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astra-53 [7]
3 years ago
11

Which of the following are inertial reference frames? A. A car driving at steady speed on a straight and level road. B. A car dr

iving at steady speed up a 10∘ incline. C. A car speeding up after leaving a stop sign. D. A car driving at steady speed around a curve.

Physics
2 answers:
Aloiza [94]3 years ago
8 0

Answers:

A. A car driving at steady speed on a straight and level road.

B. A car driving at steady speed up a 10∘ incline.

Explanation:

An object is said to be in an inertial reference frame if the net force acting on the object is zero. According to Newton's second law, this also means that the acceleration of the object is also zero:

F=ma

Since F=0, a=0 as well.

Let's now analyze each case.

A. A car driving at steady speed on a straight and level road. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

B. A car driving at steady speed up a 10∘ incline. --> YES: this is an inertial reference frame, because the car is keeping a constant speed and a constant direction, so its velocity is not changing, and its acceleration is zero.

C. A car speeding up after leaving a stop sign. --> NO: this is not an intertial reference frame, because the car is speeding up, so it is accelerating.

D. A car driving at steady speed around a curve. --> NO: this is not an inertial reference frame, because the car is changing direction, therefore its velocity is changing and so the car is accelerating.

So the only two choices which are correct are A and B.

MrRissso [65]3 years ago
3 0

Answer:

I’s B

Explanation:

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Which of the following equations is balanced?
Gnesinka [82]

Answer: c

Explanation:

The way to check which one is the correct one is to simply multiply and see if there are the same number of atoms in both sides for each element.

a. 2×2 atoms of Al ≠ 3×1 atoms of Al

2×3 atoms of O = 3×2 atoms of O

BOTH MUST BE EQUAL FOR IT TO BE ADJUSTED!!!!!

b. 3×2 atoms of Al ≠ 3×1 atoms of Al

3×3 atoms of O ≠ 2×2 atoms of O

c. 2×2 atoms of Al = 4×1 atoms of Al

2×3 atoms of O = 3×2 atoms of O

BOTH ARE EQUAL, CORRECT ANSWER!!!

d. 2×2 atoms of Al ≠ 1×1 atoms of Al

2×3 atoms of O = 3×2 atoms of O

4 0
3 years ago
In which situation would an object weigh the LEAST? (assume all the objects have the same mass)
inessss [21]

Answer:

An object on the moon would weigh the LEAST among these. So correct answer is B.

Explanation:

  • Weight of an object on any place is given by:

W = Mass * Acceleration due to gravity(g)

  • It means when masses of different objects those are in different places are same, the weight of  those objects depends upon the 'g' of that particular place.
  • As we know, acceleration due to gravity on surface of moon (g') is 6 times weaker than the acceleration on surface of earth (g), which is due to the large M/R^2 of the earth than the moon.

i.e. g' = g/6 so W' = W/6

  • And in the space between the two, the object is weightless.
8 0
3 years ago
Read 2 more answers
The parking brake on a 1200kg automobile has broken, and the vehicle has reached a momentum of 7800kg.M/s. What is the velocity
AysviL [449]
Ok so the equation for momentum is:
v=p/m

So you would do:
7800/1200=6.5

So the answer is:
6.5 m/s

Hope this helps :)
6 0
3 years ago
The frequency of a physical pendulum comprising a nonuniform rod of mass 1.15 kg pivoted at one end is observed to be 0.658 Hz.
S_A_V [24]

Answer:

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

Explanation:

The angular frequency of a physical pendulum is measured by the following expression:

\omega = \sqrt{\frac{m\cdot g \cdot d}{I_{o}} }

Where:

\omega - Angular frequency, measured in radians per second.

m - Mass of the physical pendulum, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

d - Straight line distance between the center of mass and the pivot point of the pendulum, measured in meters.

I_{O} - Moment of inertia with respect to pivot point, measured in kg\cdot m^{2}.

In addition, frequency and angular frequency are both related by the following formula:

\omega =2\pi\cdot f

Where:

f - Frequency, measured in hertz.

If f = 0.658\,hz, then angular frequency of the physical pendulum is:

\omega = 2\pi \cdot (0.658\,hz)

\omega = 4.134\,\frac{rad}{s}

From the formula for the physical pendulum's angular frequency, the moment of inertia is therefore cleared:

\omega^{2} = \frac{m\cdot g \cdot d}{I_{o}}

I_{o} = \frac{m\cdot g \cdot d}{\omega^{2}}

Given that m = 1.15\,kg, g = 9.807\,\frac{m}{s^{2}}, d = 0.425\,m and \omega = 4.134\,\frac{rad}{s}, the moment of inertia associated with the physical pendulum is:

I_{o} = \frac{(1.15\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.425\,m)}{\left(4.134\,\frac{rad}{s} \right)^{2}}

I_{o} = 0.280\,kg\cdot m^{2}

The rotational inertia of the pendulum around its pivot point is 0.280\,kg\cdot m^{2}.

8 0
3 years ago
Given two vectors A=4i^+3j^ and vector<br> B=5i^-2j^.find the magnitude of each vector
Flura [38]

Answer:

<em>Magnitude of A=5</em>

<em>Magnitude of B=5.39</em>

Explanation:

<u>The magnitude of Vectors in Rectangular Form</u>

Given a vector v in its rectangular form:

\mathbf{v}=x\hat i+y\hat j

The magnitude of v is:

\mid\mid\mathbf{v}\mid \mid=\sqrt{x^2+y^2}

We are given the vectors

\mathbf{A}=4\hat i+3\hat j

\mathbf{B}=5\hat i-2\hat j

Their magnitudes are:

\mid\mid\mathbf{A}\mid \mid=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}

\mid\mid\mathbf{A}\mid \mid=5

\mid\mid\mathbf{B}\mid \mid=\sqrt{5^2+(-2)^2}=\sqrt{25+4}=\sqrt{29}

\mid\mid\mathbf{B}\mid \mid=\sqrt{29}=5.39

4 0
3 years ago
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