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marysya [2.9K]
3 years ago
9

I really need help with this it is my first grade so if I don’t get it right I will fail please only answer if you know (Image a

ttached)

Physics
1 answer:
hram777 [196]3 years ago
8 0

Answer:

distance travelled 7 metres and

displacement 3 metres

Explanation:

Distance travelled is the total diatance travelled by you which is 2m + 5m = 7m

Displacement is the shortest distance between two points start and stop which is 3m.

You might be interested in
If a 25 kg lawnmower produces 347 w and does 9514 J of work, for
Igoryamba

Steps 1 and 2)

The variables are W = work, P = power, and t = time. In this case, W = 9514 joules and P = 347 watts.

The goal is to solve for the unknown time t.

-----------------------

Step 3)

Since we want to solve for the time, and we have known W and P values, we use the equation t = W/P

-----------------------

Step 4)

t = W/P

t = 9514/347

t = 27.4178674351586

t = 27.4 seconds

-----------------------

Step 5)

The lawn mower ran for about 27.4 seconds. I rounded to three sig figs because this was the lower amount of sig figs when comparing 9514 and 347.

-----------------------

Note: we don't use the mass at all

6 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
Three. Grams of the same substance in different phases are stored in three different containers. Each container is barely large
zzz [600]

Complete question is;

Three grams of the same substance in different phases are stored in three different containers. Each container is barely large enough to hold the substance. Which statement can be known about the behavior of the atoms in each container?

A) The atoms in the solid would be moving vigorously around the container.

B) The atoms in the liquid would be vibrating in position.

C) The atoms in the solid would be vibrating in position.

D) The atoms in the gas would be moving slowly around the container.

Answer:

Option C - The atoms in the solid would be vibrating in position

Explanation:

Option A is wrong because the molecules in solids are held together by strong inter molecular forces and therefore can't move about freely but instead will vibrate in their position.

Option B is wrong because the atoms in liquid are more free than those in solids. Thus they will not have room to vibrate.

Option C is correct from the explanation from option A.

Option D is wrong because the atom of the gas would be very loose and free and would therefore be moving very fast around the container

5 0
3 years ago
Please answer my question!,
Karolina [17]

Answer:

Hope this helps! Mark as brainliest if liked thanks!

Explanation:

Your reasoning that the shadow is the shortest at mid-day is spot-on!

The wording of the question is the key to the answer. It says that the measurements were made in Summer. So this means that British Summer Time (BST) is being applied. BST is one hour ahead of Greenwich Mean Time and so what looks like 1pm is really 12 noon.

The safest sort of answer is to say that the shadow is shortest when the sun is at its highest point, and in this particular question that is at 1 pm because it is BST.

3 0
3 years ago
Apples are stored in a container (width = length = 3.5 ft) that is filled to a depth of 2.75 feet. If the unit density of an app
Agata [3.3K]

Answer:

The mass of the apples in the box are 1145.95 lb.

Explanation:

Given that,

Width = 3.5 ft

length = 3.5 ft

Depth = 2.75 feet

Density = 49.3 lb/ft³

We need to calculate the volume

Using formula of volume

V=l\times h\times b

Put the value

V=3.5\times3.5\times2.75

V=33.6875\ ft^3

We need to calculate the mass

Using formula of density

\rho=\dfraxc{m}{V}

m=\rho\timesV

Put the value into the formula

m=49.3\times33.6875

m=1660.79\ lb

But porosity is 31 % so apples is 69% mass of total container.

So, Total mass in pounds is

M=\dfrac{69}{100}\times1660.79

M=1145.95\ lb

Hence, The mass of the apples in the box are 1145.95 lb.

8 0
3 years ago
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