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3 years ago

Una locomotiva traina un solo vagone, che ha la sua stessa massa, con la forza di

Physics

1 answer:

gregori [183]3 years ago

3 0

Answer:

c) F = 16000 N

Explanation:

For this exercise we use Newton's second law

F = ma

they tell us that adding the other wagons the acceleration of the locomotive must be maintained

F = m a

by adding the other four wagons

mass = 4 no

therefore to maintain the force you must also raise the same factor

Fe = 4Fo

Fe = 4 4000

F = 16000 N

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Afina-wow [57]

Answer:

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2 years ago

What is the equation: If F=10 N, a=5 m/s², m=?

Romashka-Z-Leto [24]

Answer:

2 kg

Explanation:

Remember:

F = m * a re-arrange to

F/a = m substitute in the given values

10 / 5 = 2 kg

8 0

2 years ago

Which property does not apply to a motor? A. It uses an electromagnet. B. The direction of current changes twice during the rota

Rufina [12.5K]

Answer:

Explanation:

A motor converts electrical energy to mechanical energy. A current carrying coil is placed between electromagnets. A magnetic force is exerted on the coil which makes it rotate. The direction of current changes twice during the rotation of the coil. An outside source of electric current is used.

Thus, the wrong statement is D. It converts mechanical energy into electrical energy. It is the generator which converts mechanical energy into electrical energy.

3 0

3 years ago

Read 2 more answers

The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$