All categories

3 years ago

What is an output force?

Physics

2 answers:

lbvjy [14]3 years ago

7 0

I believe d is the answer tell me if i am right

Vinil7 [7]3 years ago

5 0

The answer would be D

You might be interested in

Pls helpppppp i’ll give brainliest

Tema [17]

Answer:

Explanation:

The answer is Niels Bohr's planetary model, the difference between this model and all of the other models is that the Bohr's PM Is more of layers of

Nucleus - Protons and Neutrons

Electron Orbital - Period 1 Elements

2 electrons

Electron Orbital - Period 2 Elements

8 electrons

Electron Orbital - Period 3 Elements

8 electrons

If that made sense-

3 0

2 years ago

Read 2 more answers

A positive charge of 8.0 × 10-4 C is in an electric field that exerts a force of 3.5 × 10-4 N on it. What is the strength of the

Gennadij [26K]

Answer:

E = 0.437 N/C

Explanation:

Given that,

Charge, $q=8\times 10^{-4}\ C$

Electric force, $F=3.5\times 10^{-4}\ N$

Let the strength of the electric field is E. We know that, the electric force is given by :

F = qE

Where

E is the electric field strength

$E=\dfrac{F}{q}\\\\E=\dfrac{3.5\times 10^{-4}}{8\times 10^{-4}}\\E=0.437\ N/C$

So, the strength of the electric field is equal to 0.437 N/C.

6 0

3 years ago

At takeoff, a commercial jet has a speed of 72 m/s. Its tires have a diameter of 0.89 m. Part (a) At how many rev/min are the ti

34kurt

Answer:

a) Revolutions per minute = 2.33

b) Centripetal acceleration = 11649.44 m/s²

Explanation:

a) Angular velocity is the ratio of linear velocity and radius.

Here linear velocity = 72 m/s

Radius, r = 0.89 x 0. 5 = 0.445 m

Angular velocity

$\omega =\frac{72}{0.445}=161.8rad/s$

Frequency

$f=\frac{2\pi}{\omega}=\frac{2\times \pi}{161.8}=0.0388rev/s=2.33rev/min$

Revolutions per minute = 2.33

b) Centripetal acceleration

$a=\frac{v^2}{r}$

Here linear velocity = 72 m/s

Radius, r = 0.445 m

Substituting

$a=\frac{72^2}{0.445}=11649.44m/s^2$

Centripetal acceleration = 11649.44m/s²

6 0

2 years ago

1. Viruses are unique among infectious agents because they are

yuradex [85]

1,) C

2,) C

Hope this helps

4 0

3 years ago

A man pushes on a piano with mass 190 kgkg ; it slides at constant velocity down a ramp that is inclined at 18.0 ∘∘ above the ho

creativ13 [48]

Answer:

The magnitude of applied force,parallel to the incline is 575.38 N and parallel to the floor is 605 N.

Explanation:

Given:

Mass of the piano $(m)$ = 190 kg

Inclined angle $(\theta)$ = 18 degree

Considering gravity, $g$ = 9.8 $ms^-^2$

And

Using, $sin(18) =0.30$ and $cos(18)=0.95$

<em>FBD diagram is attached with all the force acting on the floor and and the inclined. </em>

We have to find the magnitude of forces,when the man pushes it parallel to the incline and to the floor.

When the man pushes it parallel to the incline.

Balancing the forces as $\sum F=0$ .

⇒ $F+mgsin(\theta) =0$

⇒ $F=-mgsin(\theta)$

⇒ Here it is negative as the force is acting downward.

⇒ Plugging the values of mass $(m)$ and angle $(\theta)$ .

⇒ $F=190\times 9.8\times sin(18)$

⇒ $F=575.38$ N

When the force is parallel to the floor.

⇒ $Fcos(\theta)=mgsin(\theta)$

⇒ $F=\frac{mgsin(\theta)}{cos(\theta)}$

⇒ Plugging the values.

⇒ $F=\frac{190\times 9.8\times sin(18)}{cos(18)}$

⇒ $F=605$ N

So,

The magnitude of applied force in inclined direction is 575.38 Newton and parallel to the floor is 605 N.

6 0

3 years ago