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3 years ago

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while doing a lab , a student found the density of a piece of pure aluminum to be 2.80 g/cm . The accepted value for the density

of aluminum is 2.70 g/cm . What was the students percent error ?

1 answer:

postnew [5]3 years ago

4 0

Look at the attachment or go online and search the question to get the answers you need

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can someone describe how someone is moving-- have examples of positive velocity negative velocity and acceleration and the perso

Rama09 [41]

Assume the motion when you are in the car or in the school bus to go to the school.

To describe the motion the first thing you need is a point of reference. Assume this is your house.

This should be a description:

When you are sitting and the car has not started to move you are at rest.

The car starts moving from rest, gaining speed, accelerating. You start to move away from your house, with a positive velocity (from you house to your school) and positive acceleration (velocity increases).

The car reaches a limit speed of 40mph, and then moves at constant speed. The motion is uniform, the velocity is constant, positive, since you move in the same direction), and the acceleration is zero.

When the car approaches the school, the driver starts to slow down. Then, you speed is lower but yet the velocity is positive, as you are going in the same direction. The acceleration is negative because it is in the opposite direction of the motion.

When the car stops, you are again at rest: zero velocity and zero acceleration.

In all the path your velocity was positive, constant at times (zero acceleration) and variable at others (accelerating or decelerating).

When you comeback home, then you can start to compute negative velocities, as you will be decreasing the distance from your point of reference (your house).

4 0

2 years ago

A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figur

morpeh [17]

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

<h3>Angular Speed of the pulley </h3>

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

$\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\$

$\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\$

Substitute the given parameters and solve for the angular speed;

$\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s$

<h3>Linear speed of the block</h3>

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: brainly.com/question/24772394

5 0

1 year ago

In a local bar, a customer slides an empty beer mug down the counter for a refill. The height of the counter is 1.42 m. The mug

telo118 [61]

Answer:

a) $V_{x}=3.72m/s$ , b) ∠=-54.83°

Explanation:

In order to solve this problem, we must start with a drawing of the situation, this will help us visualize the problem better. (See picture attached).

a)

Now, the idea is that the beer mug has a horizontal speed and no vertical speed at initial conditions. So knowing this, we can start finding the initial velocity of the mug.

In order to do so, we need to find the time it takes for the mug to reach the ground. We can find it by using the following equation:

$y=y_{0}+V_{y0}t+\frac{1}{2}a_{y}t^{2}$

We can see from the drawing that y and the initial velocity in y are zero, so we can simplify our formula:

$0=y_{0}+\frac{1}{2}a_{y}t^{2}$

so we can solve for t, so we get:

$t=\sqrt{\frac{-(2)y_{0}}{a}}$

so now we can substitute the known values, so we get:

$t=\sqrt{\frac{-(2)(1.42)}{-9.8}}$

which yields:

t=0.538s

So we can use this value to find the velocity in x:

$V_{x}=\frac{x}{t}$

When substituting we get:

$V_{x}=\frac{2m}{0.538s}$

which yields:

$V_{x}=3.72m/s$

b)

In order to solve part b, we need to find the y-component of the velocity, for which we can use the following formula:

$\Delta y=\frac{V_{f}^{2}-V_{0}^{2}}{2a}$

We know that $V_{0}$ is zero, so we can simplify the expression:

$\Delta y=\frac{V_{yf}^{2}}{2a}$

So we can solve the equation for $V_{yf}^{2}$ so we get:

$V_{yf}=\sqrt{2\Delta y a}$

and when substituting the known values we get:

$V_{yf}=\sqrt{2(-1.42m)(-9.8m/s^{2})}$

which yields:

$V_{yf}=-5.28m/s$

Once we got the final velocity in y, we can use it together with the velocity in x to find the angle.

So we can use the following formula:

$tan \theta =\frac{V_{y}}{V_{x}}$

when solving for theta we get:

$\theta = tan^{-1}(\frac{V_{y}}{V_{x}})$

We can substitute so we get:

$\theta = tan^{-1}(\frac{-5.28m/s}{3.72m/s})$

which yields:

$\theta = -54.83^{o}$

7 0

3 years ago

A 100 V battery is connected across a

Sergio039 [100]

1 mA = 0.001 A

Therefore, 5 mA = 0.001 * 5

=0.005 A

Resistance = voltage / current

= 100 / 0.005

= 20000 ohms

Current = voltage / resistance

= 25 / 20000

= 0.00125 A (or) 1.25 mA

5 0

3 years ago

A 0.70-kg basketball dropped on a hardwood floor rises back up to 65% of its original height. (a) If the basketball is dropped f

dem82 [27]

Answer:

Part a)

$Loss = 3.6 J$

Part b)

$Loss = 0.99 J$

Part C)

This is loss in terms of thermal energy due to collision with the floor

Explanation:

Part a)

Since we know that the ball rises up by 65% of initial height

so after first bounce it will lose 35% of its initial energy

so we will have

$U = mgH$

Energy Loss = 0.35 mgH[/tex]

$Loss = 0.35(0.70)(9.81)(1.5)$

$Loss = 3.6 J$

Part b)

Energy of the ball after first bounce

$U_1 = 0.65 mgH$

energy of ball after 2nd Bounce

$U_2 = 0.65(0.65 mgH)$

energy of the ball after 3rd bounce

$U_3 = (0.65)(0.65^2)mgH$

$U_3 = 0.65^3(0.70)(9.81)(1.5)$

$U_3 = 2.83 J$

Now we will have energy loss in fourth bounce given as

$Loss = 0.35 U_3$

$Loss = 0.35(2.83)$

$Loss = 0.99 J$

Part C)

This is loss in terms of thermal energy due to collision with the floor

7 0

3 years ago