrow slits, and the interference pattern is observed on a screen

weeeeeb [17]

The position of bright fringes Y_m on screen in double slit experiment is given by following expressión

$y_m={m\lambdaD}{d}$

We can solve for d, clearing the variables, so

$d=\frac{m\lambda D}{y_m}$

So making the substitution for λ= 594nm, D=3m, y_1=4.84 and m=1, we have

$d=\frac{594*10^{-9}*3*1}{4.84*10^{-3}}\\d=3.7*10^{-4}m$

Through this value we can find the position of dark fringes y_m on screen. The following expressión can approach better,

$y_m=(m+\frac{1}{2})\frac{\lambda D}{d}$

To solve the wavelength m is equal to 0,

$y_0=\frac{\lambda D}{2d}$

clearing for \lambda

$\lambda = \frac{2y_0 d}{D}$

Making the substitution for $y_0=4.84mm, d=3.7*10^{-4}, D=3m$

$\lambda=\frac{2(4.84*10^{-3})(3.7*10^{-4})}{3}\\\lambda=1.19 \mu m$

So we have that the wavelength required is $1.19 \mu m$

4 0

3 years ago