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2 years ago

In Figure 2, which object (A, B, or C) has the highest density? Why?

Physics

2 answers:

pishuonlain [190]2 years ago

3 0

Answer:

Explanation:

the particles are closer together, meaning it would have a higher density because density=mass/volume

jek_recluse [69]2 years ago

3 0

Answer:

Explanation:

We just have to remember that density of an object is given by the quantity of matter that is located in a certain volume, so the more matter in the less volume the greater the density, so as we can see the B image is the one that has the most volume so that wouldn´t be the one with the highest density, the A has the same volume as the C but has less particles, so the one with the highest density would be C because has the less volume and the most particles.

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Suppose a tank filled with water has a liquid column with a height of 19 meter. If the area is 2 square meters 2m squared, what’

ycow [4]

Answer:

372,400 N

Explanation:

The volume of the column is ...

V = Bh = (2 m^2)(19 m) = 38 m^3

If we assume the density is 1000 kg/m^3, then the mass of the water is ...

M = ρV = (1000 kg/m^3)(38 m^3) = 38,000 kg

The force of gravity on that mass is ...

F = Mg = (38,000 kg)(9.8 m/s^2) = 372,400 N

4 0

2 years ago

What is the particle arrangement in a liquid

natali 33 [55]

Answer:

the particle arrangement in liquid are close together with no regular arrangement

8 0

2 years ago

A 2000 kg car moving at 100 km/h crosses the top of a hill with a radius of curvature of 100 m. What is the normal force exerted

Tpy6a [65]

Answer:

The normal force the seat exerted on the driver is 125 N.

Explanation:

Given;

mass of the car, m = 2000 kg

speed of the car, u = 100 km/h = 27.78 m/s

radius of curvature of the hill, r = 100 m

mass of the driver, = 60 kg

The centripetal force of the driver at top of the hill is given as;

$F_c = F_g - F_N$

where;

Fc is the centripetal force

$F_g$ is downward force due to weight of the driver

$F_N$ is upward or normal force on the drive

$F_N = F_g-F_c\\\\F_N = mg - \frac{mv^2}{r} \\\\F_N = (60 \times 9.8) -\frac{60 \ \times \ 27.78^2 \ }{100} \\\\F_N = 588 \ N - 463 \ N\\\\F_N = 125 \ N$

Therefore, the normal force the seat exerted on the driver is 125 N.

6 0

2 years ago

2 Points

Tom [10]

The answers are A and E!

6 0

2 years ago

A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 63.0°-angle with the horizontal.

crimeas [40]

Answer:

Explanation:

Given:

length of ladder $r_L = 14m$

weight of ladder $F_L = 490N$

position of firefighter $r_F = 3.8m$

weight of firefighter $F_F = 820N$

angle of ladder $\alpha = 63$

Unknown:

force of the wall on the ladder $F_W$

force of friction on base of ladder $F_R$

normal force on base of ladder $F_N$

From the free body diagram of the sketch you get 3 equations:

$F_x = ma_x = F_W - F_R = 0\\ F_y = ma_y = F_N - F_F - F_L = 0\\ \tau _P = \overrightarrow{r} \times \overrightarrow{F} = r_FF_Fcos\alpha + \frac{1}{2}r_LF_Lcos\alpha - r_LF_Wsin\alpha = 0$