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GuDViN [60]
3 years ago
12

What are used for manufacturing paper along with wood and chemicals

Physics
1 answer:
Tju [1.3M]3 years ago
8 0

Answer:

Paper and pulp are made from cellulosic fibers and other plant materials. Some synthetic materials may be used to impart special qualities to the finished product. Paper is made from wood fibers, but rags, flax, cotton linters, and bagasse are also used in some papers.

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At the bottom of its path, the ball strikes a 2.30 kg steel block initially at rest on a frictionless surface. The collision is
Triss [41]

Answer:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

\dfrac{1}{2}mv^2=mgl

v=\sqrt{2gl}

Put the value into the formula

u=\sqrt{2\times9.8\times50.0\times10^{-2}}

u=3.13\ m/s

The initial speed of the ball u_{1}=3.13\ m/s

The initial speed of the block u_{2}=0

(a). We need to calculate the speed of the ball after collision

Using formula of collision

v_{1}=(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{1}+(\dfrac{2m_{2}}{m_{1}+m_{2}})u_{2}

Put the value into the formula

v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13

v_{1}=-2.01\ m/s

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

v_{2}=(\dfrac{2m_{1}}{m_{1}+m_{2}})u_{1}+(\dfrac{m_{1}-m_{2}}{m_{1}+m_{2}})u_{2}

Put the value into the formula

v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0

v_{2}=1.11\ m/s

Hence, (a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

8 0
3 years ago
Two 1.0 kg masses are 4.0 m apart on a frictionless table. Each has 1.0μC of charge. Part A What is the magnitude of the electri
9966 [12]

Force between two charges is given by

F =\frac{kq_1q_2}{r^2}

F =\frac{9*10^9* 1* 10^{-6}* 1 * 10^{-6}}{4^2}

F = 5.625 * 10^{-4} N

Now in order to find the acceleration of each mass

we can use

F = ma

5.625 * 10^{-4} = 1 * a

a= 5.625 * 10^{-4} m/s^2

8 0
3 years ago
Read 2 more answers
An object’s motion remains constant when acted upon by what?
igomit [66]

Answer:

An outside force

Explanation:

Newton's law an object in motion stays in motion an object at rest stays at rest unless acted on by an outside force.

6 0
2 years ago
A sound wave has a speed of 330m/s and a wavelength of 0.372 m. what is the frequency of the wave?
Alja [10]

Answer:

887.1Hz

Explanation:

Given parameters:

Speed of sound wave  = 330m/s

Wavelength  = 0.372m

Unknown:

Frequency  = ?

Solution:

To solve this problem, we use the expression below:

             Speed  = Frequency x wavelength

            330  = Frequency x 0.372

   Frequency  = 887.1Hz

5 0
3 years ago
Which statement explains why a chemical equation must be balanced?
Katena32 [7]
The answer to this question is b
7 0
2 years ago
Read 2 more answers
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