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3 years ago

15

A container is filled with liquid the depth of the liquid is 60 CM if exertingc in pressure is 2000pa. calculator the density of

a liquid

1 answer:

riadik2000 [5.3K]3 years ago

8 0

Answer:

The density of the liquid is 340.136 kg/m³

Explanation:

Given;

depth of liquid in the container, h = 60 cm = 0.6 m

pressure of the liquid, P = 2000 Pa = 2000 N/m²

The pressure of the liquid is calculated as ;

P = ρgh

where;

ρ is density of the liquid

g is acceleration due to gravity = 9.8 m/s²

h is depth of the liquid

Make the density (ρ) the subject of the formula;

$\rho = \frac{P}{gh} \\\\\rho = \frac{2000}{9.8 \times 0.6}\\\\\rho = 340.136 \ kg/m^3$

Therefore, the density of the liquid is 340.136 kg/m³

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How many significant figures are there in : (a) 0.000054 (b) 3.001 x 10^5 (c) 5.600

melomori [17]

Answer:

(a) 2 (b) 4 (c) 4

Explanation:

Significant figures : The figures in a number which express the value -the magnitude of a quantity to a specific degree of accuracy is known as significant digits.

Rules for significant figures:

Digits from 1 to 9 are always significant and have infinite number of significant figures.
All non-zero numbers are always significant. For example: 654, 6.54 and 65.4 all have three significant figures.
All zero’s between integers are always significant. For example: 5005, 5.005 and 50.05 all have four significant figures.
All zero’s preceding the first integers are never significant. For example: 0.0078 has two significant figures.
All zero’s after the decimal point are always significant. For example: 4.500, 45.00 and 450.0 all have four significant figures.
All zeroes used solely for spacing the decimal point are not significant. For example : 8000 has one significant figure.

As per question,

0.000054 has 2 significant figures.

3.001 x 10⁵ has 4 significant figures.

5.600 has 4 significant figures.

4 0

3 years ago

A girl throws a tennis ball upward with an initial velocity of 4 m/s. What is the maximum displacement of the ball?

Jobisdone [24]

Answer:

0.82 m

Explanation:

The ball is in free fall - uniform accelerated motion with constant acceleration downward, $a=g=-9.8 m/s^2$ (acceleration of gravity). So we can use the following suvat equation to solve the problem:

$v^2-u^2=2as$

where

v is the final velocity

u = 4 m/s is the initial velocity

a is the acceleration

s is the displacement

At the maximum displacement, v = 0 (the velocity becomes zero). Substituting and solving for s, we find:

$s=-\frac{u^2}{2a}=-\frac{4^2}{2(-9.8)}=0.82 m$

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3 years ago

A 600 N astronaut travels to an asteroid where the gravitational force is one-hundredth that of Earth. What is the astronaut's w

Ganezh [65]

Answer:

A) 6N

Explanation:

the weight of the astronaut on earth can be calculated with the formula

Weight = mass * gravity of earth

Since the gravitational force is one-hundredth of Earth, the formula should be

Weight = mass * (gravity of earth / 100)

Weight = (mass * gravity of earth) / 100

Since you already know the weight, you only need to divide by 100

Weight = (mass * gravity of earth) / 100

Weight = 600N / 100

Weight = 6N

4 0

3 years ago

A metal disk of radius 6.0 cm is mounted on a frictionless axle. Current can flow through the axle out along the disk, to a slid

Galina-37 [17]

Answer:

0.09 N

Explanation:

We are given that

Radius of disk,r=6 cm= $\frac{6}{100}=0.06 m$

1 m=100 cm

B=1 T

Current,I=3 A

We have to find the frictional force at the rim between the stationary electrical contact and the rotating rim.

$dF=IBdr$

$dF=IBdr$

$\tau=rdF=IBrdr$

$\tau=\int_{0}^{R}IBr dr$

$\tau=IB(\frac{R^2}{2}$

Torque due to friction

$\tau=R\times F$

Where friction force=F

$R\times F=\frac{IBR^2}{2}$

$F=\frac{IBR}{2}$

Substitute the values

$F=\frac{3\times 1\times 0.06}{2}$

$F=0.09 N$

7 0

3 years ago

A horizontal spring with spring constant 130 N/m is compressed 17 cm and used to launch a 2.8 kg box across a frictionless, hori

olasank [31]

Explanation:

The given data is as follows.

k = 130 N/m, $\Delta x$ = 17 cm = 0.17 m (as 1 m = 100 cm)

mass (m) = 2.8 kg

When the spring is compressed then energy stored in it is as follows.

Energy = $\frac{1}{2}kx^{2}$

Now, spring energy gets converted into kinetic energy when the box is launched.

So, $\frac{1}{2}kx^{2}$ = $\frac{1}{2}mv^{2}$

$\frac{1}{2} \times 130 \times (0.17)^{2}$ = $\frac{1}{2} \times 2.8 \times v^{2}$

$v^{2} = \frac{3.757}{2.8}$

= 1.34

v = 1.15 m/sec

Now,

Frictional force = $\mu \times mg$

= $0.15 \times 2.8 \times 9.8$

= 4.116 N

Also, Kinetic energy = work done by friction

$\frac{1}{2}mv^{2} = F_{f} \times d$

$\frac{1}{2} \times 2.8 \times (1.15)^{2} = 4.116 \times d$

1.8515 = $4.116 \times d$

d = 0.449 m

Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.

8 0

3 years ago