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2 years ago

Distinguish between force and friction. class 8

1 answer:

7 0

Answer: Friction is the resistance to motion of one object moving relative to another. For example, when you try to push a book along the floor, friction makes this difficult. Force: Force is essentially a push, or a pull action, that can lead to certain outcomes.

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A leaky 10-kg bucket is lifted from the ground to a height of 10 m at a constant speed with a rope that weighs 0.9 kg/m. Initial

sladkih [1.3K]

Answer:

The work done shall be 14715 Joules

Explanation:

The work done by a force 'F' in a displacement 'dy' is given by

$W=m(y)g\times dy$

At any position 'y' the weight shall be sum of weft of water and weight of string

$\therefore m(y)=m_{water}(y)+m_{string}(y)\\\\m(y)=30(1-\frac{y}{10})+0.9y$

Thus applying values we get

$W=\int m(y)g\times dy\\\\W=\int_{0}^{10}(30(1-\frac{y}{10}+0.9y)\times g)dy\\\\Solving\\W=294.3\int_{0}^{10}(1-\frac{y}{10}+0.9y)dy\\\\W=14715J$

8 0

2 years ago

You have three objects of varying shapes and sizes: Object 1 is a rectangular block of tin. Object 2 is a cube of aluminum. Obje

Over [174]

Answer: a. m = 7.7 kg

b. V = 435.52 in³

c. m = 1927 kg

d. V = 335.37 cm³

e. m = 3 kg

Explanation: <u>Density</u> is the ratio of mass per volume, i.e., it's the measure of an object's compactness. Its representation is the greek letter ρ.

The formula for density is

$\rho=\frac{m}{V}$

Density's unit in SI is kg/m³, but it can assume lots of other units.

Some unit transformations necessary for the resolution of the question:

1 L = 1 dm³ = 1000 cm³

1 in³ = 16.3871 cm³

1 g = 0.001 kg

a. V = 1.34 L = 1340 cm³

$\rho=\frac{m}{V}$

$m=\rho.V$

m = 5.75 * 1340

m = 7705 g => 7.705 kg

Mass of object 1 with volume 1.34L is 7.7 kg.

b. A cube's volume is calculated as V = side³

V = 7.58³

V = 435.52 in³

Volume of object 2 is 435.52 in³.

c. Using 1 in³ = 16.3871 cm³ to change units:

V = 435.52 * 16.3871

V = 713689.4 cm³

Then, mass will be

$m=\rho.V$

m = 2.7 * 713689.4

m = 1926961.4 g => 1927 kg

Mass of object 2 is 1927 kg.

d. Volume of a sphere is calculated as $V=\frac{4}{3}.\pi.r^{3}$

Diameter is twice the radius, then r = 4.31 cm.

Volume is

$V=\frac{4}{3}.\pi.(4.31)^{3}$

V = 335.37 cm³

Volume of object 3 is 335.37 cm³.

e. $m=\rho.V$

m = 8.96 * 335.37

m = 3004.91 g => 3 kg

Mass of object 3 is 3 kg.

4 0

2 years ago

A catcher "gives" with a baseball when catching it. If the baseball exerts a force of 437 N on the glove, so that the glove is d

sergiy2304 [10]

437x9
is ur answer. I'm not sure tho hope it helps

5 0

3 years ago

A photon ionizes a hydrogen atom from the ground state. The liberated electron 11. recombines with a proton into the first excit

anygoal [31]

Answer:

a) 23.2 e V

b) energy of the original photon is 36.8 eV

Explanation:

given,

energy at ground level = -13.6 e V

energy at first exited state = - 3.4 e V

A photon of energy ionized from ground state and electron of energy K is released.

h ν₁ - 13.6 = K

K combine with photon in first exited state giving out photon of energy

$h\nu_2 =\dfrac{hc}{\lambda}=\dfrac{12400}{466}$

= 26.6 e V

h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°

K + ( 3.4 ) = 26.6 e V

a) energy of free electron

K = 26.6 - 3.4 = 23.2 e V

b) energy of the original photon

h ν₁ - 13.6 = K

h ν₁ = 23.2 + 13.6

= 36.8 e V

energy of the original photon is 36.8 eV

3 0

3 years ago

Find the fundamental frequency and the next three frequencies that could cause standing-wave patterns on a string that is 30.0 m

maksim [4K]

Answer:

0.786 Hz, 1.572 Hz, 2.358 Hz, 3.144 Hz

Explanation:

The fundamental frequency of a standing wave on a string is given by

$f=\frac{1}{2L}\sqrt{\frac{T}{\mu}}$

where

L is the length of the string

T is the tension in the string

$\mu$ is the mass per unit length

For the string in the problem,

L = 30.0 m

$\mu=9.00\cdot 10^{-3} kg/m$

T = 20.0 N

Substituting into the equation, we find the fundamental frequency:

$f=\frac{1}{2(30.0)}\sqrt{\frac{20.0}{(9.00\cdot 10^{-3}}}=0.786 Hz$

The next frequencies (harmonics) are given by

$f_n = nf$

with n being an integer number and f being the fundamental frequency.

So we get:

$f_2 = 2 (0.786 Hz)=1.572 Hz$

$f_3 = 3 (0.786 Hz)=2.358 Hz$

$f_4 = 4 (0.786 Hz)=3.144 Hz$

6 0

3 years ago

Distinguish between force and friction. class 8​

Distinguish between force and friction. class 8