Ask question

Ask question

All categories

3 years ago

12

According to Newtons second law of motion, which is equal to the acceleration of an object?A: net force + massB: net force x mas

sC: net force - massD: net force divides mass

2 answers:

Assoli18 [71]3 years ago

5 0

D: net force divides mass

Application of F = ma.

mart [117]3 years ago

4 0

Answer:

The acceleration of an object is equal to the net force divided by the mass of the object.

(D) is correct option.

Explanation:

According to Newton's second law of motion

The net force is equal to the product of the mass of the object and the acceleration of the object.

In mathematically terms,

$F = ma$

The acceleration is

$a = \dfrac{F}{m}$

Where, a = acceleration

F = net force

m = mass

Hence, The acceleration of an object is equal to the net force divided by the mass of the object.

You might be interested in

According to one set of measurements, the tensile strength of hair is 196 MPa , which produces a maximum strain of 0.380 in the

slega [8]

Answer:

(a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

Explanation:

Given that,

Tensile strength = 196 MPa

Maximum strain = 0.380

Diameter = 50.0 μm

Length = 12.0 cm

We need to calculate the area

Using formula of area

$A=\dfrac{\pi}{4}\times d^2$

Put the value into the formula

$A=\dfrac{\pi}{4}\times(50.0\times10^{-6})^2$

$A=1.96\times10^{-9}\ m^2$

We need to calculate the magnitude of the force

Using formula of force

$F=\sigma A$

Put the value into the formula

$F=196\times10^{6}\times1.96\times10^{-9}$

$F=0.38416\ N$

(b). If the length of a strand of the hair is 12.0 cm at its breaking point

We need to calculate the unstressed length

Using formula of strain

$strain=\dfrac{\Delta l}{l_{0}}$

$\Delta l=strain\times l_{0}$

Put the value into the formula

$\Delta l=0.380\times l_{0}$

Length after expansion is 12 cm

We need to calculate the original length

Using formula of length

$l=l_{0}+\Delta l$

Put the value into the formula

$I=l_{0}+0.380\times l_{0}$

$l=1.38l_{0}$

$l_{0}=\dfrac{l}{1.38}$

$l_{0}=\dfrac{12\times10^{-2}}{1.38}$

$l_{0}=0.0869\ m$

Hence, (a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

4 0

3 years ago

A classroom is about 3 meters high, 20 meters wide and 30 meters long. If the density of air is 1.29 kg/m3, what is the mass of

Deffense [45]

Answer:

the mass of the air in the classroom = 2322 kg

Explanation:

given:

A classroom is about 3 meters high, 20 meters wide and 30 meters long.

If the density of air is 1.29 kg/m3

find:

what is the mass of the air in the classroom?

density = mass / volume

where mass (m) = 1.29 kg/m³

volume = 3m x 20m x 30m = 1800 m³

plugin values into the formula

1.29 kg/m³ = <u> mass </u>

1800 m³

mass = 1.29 kg/m³ ( 1800 m³ )

mass = 2322 kg

therefore,

the mass of the air in the classroom = 2322 kg

8 0

3 years ago

Read 2 more answers

Energy is to be stored in a flywheel in the shape of a uniform solid disk with a radius of R = 1.22 m and a mass of 67.0 kg . To

eduard

Answer: 71.7 KJ

Explanation:

The rotational kinetic energy of a rotating body can be written as follows:

Krot = ½ I ω2

Now, any point on the rim of the flywheel, is acted by a centripetal force, according to Newton’s 2nd Law, as follows:

Fc = m. ac

It can be showed that the centripetal acceleration, is related with the angular velocity and the radius, as follows:

ac = ω2 r

We know that this acceleration has a limit value, so , we can take this limit to obtain a maximum value for the angular velocity also.

As the flywheel is a solid disk, the rotational inertia I is just ½ m r2.

Replacing in the expression for the Krot, we have:

Krot= ½ (1/2 mr2.ac/r) = ¼ mr ac = ¼ 67.0 Kg. 1.22 m . 3,510 m/s2 = 71. 7 KJ

5 0

3 years ago

An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where ther

jeka94

Answer:

T = 188.5 s, correct is C

Explanation:

This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved

initial instant. Before the crash

L₀ = r m v₀ + I₀ w₀

the angular speed of the fan is zero w₀ = 0

final instant. After the crash

L_f = I₀ w + m r v

L₀ = L_f

m r v₀ = I₀ w + m r v

angular and linear velocity are related

v = r w

w = v / r

m r v₀ = I₀ v / r + m r v

m r v₀ = (I₀ / r + mr) v

v = $\frac{m}{\frac{I_o}{r} +mr} \ r v_o$

let's calculate

v = $\frac{0.020}{\frac{1.4}{0.6 } + 0.020 \ 0.6 } \ 0.6 \ 4$

v = $\frac{0.020}{2.345} \ 2.4$

v = 0.02 m / s

To calculate the time of a complete revolution we can use the kinematics relations of uniform motion

v = x / T

T = x / v

the distance of a circle with radius r = 0.6 m

x = 2π r

we substitute

T = 2π r / v

let's calculate

T = 2π 0.6/0.02

T = 188.5 s

reduce

t = 188.5 s ( 1 min/60 s) = 3.13 min

correct is C

6 0

3 years ago

A person of mass m is standing on the surface of the Earth, of mass M E . What is the acceleration that the Earth experiences du

Lana71 [14]

Answer:

$a_E=\dfrac{Gm}{r^2}$

Explanation:

$M_E$ = Mass of the Earth = 5.972 × 10²⁴ kg

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Earth = 6371000 m

m = Mass of person

The force on the person will balance the gravitational force

$M_Ea_E=\dfrac{GmM_E}{r^2}\\\Rightarrow a_E=\dfrac{Gm}{r^2}$

The acceleration that the Earth will feel is $a_E=\dfrac{Gm}{r^2}$

7 0

3 years ago