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3 years ago

Help please with question 2bii The one that starts with calculate the average ....

Physics

1 answer:

Anna11 [10]3 years ago

4 0

Answer:

add text fully or atleast add book name edition . solution is in picture. chk pictures ignore b part

Explanation:

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Which of the following is an example of wernickes aphasia?

Margarita [4]

An example of wernickes aphasia is someone making a speech which does not make sense.

It majorly caused by loss of flow of blood to the brain. Other causes include

Head injury
Brain inflammation
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Stroke. e.t.c

<h3>What is wernickes aphasia?</h3>

Wernickes aphasia can simply be defined as a language disorder that makes it very difficult for an individual to understand words and communicate with other people.

So therefore; an exam of wernickes aphasia is someone making a speech which does not make sense.

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2 years ago

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4 years ago

The scientist who discovered that an electric current could be

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3 years ago

) The magnitude of scalar product of two unit vectors perpendicular to each other is

fenix001 [56]

This question involves the concept of the scalar product.

The magnitude of the scalar product will be "0".

<h3>SCALAR PRODUCT</h3>

The scalar product, also known as the dot product of the two vectors is given by the following formula:

$A.B = |A||B|Cos\theta$

where,

A.B = Scalar product = ?
|A| = Mangnitude of vector A = 1 unit
|B| = Magnitude of Vector B = 1 unit
θ = Angle between vectors = 90°

Therefore,

$A.B = (1)(1)Cos90^o = (1)(1)(0)$

A.B = 0

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3 years ago

P5. A bullet with an initial velocity of 280 m/s in the x-direction penetrates an initially stationary block of mass 11 kg and e

aliya0001 [1]

Answer:

the mass of the bullet is 10.5 g

Explanation:

Given;

initial velocity, u₁ = 280 m/s

final velocity of the bullet, v₁ = 70 m/s

final velocity of the block, v₂ = 0.2 m/s

mass of the block, m₂ = 11 kg

initial velocity of the block, u₂ = 0

let the mass of the bullet = m₁

Apply the principle of conservation of linear momentum for elastic collision to calculate the mass of the bullet.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

280m₁ + 11(0) = 70m₁ + 11 x 0.2

280m₁ = 70m₁ + 2.2

280m₁ - 70m₁ = 2.2

210m₁ = 2.2

m₁ = 2.2/210

m₁ = 0.0105 kg

m₁ = 10.5 g

Therefore, the mass of the bullet is 10.5 g

7 0

3 years ago