All categories

2 years ago

A student walks 4 block east, 7 blocks west, 1 blocks east then 2 blocks west in an hour. Her average speed is____ block/hour

Physics

1 answer:

Gemiola [76]2 years ago

4 0

Average speed = (total distance) / (total time)

Average speed = (4+7+1+2 blox) / (1 hour)

<em>Average speed = 14 blocks/hour</em>

I'm gonna go out on a limb here and take a wild guess:

I'm guessing that there's another question glued onto the end of this one, and it asks you to find either her displacement or her average velocity. I'm so sure of this that I'm gonna give you the solution for that too. If there's no more question, then you won't need this, and you can just discard it. I won't mind.

Average velocity = (displacement) / (time for the displacement)

"Displacement" = distance and direction from the start point to the end point, regardless of how she got there.

Displacement = (4E + 7W + 1E + 2W)

Displacement = (5E + 9W)

<em>Displacement = 4 blocks west</em>

Average velocity = (4 blocks west) / (1 hour)

<em>Average velocity = 4 blocks/hour West</em>

You might be interested in

If you given volume. for example 200 cm³ how can you change it to area m²

PIT_PIT [208]

Answer:

move the decimal 6 places to the left.

Explanation:

um I assume you meant to say area m^3

7 0

3 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

2 years ago

Pure water is a(n) _____.

Mashutka [201]

The correct answer would be compound

6 0

3 years ago

Ryan is driving his car to band practice. His speed is 55 mph. Splat! A bug smashes against the windshield, and bug "guts" are e

MrRa [10]

Answer:

The answer is C.

Explanation:

Every point mass attracts every single other point mass by a force acting along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them. In magnitude, the force they apply each other is the same. Therefore, the force that the windshield exerts on the bug and the force that the bug exerts on the windshield are the same magnitude.

3 0

3 years ago

Read 2 more answers

What type of force act on a man at room temperature and in a ground?

lakkis [162]

Isaac Newton’s second law of motion

6 0

3 years ago