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matrenka [14]
3 years ago
13

An object at rest cannot remain at rest unless which of the following holds?View Available Hint(s)An object at rest cannot remai

n at rest unless which of the following holds?The net force acting on it is zero.The net force acting on it is constant and nonzero.There are no forces at all acting on it.There is only one force acting on it.
Physics
1 answer:
Leni [432]3 years ago
5 0

Answer:

True The net force must be zero for the acceleration to be zero

Explanation:

In order to analyze the statements of this problem we propose your solution.

First let's look at Newton's first, which stable that every object is at rest or with constant speed unless something takes it out of this state (acceleration)

Now let's look at the second postulate, which says that force is related to the product of the mass of a body and its acceleration.

As a result of these two laws, for a body is a constant velocity the summation force on it must be zero.

Now we can analyze the statements given.

True The net force must be zero for the acceleration to be zero

False. If the force is different from zero, there is acceleration that changes the speeds

False. There may be forces, but the sum of them must be zero

False. If a force acts, the acceleration is different from zero and the speed changes

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An aircraft flying in a straight
Lelechka [254]

Answer:

t = 10.1 s

d = 2020 m

Explanation:

Time to drop from vertical rest

h = ½gt²

t = √(2h/g) = √(2(500)/9.8) = 10.1 s

d = vt = 200(10.1) = 2020 m  

3 0
3 years ago
If the earth shrank until its radius were only one-quarter its present size without changing its mass what would a 20 n object w
Dahasolnce [82]

Basing on the information given, we can calculate the new weight of the object by the following given:current weight = 20 Ng = 10m/s2

20N/4 = 5N

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5 0
3 years ago
Which of the following statements about matter is true? Mass and matter are always the same. Matter is made up of atoms and has
Zepler [3.9K]
The correct answer is Matter is made up of atoms and has mass.
7 0
3 years ago
A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w
cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

\displaystyle P=\frac{W}{t}

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

W=F.x

The second Newton's law gives us the net force as

F=m.a

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

v_f=v_o+at

v_f=at

Solving for a

\displaystyle a=\frac{v_f}{t}={5.4}{1}

a=5.4\ m/s^2

The distance traveled in the interval is given by

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Since vo=0

\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}

x=2.7\ m

The force is given by

F=m.a

We don't know the value of m, so the force is

F=2.7m

Computing the work done by the sprinter

W=F.x=2.7m(5.4)

W=14.58m

The power is finally computed

\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}

P=14.58m

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}

a=8.8\ m/s^2

The distance is

\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}

x=3.8\ m

The net force is

F=m(8.8)=8.8m

The work done by the sprinter is now computed as

W=8.8m(3.8)=33.44m

At last, the output power is

\displaystyle P=\frac{33.44m}{0.5}=66.88m

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0
3 years ago
A Carnot engine whose high-temperature reservoir is at 464 K has an efficiency of 25.0%. By how much should the temperature of t
nexus9112 [7]

Answer

given,

high temperature reservoir (T_c)= 464 K

efficiency  of reservoir (ε)= 25 %

temperature to decrease = ?

increase in efficiency = 42 %

now, using equation

 \epsilon = 1 - \dfrac{T_C}{T_H}

 0.25 = 1 - \dfrac{T_C}{464}

 T_C= (1 - 0.25) \times 464

 T_C= 0.75 \times 464

      T_C = 348 K

now,

if the efficiency is equal to 42$ = 0.42

 T_C= (1 - 0.42) \times 464

 T_C= 0.58 \times 464

 T_C= 269.12\ K

8 0
3 years ago
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