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3 years ago

The net Forward force on the propeller of a 3.2 KG model airplane is 7.0 N. What is the acceleration of the airplanes

Physics

1 answer:

Alina [70]3 years ago

3 0

Answer:

$2.1875 ms^{-2}$

Explanation:

Newton's second law: $F=ma$ Let's substitute and solve for whatever is left behind: $7.0N= 3.2kg *a[tex]a =\frac {7.0}{3.2} \frac{N}{kg} = 2.1875 \frac{kg\frac{m}{s^2}}{kg}=2.1875 \frac{m}{s^2}$

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A force of 5.00 N to the left causes a 1.35 kg book to have a net acceleration of 0.76 m/s2 to the left. What is the frictional

Nuetrik [128]

Answer:

4.0 N

Explanation:

Sum the forces in the x direction:

∑F = ma

F − Fr = ma

Fr = F − ma

Fr = 5.00 N − (1.35 kg) (0.76 m/s²)

Fr = 4.0 N

8 0

2 years ago

Hello everyone.Roads are made winding in hilly regions why?Explain

kicyunya [14]

Answer:

Because winding roads have a gentle slope on hills, so it's easy to climb it than a steepy.

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2 years ago

Which has greater kinetic energy, a car traveling at 40 mph or a half-as-massive car traveling at 80 mph?

ziro4ka [17]

Answer:

The 80 mph car

Because the formula says 1/2 mass but for the velocity it is squared

8 0

2 years ago

A woman holds a book by placing it between her hands such that she presses at right angles to the front and back covers. The boo

In-s [12.5K]

Answer:

Explanation:

In order to solve this problem we need to make a free body diagram of the book and the forces that interact on it. In the picture below you can see the free body diagram with these forces.

The person holding the book is compressing it with his hands, thus exerting a couple of forces of equal magnitude and opposite direction with value F.

Now the key to solving this problem is to analyze the equilibrium condition (Newton's third law) on the x & y axes.

To find the weight of the book we simply multiply the mass of the book by gravity.

W = m*g

W = 1.3[kg] * 9.81[m/s^2]

W = 12.75 [N]

7 0

3 years ago

The force of attraction between a -165.0 uC and +115.0 C charge is 6.00 N. What is the separation between these two charges in m

Simora [160]

Answer:

The distance between the charges is 5,335.026 m

Explanation:

To obtain the forces between the particles, we can use Coulomb's Law in scalar form, this is, the force between the particles will be:

$F = k \frac{q_1 q_2}{d^2}$

where k is Coulomb's constant, $q_1$ and $q_2$ are the charges and d is the distance between the charges.

Working a little the equation, we can take:

$d^2 = k \frac{q_1 q_2}{F}$

$d = \sqrt{ k \frac{q_1 q_2}{F}}$

And this equation will give us the distance between the charges. Taking the values of the problem

$k= 9.00 \ 10^9 \frac{N \ m^2}{C^2} \\q_1 = 165.0 \mu C \\q_2 = 115.0 C\\F=- 6.00$

(the force has a minus sign, as its attractive)

$d = \sqrt{ 9.00 \ 10^9 \frac{N \ m^2}{C^2} \frac{(165.0 \mu C) (115.0 C)}{- 6.00 \ N}}$

$d = \sqrt{ 28,462,500 \ m^2}}$

$d = 5,335.026 m$

And this is the distance between the charges.

3 0

3 years ago