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Norma-Jean [14]
2 years ago
6

Four students measured the acceleration of gravity. The accepted value for

Physics
1 answer:
Ratling [72]2 years ago
7 0
Answer: The student with the largest measurement of percentage error is student 4. Option A is correct.The percent error shows the disparity between the observed value and the actual value. It is given by the expression.

Therefore, we can conclude that the student's measurement that has the largest percent error is student 4.







Explanation:

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A driver averaged 64 mph and took 3½ hours to travel from st. Louis to chicago. Based on this, what is the distance between st.
sammy [17]

Average speed of the driver is given as

v = 64 mph

if he moved for total time t = 3.5 hours

so the distance between the tow is given as

d = v* t

d = 64 * 3.5

d = 224 miles

so the distance between St. Louis and Chicago is 224 miles

6 0
3 years ago
A uniform marble rolls down a symmetrical bowl, start- ing from rest at the top of the left side. The top of each side is a dist
Paha777 [63]

Answer:

Part a)

h' = \frac{10}{14} h

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

h' = \frac{10}{14} h

so on smooth it will go to lower height

Explanation:

As we know by energy conservation the total energy at the bottom of the bowl is given as

\frac{1}{2} mv^2 + \frac{1}{2}I\omega^2 = mgh

here we know that on the left side the ball is rolling due to which it is having rotational and transnational both kinetic energy

now on the right side of the bowl there is no friction

so its rotational kinetic energy will not change and remains the same

so it will have

\frac{1}{2}mv^2 = mgh'

now we know that

I = \frac{2}{5}mr^2

\omega = \frac{v}{r}

so we have

\frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2 = mgh

\frac{1}{2}mv^2 + \frac{1}{5}mv^2 = mgh

\frac{7}{10}mv^2 = mgh

\frac{1}{2}mv^2 = \frac{10}{14}mgh

so the height on the smooth side is given as

h' = \frac{10}{14} h

Part b)

if both sides are rough then it will reach the same height on the other side because the energy is being conserved.

Part c)

Since marble will go to same height when it is rough while when it is smooth then it will go to the height

h' = \frac{10}{14} h

so on smooth it will go to lower height

6 0
3 years ago
Read 2 more answers
When a scientist contrasts two or more objects what is he or she looking for?
tekilochka [14]
When your Contrasting multiple objects, your looking for any differences that may or may not be their. Think Compare and Contrast if that helps, The answer to this is A
5 0
2 years ago
Read 2 more answers
A man 2 m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 23 m tall. When the man is 10 m from th
Evgen [1.6K]

Answer:

\dfrac{d\theta}{dt}=0.038\ rad/s

Explanation:

Given that

\dfrac{dx}{dt}= -1\ m/s

From the diagram

tan\theta=\dfrac{21}{x}

By differentiating with time t

sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}

When x= 10 m

tan\theta=\dfrac{21}{10}

θ = 64.53°

Now by putting the value in equation

sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}

sec^264.53^{\circ} \dfrac{d\theta}{dt}=-\dfrac{21}{10^2}\times (-1)

\dfrac{d\theta}{dt}=0.038\ rad/s

Therefore rate of change in the angle is 0.038\ rad/s

8 0
2 years ago
A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s. Assume in this problem that air drag i
stiks02 [169]

Answer:

P.E = 0.068 J = 68 mJ

Explanation:

First we need to find the height attained by the ball toy. For this purpose, we will be using 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = -9.8 m/s²  (negative sign due to upward motion)

h = height attained by the ball toy = ?

Vf = Final Velocity = 0 m/s (since it momentarily stops at the highest point)

Vi = Initial Velocity = 3 m/s

Therefore,

2(-9.8 m/s²)h = (0 m/s)² - (3 m/s)²

h = (9 m²/s²)/(19.6 m/s²)

h = 0.46 m

Now, the gravitational potential energy of ball at its peak is given by the following formula:

P.E = mgh

P.E = (0.015 kg)(9.8 m/s²)(0.46 m)

<u>P.E = 0.068 J = 68 mJ</u>

4 0
3 years ago
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