All categories

3 years ago

An objest with the mass of 2 kg is accelerated at 4 m/s2. the net force acting on the object is

Physics

1 answer:

poizon [28]3 years ago

5 0

From Newton’s 2nd law of motion
F=ma
= 2*4
=8N

8N should be the answer

You might be interested in

A string has its 4th harmonic at 31.5 Hz. What is the fundamental frequency?

seropon [69]

Given data

*The given 4th harmonic frequency is 31.5 Hz

The fundamental frequency is calculated as

$\begin{gathered} f_n=\frac{31.5}{4} \\ =7.875\text{ Hz} \end{gathered}$

Hence, the fundamental frequency is 7.875 Hz

5 0

1 year ago

A car generates 2,552 N and weighs 2,250 pounds. what is the rate of acceleration

Lilit [14]

Ans: a = 2.50 m/s^2

Explanation:
First convert the mass in its standard unit i.e. kilogram(kg):
2250 lbs = 1020.583kg

Next use Newton's Second law:
F = ma

Where F = 2552N
m = 1020.583kg

=> a = (2552/1020.583)
a = 2.50 m/s^2

6 0

3 years ago

In thermodynamics, work is typically done by

kondor19780726 [428]

The answer is gases

3 0

3 years ago

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. What is the minimum coef

navik [9.2K]

An advertisement for an all-terrain vehicle (ATV) claims that the ATV can climb inclined slopes of 35°. The minimum coefficient of static friction needed for this claim to be possible is 0.7

In an inclined plane, the coefficient of static friction is the angle at which an object slide over another.

As the angle rises, the gravitational force component surpasses the static friction force, as such, the object begins to slide.

Using the Newton second law;

$\sum F_x = \sum F_y = 0$

$\mathbf{mg sin \theta -f_s= N-mgcos \theta = 0 }$

So; On the L.H.S

$\mathbf{mg sin \theta =f_s}$

$\mathbf{mg sin \theta =\mu_s N}$

On the R.H.S

N = mg cos θ

Equating both force component together, we have:

$\mathbf{mg sin \theta =\mu_s \ mg \ cos \theta}$

$\mathbf{sin \theta =\mu_s \ \ cos \theta}$

$\mathbf{\mu_s = \dfrac{sin \theta }{ cos \theta}}$

From trigonometry rule:

$\mathbf{tan \theta= \dfrac{sin \theta }{ cos \theta}}$

∴

$\mathbf{\mu_s =\tan \theta}}$

$\mathbf{\mu_s =\tan 35^0}}$

$\mathbf{\mu_s = 0.700}}$

Therefore, we can conclude that the minimum coefficient of static friction needed for this claim to be possible is 0.7

Learn more about static friction here:

brainly.com/question/24882156?referrer=searchResults

8 0

3 years ago

A 15.0-μF capacitor is charged by a 130.0-V power supply, then disconnected from the power and connected in series with a 0.280-

SVETLANKA909090 [29]

The resonant frequency of a circuit is the frequency $\omega_0$ at which the equivalent impedance of a circuit is purely real (the imaginary part is null).

Mathematically this frequency is described as

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})$

Where

L = Inductance

C = Capacitance

Our values are given as

$C = 15*10^{-6}\mu F$

$L = 0.280*10^{-3}mH$

Replacing we have,

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{LC}})$

$f = \frac{1}{2\pi}(\sqrt{\frac{1}{(15*10^{-6})(0.280*10^{-3})}})$

$f= 2455.81Hz$

From this relationship we can also appreciate that the resonance frequency infers the maximum related transfer in the system and that therefore given an input a maximum output is obtained.

For this particular case, the smaller the capacitance and inductance values, the higher the frequency obtained is likely to be.

7 0

3 years ago