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2 years ago

The weight of a metal bracelet is measured to be 0.100 N in air and 0.092 N when immersed in water. Find its density

1 answer:

max2010maxim [7]2 years ago

4 0

If you subtract the two weights, 0.100 N and 0.092 N, we will get 0.008. So, the density between 0.100 and 0.092 is 0.008. :)

Answer: 0.008

I hope this helped!:)

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Plastic foam is about 0.10 times as dense as water. What weight of bricks could you stack on a 1m x 1m x 0.10m slab of foam, so

goblinko [34]

Answer: Weight = 98.1N

Explanation:

Density of water = 1000 kg/m^3

Given that the Plastic foam is about 0.10 times as dense as water. That is,

Density of plastic foam = 0.1 × 1000 = 100kg/m^3

The volume V = 1 ×1×0.1 = 0.1 m^3

Density is the ratio of mass to volume

Density = mass/volume

Let us substitute for density and volume to get mass.

100 = M/0.1

Make M the subject of formula

M = 100 × 0.1 = 10 kg

Weight = mg

Where g = 9.81 m/s

Substitute the M and g into the formula

Weight = 10 × 9.81 = 98.1 N

Therefore, the weight of the brick is 98.1 N

4 0

2 years ago

The population in the United States in 2015 was 321 million people. It is projected to increase to 438 million people by the yea

velikii [3]

It increases by 35% ......

4 0

2 years ago

Read 2 more answers

While ice skating, you unintentionally crash into a person. Your mass is 60 kg, and you are traveling east at 8.0 m/s with respe

kaheart [24]

Answer:

6.18 m/s

Explanation:

Roller skate collision

The final direction of the system (me=M + person=P) velocity vector is at an angle; Ф, to the direction running south to north. Apply the component form of the impulse-momentum equation, firstly;

x-axis component form (+x east);

$P_{Miy}$ + $p_{Piy}$ + $j_{y}$ = $P_{Mfy}$ + $P_{pfy}$

$m_{Mu_{Miy}+ m_{pu_{piy}}+0=(m_{M}+m_{p})V_{f} sin$ Ф

60 ·8 + 0 = (60 + 80) $V_{f}sin$ Ф

480 = 140 $V_{f} sin$ Ф................. (I)

y-axis component form (+y north);

$P_{Mix}$ + $p_{Pix}$ + $j_{x}$ = $P_{Mfx}$ + $P_{pfx}$

$m_{Mu_{Mix}+ m_{pu_{pix}}+0=(m_{M}+m_{p})V_{f} cos$ Ф

0 + 80.9 = (60 + 80) $V_{f}cos$ Ф

720= 140 $V_{f}cos$ Ф

140Vf= $\frac{720}{cos}$ Ф......................................(2)

Substituting (2) into (1) to give the angle;

480 = 720tan Ф

Ф = arctan(0.67) =33.69°.......................(3)

Evaluating (1) with (3) gives the velocity magnitude

480 = 140Vfsin 33.69°

Vf=6.18 m/s

note 1:

This angle corresponds to a direction; 90° - 33.69° = 56.31° north of east.

7 0

2 years ago

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane.

myrzilka [38]

Answer:

a. The plane speeds up but the cargo does not change speed.

Explanation:

Just to make it clear, the question is as follows from what I understand.

A small glider is coasting horizontally when suddenly a very heavy piece of cargo falls out of the bottom of the plane. You can neglect air resistance.

Just after the cargo has fallen out:

a. The plane speeds up but the cargo does not change speed.

b. The cargo slows down but the plane does not change speed.

c. Neither the cargo nor the plane change speed.

d. The plane speeds up and the cargo slows down.

e. Both the cargo and the plane speed up.

And we are requested to choose the right answer under the given conditions. We know the glider has no motor, then it must be in free fall movement, then it is experiencing some force that pulls it to the from due to the gravity effect on it, and a force in general is calculated by

F=m*a, m:= mass of the object, a:= acceleration.

Here we are only considering the horizontal effect of the forces, then since the mass is reduced the acceleration must increase to compensate and maintain the equilibrium of the forces, then the glider being lighter can travel faster due to the acceleration. On the other hand by the time the cargo left the glider there was no acceleration and the speed it had at the moment he left the plane continues, then the cargo does not change its speed, then horizontally speaking the answer would be a. The plane speeds up but the cargo does not change speed.

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2 years ago

Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat

Alja [10]

Answer:

Explanation:

Given that, the distance between the electrode is d.

The electron kinetic energy is Ek when the electrode are at distance "d" apart.

So, we want to find the K.E when that are at d/3 distance apart.

K.E = ½mv²

Note: the mass doesn't change, it is only the velocity that change.

Also,

K.E = Work done by the electron

K.E = F × d

K.E = W = ma × d

Let assume that if is constant acceleration

Then, m and a is constant,

Then,

K.E is directly proportional to d

So, as d increase K.E increase and as d decreases K.E decreases.

So,

K.E_1 / d_1 = K.E_2 / d_2

K.E_1 = E_k

d_1 = d

d_2 = d/3

K.E_2 = K.E_1 / d_1 × d_2

K.E_2 = E_k × ⅓d / d

Then,

K.E_2 = ⅓E_k

So, the new kinetic energy is one third of the E_k

7 0

2 years ago