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maxonik [38]
3 years ago
13

Newton's second law states that the force produced by a moving object depends on its mass and acceleration. Which is an example

of this? Two boulders move at the same rate of acceleration; the boulder with the smaller mass produces a greater force. Two boulders move at the same rate of acceleration; the boulder with the greater mass produces a greater force. A runner accelerates at a rate of 1 meter per second per second during a race.
Physics
1 answer:
Nutka1998 [239]3 years ago
5 0
Newton's second law is NOT talking about the force 'produced' by
a moving object.  The law is saying that if you WANT to accelerate
that mass at that rate, then THAT's the force YOU'll have to apply
in order to accomplish the job.

The only choice that kind-of illustrates Newton's law is the last one.
The runner is able to deliver enough force from the muscles in his
legs to accelerate his own mass at the rate of 1 meter per second².
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Serggg [28]

Answer:

A., an action which causes movement and an opposite reaction

Explanation:

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3 years ago
Find equivalent resistance. <br><br>Answer asap and please, please don't spam.​
Alinara [238K]

Answer:

R = 4.77 ohms

Explanation:

Four resistors are given such that,

R₁ = 2 ohms

R₂ = 3 ohms

R₃ = 5 ohms

R₄ = 10 ohms

Here, R₁ and R₂ in series. The equivalent is given by :

R₁₂ = R₁ + R₂

= 2 + 5

R₁₂ = 7 ohms

Similarly, R₃ and R₄ are in series. so,

R₃₄ = R₃ + R₄

= 10+5

R₃₄ = 15 ohms

Now, R₁₂ and R₃₄ are in parallel. So,

\dfrac{1}{R}=\dfrac{1}{R_{12}}+\dfrac{1}{R_{34}}\\\\\dfrac{1}{R}=\dfrac{1}{7}+\dfrac{1}{15}\\\\R=4.77\ \Omega

So, the equivalent resistance s 4.77 ohms.

4 0
3 years ago
A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130
konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

W=Fd      (1)

F: applied force = 130N

d: distance = 5.0m

W=(130N)(5.0m)=650J

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

W_f=F_fd=\mu Mgd       (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

W = 0J

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

\Delta K=W_T         (3)

You calculate the total work:

W_T=Fd-F_fd=(F-F_f)d     (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N

Next, you replace the values of all parameters in the equation (4):

W_T=(130N-105.84N)(5.00m)=120.80J

\Delta K=120.80J

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

W_T=\frac{1}{2}M(v_2^2-v_1^2)       (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}

The final speed of the box is 2.58m/s

5 0
3 years ago
15 points! Middle School Physical Science.
solniwko [45]

Answer:

1st t=d/s

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are all of the variations I came up with

Explanation:

the original formula for speed is s=d/t

then, I created a ratio of s/1=d/t and cross multiplied to find d=st

then, I isolated the t in one side by dividing by s on both sides.

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Because chemical changes result in different substances, they often cannot be undone. Some chemical changes can be reversed, but only by other chemical changes.
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