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4 years ago

How to rearrange the formula f = ma and solve for m

Physics

2 answers:

GuDViN [60]4 years ago

7 0

F /a = m

is the answer

Svetach [21]4 years ago

4 0

Answer:

m = F/a

Explanation:

From newtons second law of motion, the force applied to mass (m) of an object is directly to its acceleration(a).

Mathematically, Force = mass×acceleration

F = ma

To derive the mass m from the formula, we will divide both sides of the equation acceleration (a)

F/a = ma/a

F/a=m

Rearranging the formula for 'm' will therefore give us;

m = F/a

Mass = Force/ acceleration

This shows that the ratio of the apples force to the acceleration of an object is equivalent to its mass.

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suppose a car manufacturer tested its cars for front end collsion by hauling them up on a crane and dropping them from a certain

IRINA_888 [86]

Initial height: 66.5 m

Explanation:

The problem can be solved by using the principle of conservation of energy.

If we neglect air resistance, the total mechanical energy of the car is conserved during the fall, therefore we can write:

$K_i + U_i = K_f + U_f$

where :

$K_i = 0$ is the kinetic energy of the car at the top (it starts from rest)

$U_i = mgh$ is the gravitational potential energy of the car at the top, with:

m = the mass of the car

g = the acceleration of gravity

h = the heigth of the car

$K_f = \frac{1}{2}mv^2$ is the kinetic energy of the car just before hitting the ground, with

v = 130 km/h final speed of the car

$U_f = 0$ is the gravitational potential energy of the car at the bottom

Re-arranging the equation, we find

$mgh=\frac{1}{2}mv^2$

and we have:

$g=9.8 m/s^2\\v = 130 km/h = 36.1 m/s$

Solving for h, we find the initial height of the car:

$h=\frac{v^2}{2g}=\frac{36.1^2}{2(9.8)}=66.5 m$

Learn more about kinetic energy and potential energy:

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5 0

3 years ago

The energy stored per unit volume of the inductor is called the energy density of the magnetic field. Why?

Marrrta [24]

Answer:

yes because physics want that

7 0

2 years ago

A roller coaster car may be approximated by a block of mass m. Thecar, which starts from rest, is released at a height h above t

elena55 [62]

Answer:

The first part can be solved via conservation of energy.

$mgh = mg2R + K\\K = mg(h-2R)$

For the second part,

the free body diagram of the car should be as follows:

- weight in the downwards direction

- normal force of the track to the car in the downwards direction

The total force should be equal to the centripetal force by Newton's Second Law.

$F = ma = \frac{mv^2}{R}\\mg + N = \frac{mv^2}{R}$

where $N = 0$ because we are looking for the case where the car loses contact.

$mg = \frac{mv^2}{R}\\v^2 = gR\\v = \sqrt{gR}$

Now we know the minimum velocity that the car should have. Using the energy conservation found in the first part, we can calculate the minimum height.

$mgh = mg2R + \frac{1}{2}mv^2\\mgh = mg2R + \frac{1}{2}m(gR)\\gh = g2R + \frac{1}{2}gR\\h = 2R + \frac{R}{2}\\h = \frac{5R}{2}$

Explanation:

The point that might confuse you in this question is the direction of the normal force at the top of the loop.

We usually use the normal force opposite to the weight. However, normal force is the force that the road exerts on us. Imagine that the car goes through the loop very very fast. Its tires will feel a great amount of normal force, if its velocity is quite high. By the same logic, if its velocity is too low, it might not feel a normal force at all, which means losing contact with the track.

7 0

3 years ago

Suppose the student in (Figure 1) is 68kg, and the board being stood on has a 12kg mass. What is the reading on the left scale?

lesantik [10]

The equilibrium conditions allow to find the results for the balance forces are:

F₁ = 225.4 N

F₂ = 558.6 N

When the acceleration is zero we have the equilibrium conditions for both linear and rotational motion.

∑ F = 0

∑ τ = 0

Where F are the forces and τ the torques.

The torque is the product of the force and the perpendicular distance to the point of support,

The free-body diagrams are diagrams of the forces without the details of the bodies, see attached for the free-body diagram of the system.

We write the translational equilibrium condition.

F₁ - W₁ - W₂ + F₂ = 0

We write the equation for the rotational motion, set our point of origin at scale 1, and the counterclockwise turns are positive.

F₂ 2 - W₁ 1 - W₂ 1.5 = 0 $\frac{W_1 \ 1 + W_2 \ 1.5}{2}$