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3 years ago

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A 180 cm length of string has a mass of 5.0 g. it is stretched with a tension of 8.6 n between fixed supports. (a) what is the w

ave speed for this string

1 answer:

Svetlanka [38]3 years ago

5 0

P U S S Y

<span>Joy is planning to purchase a sweater that costs $30 dollars at her local department store. The sweaters are on sale for 20% off. Which steps are needed to find the sale price of the sweater?</span>

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A 1.5kw vacuum cleaner is used for half an hour. Calculate its energy usage.

juin [17]

Answer:

its energy usage is 0.05

Explanation:

1.5kw/30=0.05

8 0

3 years ago

A 20-kilogram child is riding on a 10-kg sled over a frictionless icy surface at 8.0 meters per second. Calculate the kinetic en

Veseljchak [2.6K]

Answer:

K = 960 J

Explanation:

Given that,

Mass of a child = 20 kg

Mass of a sled = 10 kg

Speed of child on sled = 8 m/s

We need to find the kinetic energy of the sled with the child.

The total mass of child and the sled = 20 kg + 10 kg

= 30 kg

The formula for the kinetic energy of an object is given by :

$K=\dfrac{1}{2}mv^2\\\\K=\dfrac{1}{2}\times 30\times (8)^2\\\\K=960\ J$

Hence, the kinetic energy of the sled with the child is 960 J.

6 0

3 years ago

A table tennis ball has a mass of about 2.45 g. Suppose the ball is hit across the table with a speed of 4.0 m/s. What is it’s k

cluponka [151]

Answer:

0.0196 j

Explanation:

i) The formula for kinetic energy is as follows: 0.5*m*v^2

ii) Since we have all the values all that's left is to plug them into the equation

iii) First, WE MUST, Convert grams into kgs as this is the SI unit of mass so 2.45/1000

iv) All that's left now is to plug it into the equation so:

0.5* (s.45/1000)*(4^2)

v) Lastly we add the unit joules at the end as we're talking about energy

Hope this was useful! :)

3 0

2 years ago

A block of mass 27.00 kg sits on a horizontal surface with, coefficient of kinetic

zhannawk [14.2K]

Answer:

The force is $F = 172 \ N$

Explanation:

From the question we are told that

The mass of the block is $m_b = 27.0 \ kg$

The coefficient of static friction is $\mu_s = 0.65$

The coefficient of kinetic friction is $\mu_k = 0.50$

The normal force acting on the block is

$N = m * g$

substituting values

$N = 27 * 9.8$

$N = 294.6 \ N$

Given that the force we are to find is the force required to get the block to start moving then the force acting against this force is the static frictional force which is mathematically evaluated as

$F_f = \mu_s * N$

substituting values

$F_f = 0.65 * 264.6$

$F_f = 172 \ N$

Now for this block to move the force require is equal to $F_f$ i.e

$F= F_f$

=> $F = 172 \ N$

5 0

3 years ago

A beam of light traveling through a liquid (of index of refraction n1 = 1.47) is incident on a surface at an angle of θ1 = 59° w

frosja888 [35]

Answer:

(a) $n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) $n_{2} = 1.349$

(c) $v_{1} = 2.04\times 10^{8}\ m/s$

(d) $v_{2} = 2.22\times 10^{8}\ m/s$

Solution:

As per the question:

Refractive index of medium 1, $n_{1} = 1.47$

Angle of refraction for medium 1, $\theta_{1} = 59^{\circ}$

Angle of refraction for medium 2, $\theta_{1} = 69^{\circ}$

Now,

(a) The expression for the refractive index of medium 2 is given by using Snell's law:

$n_{1}sin\theta_{1} = n_{2}sin\theta_{2}$

where

$n_{2}$ = Refractive Index of medium 2

Now,

$n_{2} = \frac{n_{1}sin\theta_{1}}{sin\theta_{2}}$

(b) The refractive index of medium 2 can be calculated by using the expression in part (a) as:

$n_{2} = \frac{1.47\times sin59^{\circ}}{sin69^{\circ}}$

$n_{2} = 1.349$

(c) To calculate the velocity of light in medium 1:

We know that:

$Refractive\ index,\ n = \frac{Speed\ of\ light\ in vacuum,\ c}{Speed\ of\ light\ in\ medium,\ v}$

Thus for medium 1

$n_{1} = \frac{c}{v_{1}$

$v_{1} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.47} = 2.04\times 10^{8}\ m/s$

(d) To calculate the velocity of light in medium 2:

For medium 2:

$n_{2} = \frac{c}{v_{2}$

$v_{2} = \frac{c}{n_{1} = \frac{3\times 10^{8}}{1.349} = 2.22\times 10^{8}\ m/s$

5 0

3 years ago

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