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mrs_skeptik [129]
3 years ago
7

A student is provided with a battery-powered toy car that the manufacturer claims will always operate at a constant speed. The s

tudent must design an experiment in order to test the validity of the claim. Which of the following measuring tools can the student use to test the validity of the claim?
a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on.
b. A meterstick to measure the distance of the track that the car travels on.
c. A motion detector that is oriented perpendicular to the direction that the car travels.
d. A mass balance to determine the mass of the car
Physics
1 answer:
guapka [62]3 years ago
5 0

Answer:

a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on.

b. A meterstick to measure the distance of the track that the car travels on.

Explanation:

Physics can be defined as the field or branch of science that typically deals with nature and properties of matter, motion and energy with respect to space, force and time.

In this scenario, a student is provided with a battery-powered toy car that the manufacturer claims will always operate at a constant speed. The student must design an experiment in order to test the validity of the claim.

Therefore, to test the validity of the claim, the student should use the following measuring tools;

a. Photogates placed at the beginning, end, and at various locations along the track that the car travels on. This device is typically used to measure time with respect to the rate of change of the interruption or block of an infra-red beam.

b. A meterstick to measure the distance of the track that the car travels on.

Hence, with these two devices the student can effectively measure or determine the validity of the claim.

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We push a 38.4-kg box across the floor at constant velocity. If we are pushing a horizontal force of 238 N, find the coefficient
zmey [24]

Answer: 0.62

Explanation:

Coefficient of friction is defined as the ratio of the moving force (Fm) acting on a body to the normal reaction (R).

Note that the normal reaction acts vertically on the object and is equal to the objects weight (W) i.e W=R

Since W = mg, W = 38.4 ×10

W= 384N =R

Normal reaction = 384N

The horizontal force acting on the body will be the moving force which is 238N

Coefficient of friction = Fm/R

Coefficient of friction = 238/384

Coefficient of friction = 0.62

Therefore, coefficient of kinetic friction between the box and the floor is 0.62

5 0
3 years ago
A pendulum is observed to complete 20 full cycles in 60 seconds. Determine the period and the frequency of the pendulum.
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i hope i have been useful buddy.

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5 0
3 years ago
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Which of the following is true about a planet orbiting a star in uniform circular motion? A. The direction of the velocity vecto
Luda [366]
<span>As it is uniform circular motion therefore speed is constant. Therefore we can rule out option B. Also in circular motion the direction of velocity vector changes therefore velocity can't be constant. Therefore option B is incorrect as well. Also centripetal acceleration is always towards the center so option D is wrong as well. That implies option A is correct.</span>
4 0
3 years ago
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A grindstone of mass 12 kg and radius 0.3 m is initially rotating freely at 48 rad/sec. An axe is brought into contact with the
lesya [120]

Answer:

I = 0.54\,kg\cdot m^{2}

Explanation:

1) The moment of inertia of the grindstone is:

I = \frac{1}{2}\cdot m \cdot r^{2}

I = \frac{1}{2}\cdot (12\,kg)\cdot (0.3\,m)^{2}

I = 0.54\,kg\cdot m^{2}

4 0
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Three equal 1.55-μC point charges are placed at the corners of an equilateral triangle whose sides are 0.500 m long. What is the
kati45 [8]

Answer:

0.12959085 J

Explanation:

k = Coulomb constant = 8.99\times 10^{9}\ Nm^2/C^2

q = Charge = 1.55 μC

d = Distance between charge = 0.5 m

Electric potential energy is given by

U=k\dfrac{q^2}{d}

In this system with three charges which are equidistant from each other

U=k\dfrac{q^2}{d}+k\dfrac{q^2}{d}+k\dfrac{q^2}{d}

\\\Rightarrow U=k\dfrac{3q^2}{d}\\\Rightarrow U=8.99\times 10^9\times \dfrac{3\times (1.55\times 10^{-6})^2}{0.5}\\\Rightarrow U=0.12959085\ J

The potential energy of the system is 0.12959085 J

6 0
3 years ago
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