Light waves can be easily blocked but ______ waves pass through all substances? ( fill in the blank)

A student measured the specific heat of water to be 4.29 J/g.Co. The

leonid [27]

Answer:

2.63 %.

Explanation:

Given that,

The calculated value of the specific heat of water is 4.29 J/g.C

Original value of specific heat of water is 4.18 J/g.C.

We need to find the student's percent error. The percentage error in any quantity is given by :

$P=\dfrac{|\text{original value-calculated value}|}{\text{original value}}\times 100\\\\P=\dfrac{4.29-4.18}{4.18}\times 100\\\\P=2.63\%$

So, the student's percent error is 2.63 %.

7 0

3 years ago

Listed below are sets of elements.

WINSTONCH [101]

Check the attached file for the answer.

8 0

3 years ago

Read 2 more answers

~~~NEED HELP ASAP~~~ Please solve each section and show all work for each section.

anastassius [24]

Explanation:

Forces on Block A:

Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass $m_A$ as

$x:\:\:(F_{net})_x = f_N - T = -m_Aa\:\:\:\:\:\:\:(1)$

$y:\:\:(F_{net})_y = N - m_Ag = 0 \:\:\:\:\:\:\:\:\:(2)$

Substituting (2) into (1), we get

$\mu_km_Ag - T = -m_Aa \:\:\:\:\:\:\:\:\:(3)$

where $f_N= \mu_kN$ , the frictional force on $m_A.$ Set this aside for now and let's look at the forces on $m_B$

Forces on Block B:

Let the x-axis be (+) up along the inclined plane. We can write the forces on $m_B$ as

$x:\:\:(F_{net})_x = T - m_B\sin30= -m_Ba\:\:\:\:\:\:\:(4)$

$y:\:\:(F_{net})_y = N - m_Bg\cos30 = 0 \:\:\:\:\:\:\:\:\:(5)$

From (5), we can solve for N as

$N = m_B\cos30 \:\:\:\:\:\:\:\:\:(6)$

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension T is given by

$T = m_A( \mu_kg + a)\:\:\:\:\:\:\:\:\:(7)$

Substituting (7) into (4) we get

$m_A(\mu_kg + a) - m_Bg\sin 30 = -m_Ba$

Collecting similar terms together, we get

$(m_A + m_B)a = m_Bg\sin30 - \mu_km_Ag$

or

$a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)$

Putting in the numbers, we find that $a = 1.4\:\text{m/s}$ . To find the tension T, put the value for the acceleration into (7) and we'll get $T = 21.3\:\text{N}$ . To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get $N = 50.9\:\text{N}$

8 0

3 years ago

A spring with a mass on the end of it hangs in equilibrium a distance of 0.4200 m above the floor. The mass is pulled down a dis

den301095 [7]

Answer:

0.48 m

Explanation:

I'm assuming that this takes place in an ideal situation, where we neglect a host of factors such as friction, weight of the spring and others

If the mass is hanging from equilibrium at 0.42 m above the floor, from the question, and it is then pulled 0.06 m below that particular position. This pulling is a means of adding more energy into the spring, when it is released, the weight compresses the spring and equals its distance (i.e, 0.06 m) above the height.

0.42 m + 0.06 m = 0.48 m

At the highest point thus, the height is 0.48 m above the ground.

3 0

3 years ago

A man walks 7 km East in 2 hours and then 2.5 km West in 1 hour. What is

madam [21]

<h2>How to solve?</h2>

We know that, Velocity is the rate of displacement covered. Displacement is the shortest path between the Initial and Final point covered by the body. So,

Velocity = Displacement / Time

And, when it comes to Average velocity, It is the total displacement by total time taken. So, By using this let's solve this question.....

<h2>Solution:</h2>

✏️ Refer to the attachment...

Let the body goes to point A that is 7 m East of the Initial point. Then it comes backward because West is opposite to East in perpendicular direction. It covers 1.5 m backwards in the same line to reach B which is the Final point.

So,

Displacement = Final point - Initial point

⇛ Displacement = 7 m - 1.5 m

⇛ Displacement = 5.5 m

Total time taken,

⇛ 2 hours + 1 hour

⇛ 3 hours

Finding Average displacement,

⇛ Total displacement / Total time taken

⇛ 5.5 m / 3 hours

⇛ 1.83333.... hours

So, the Final answer is,

$\huge{ \boxed{ \bold{ \purple{1.8 \bar{3}m}}}}$

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6 0

3 years ago