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1. the acceleration of an object is ______________ related to the net f…

Question

  1. the acceleration of an object is ____________ related to the net force exerted upon it and ____________ related to the mass of the object. in equation form: a = f_net / m.

a. directly, inversely b. inversely, directly c. directly, directly d. inversely, inversely

  1. use newtons second law to predict the effect of an alteration in mass or net force upon the acceleration of an object.

a. an object is accelerating at a rate of 8 m/s² when it suddenly has the net force exerted upon increased by a factor of 2. the new acceleration will be ________ m/s².
b. an object is accelerating at a rate of 8 m/s² when it suddenly has the net force exerted upon increased by a factor of 4. the new acceleration will be ________ m/s².
c. an object is accelerating at a rate of 8 m/s² when it suddenly has the net force exerted upon decreased by a factor of 2. the new acceleration will be ________ m/s².
d. an object is accelerating at a rate of 8 m/s² when it suddenly has its mass increased by a factor of 2. the new acceleration will be ________ m/s².
e. an object is accelerating at a rate of 8 m/s² when it suddenly has its mass decreased by a factor of 4. the new acceleration will be ________ m/s².
f. an object is accelerating at a rate of 8 m/s² when it suddenly has the net force exerted upon increased by a factor of 2 and its mass decreased by a factor of 4. the new acceleration will be ________ m/s².
g. an object is accelerating at a rate of 8 m/s² when it suddenly has the net force exerted upon increased by a factor of 4 and its mass increased by a factor of 2. the new acceleration will be ________ m/s².
h. an object is accelerating at a rate of 8 m/s² when it suddenly has the net force exerted upon increased by a factor of 3 and its mass decreased by a factor of 4. the new acceleration will be ________ m/s².

  1. these force diagrams depict the magnitudes and directions of the forces acting upon four objects. in each case, the down force is the force of gravity. rank these objects in order of their acceleration, from largest to smallest: ____ > __ > __ > ____

object a: 20 n, 50 n, 50 n, 50 n
object b: 50 n, 50 n, 50 n, 50 n
object c: 10 n, 40 n, 50 n, 50 n
object d: 15 n, 30 n, 50 n, 50 n

  1. a truck whose mass is 2000 kg is moving at constant speed. why do we say that the forces acting on the truck are in equilibrium? because its speed and direction arent changing
  2. a truck whose mass is 2000 kg accelerates at 20m/s². how much force is used to create that acceleration? f = 2000 kg×20 m/s² = 40,000

Explanation:

Step1: Recall Newton's second - law

Newton's second law is $a=\frac{F_{net}}{m}$, which means acceleration is directly proportional to net - force and inversely proportional to mass. So for question 1, the answer is a.

Step2: Analyze part 2a

Given $a=\frac{F_{net}}{m}$, initially $a_1 = 8m/s^2=\frac{F_{net1}}{m_1}$. When $F_{net2}=2F_{net1}$ and $m$ is constant, $a_2=\frac{F_{net2}}{m_1}=\frac{2F_{net1}}{m_1}=2a_1 = 16m/s^2$.

Step3: Analyze part 2b

When $F_{net2}=4F_{net1}$ and $m$ is constant, $a_2=\frac{F_{net2}}{m_1}=\frac{4F_{net1}}{m_1}=4a_1 = 32m/s^2$.

Step4: Analyze part 2c

When $F_{net2}=\frac{1}{2}F_{net1}$ and $m$ is constant, $a_2=\frac{F_{net2}}{m_1}=\frac{\frac{1}{2}F_{net1}}{m_1}=\frac{1}{2}a_1 = 4m/s^2$.

Step5: Analyze part 2d

Given $a_1=\frac{F_{net1}}{m_1}=8m/s^2$. When $m_2 = 2m_1$ and $F_{net}$ is constant, $a_2=\frac{F_{net1}}{m_2}=\frac{F_{net1}}{2m_1}=\frac{1}{2}a_1 = 4m/s^2$.

Step6: Analyze part 2e

When $m_2=\frac{1}{4}m_1$ and $F_{net}$ is constant, $a_2=\frac{F_{net1}}{m_2}=\frac{F_{net1}}{\frac{1}{4}m_1}=4a_1 = 32m/s^2$.

Step7: Analyze part 2f

When $F_{net2}=2F_{net1}$ and $m_2=\frac{1}{4}m_1$, $a_2=\frac{F_{net2}}{m_2}=\frac{2F_{net1}}{\frac{1}{4}m_1}=8a_1 = 64m/s^2$.

Step8: Analyze part 2g

When $F_{net2}=4F_{net1}$ and $m_2 = 2m_1$, $a_2=\frac{F_{net2}}{m_2}=\frac{4F_{net1}}{2m_1}=2a_1 = 16m/s^2$.

Step9: Analyze part 2h

When $F_{net2}=3F_{net1}$ and $m_2=\frac{1}{4}m_1$, $a_2=\frac{F_{net2}}{m_2}=\frac{3F_{net1}}{\frac{1}{4}m_1}=12a_1 = 96m/s^2$.

Step10: Analyze part 3

For object A, $F_{netA}=\sqrt{20^2}=20N$ (horizontal component as vertical forces cancel).
For object B, $F_{netB}=0N$ (all forces cancel).
For object C, $F_{netC}=\sqrt{(40 - 10)^2}=30N$ (horizontal component as vertical forces cancel).
For object D, $F_{netD}=\sqrt{15^2 + 30^2}=\sqrt{225+900}=\sqrt{1125}\approx33.5N$. Assuming mass is the same for all objects, ranking acceleration from largest to smallest: Object D > Object C > Object A > Object B.

Step11: Analyze part 5

Using $F = ma$, with $m = 2000kg$ and $a=20m/s^2$, $F=2000\times20 = 40000N$.

Answer:

  1. a. directly, inversely
  2. a. 16

b. 32
c. 4
d. 4
e. 32
f. 64
g. 16
h. 96

  1. Object D > Object C > Object A > Object B
  2. Because its speed and direction aren't changing (already given in the question)
  3. 40000N