A constant force applied to object A causes it to accelerate at 5 m/s2. The same force applied to object B causes an acceleration of 3 m/s2. Applied to object C, it causes an acceleration of 7 m/s2.
A. Which object has the largest mass?B. Which object has the smallest mass?C. What is the ratio of mass A to mass B?

Answers

Answer 1

Answer:

(A) object B has the largest mass because it has the least acceleration

(B) object C has the smallest mass because it has the largest acceleration

(C) mass A : mass B = 3 : 5

Explanation:

Given;

acceleration of object A = 5 m/s²

acceleration of object B = 3 m/s²

acceleration of object C = 7 m/s²

A constant force, F

According to Newton's second law of motion;

F = ma

m = F / a

Mass of object A:

m = F / 5

Mass of object B:

m = F / 3

Mass of object  C:

m = F / 7

(A). Which object has the largest mass:

object B has the largest mass because it has the least acceleration

(B). Which object has the smallest mass:

object C has the smallest mass because it has the largest acceleration

(C). What is the ratio of mass A to mass B;

mass A = F / 5

mass B = F / 3

[tex]mass \ A : \ mass \ B = \frac{F}{5} : \frac{F}{3} \\\\\frac{mass \ A}{mass \ B} = \frac{F}{5} * \frac{3}{F}= \frac{3}{5} \\\\mass \ A : \ mass \ B = 3: 5[/tex]

Answer 2

A. The Object B has largest mass.

B. The Object A has smallest mass.

C. The ratio of mass A to mass B is, [tex]\frac{3}{5}[/tex]

Newton second law of motion:

The second law states that the acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object.

                   [tex]F=ma\\\\m=\frac{F}{a}[/tex]

For constant force, mass is inversely proportional to acceleration of object.Given that, acceleration of object A is [tex]5m/s^{2}[/tex] and object B is [tex]3m/s^{2}[/tex]Thus, Object B has largest mass.Object A has smallest mass.the ratio of mass A to mass B is,

                     [tex]\frac{m_{A}}{m_{B}} =\frac{a_{B}}{a_{A}} =\frac{3}{5}[/tex]

Learn more about the acceleration of object here:

https://brainly.com/question/460763


Related Questions

A book of 500 leaves has a mass of 1kg if its thickness is 5cm what are the mass and thickness of each leaf

Answers

Answer:

0.002kg and 0.01cm

Explanation:

500 leaves has a thickness is 5cm

Means I leaf has a thickness of 5/500= 0.01cm

Similarly the mass of one leaf would be 1/500 =0.002kg

A ball is projected upward at time t= 0.0 s, from a point on a roof 90 m above the ground. The ball rises, then falls and strikes the ground. The initial velocity of the ball is 36.2 m/s if air resistance is negligible. The time when the ball strikes the ground is closest to:____________A. 9.0 sB. 9.4 sC. 9.7 sD. 8.7 sE. 10 s

Answers

Answer:

B. 9.4 s

Explanation:

In order to calculate the total time taken by the ball to hit the ground, we first analyze the upward motion. We will use subscript 1 for upward motion. Now, using 1st equation of motion:

Vf₁ = Vi₁ + gt₁

where,

Vf, = Final Velocity in upward motion = 0 m/s (ball stops at highest point)

Vi = Initial Velocity in upward motion = 36.2 m/s

g = - 9.8 m/s² (negative due to upward motion)

t₁ = Time taken in upward motion = ?

Therefore,

0 m/s = 36.2 m/s + (-9.8 m/s²)(t₁)

t₁ = (36.2 m/s)/(9.8 m/s²)

t₁ = 3.7 s

Now, using 2nd equation of motion:

h₁ = (Vi₁)(t₁) + (0.5)(g)(t₁)²

where,

h₁ = distance from top of building to highest point ball reaches = ?

Therefore,

h₁ = (36.2 m/s)(3.7 s) + (0.5)(-9.8 m/s²)(3.7 s)²

h₁ = 133.58 - 66.86 m

h₁ = 66.72 m

No, considering downward motion and using subscript 2, for it.

Using 2nd equation of motion:

h₂ = (Vi₂)(t₂) + (0.5)(g)(t₂)²

where,

h₂ = height of the highest point from ground = h₁ + height of building

h₂ = 66.72 m + 90 m = 156.72 m

Vi₂ = Initial Speed during downward motion = 0 m/s (ball stops for a moment at highest point)

t₂ = Time Taken in downward motion = ?

g = 9.8 m/s²

Therefore,

156.72 m = (0 m/s)(t₂) + (0.5)(9.8 m/s²)(t₂)²

t₂² = (156.72 m)/(4.9 m/s²)

t₂ = √31.98 s²

t₂ = 5.7 s

Now, the total time taken by ball to reach the ground is"

Total Time = T = t₁ + t₂

T = 3.7 s + 5.7 s

T = 9.4 s

Therefore, the correct answer is:

B. 9.4 s

A bicycle wheel has an initial angular velocity of 1.10 rad/s . Part A If its angular acceleration is constant and equal to 0.200 rad/s2 , what is its angular velocity at t = 2.50 s ? (Assume the acceleration and velocity have the same direction) Express your answer in radians per second. ω = nothing rads Request Answer Part B Through what angle has the wheel turned between t = 0 and t = 2.50 s ? Express your answer in radians. Δθ = nothing rad Request Answer Provide Feedback

Answers

Let [tex]\theta[/tex], [tex]\omega[/tex], and [tex]\alpha[/tex] denote the angular displacement, velocity, and acceleration of the wheel, respectively.

(A) The wheel has angular velocity at time [tex]t[/tex] according to

[tex]\omega=\omega_0+\alpha t[/tex]

so that after 2.50 s, the wheel will have attained an angular velocity of

[tex]\omega=1.10\dfrac{\rm rad}{\rm s}+\left(0.200\dfrac{\rm rad}{\mathrm s^2}\right)(2.50\,\mathrm s)=\boxed{1.60\dfrac{\rm rad}{\rm s}}[/tex]

(B) The angular displacement of the wheel is given by

[tex]\theta=\theta_0+\omega_0t+\dfrac\alpha2t^2\implies\Delta\theta=\omega_0t+\dfrac\alpha2t^2[/tex]

After 2.50 s, the wheel will have turned an angle [tex]\Delta\theta[/tex] equal to

[tex]\Delta\theta=\left(1.10\dfrac{\rm rad}{\rm s}\right)(2.50\,\mathrm s)+\dfrac12\left(0.200\dfrac{\rm ram}{\mathrm s^2}\right)(2.50\,\mathrm s)^2=\boxed{3.38\,\mathrm{rad}}[/tex]

You rub a balloon on your head and it becomes negatively charged. The balloon will be most attracted to what?

Answers

Answer:

To things that are positive charged

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