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3 years ago

Which is a tool that use in medicine to view internal organs and pregnancy

Physics

2 answers:

Likurg_2 [28]3 years ago

6 0

An ultrasound scan uses high-frequency sound waves to make an image of a person's internal body structures.

kolbaska11 [484]3 years ago

3 0

An ultrasound? it’s commonly used to make pictures of internal organs and to check to see if someone is pregnant

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Why is an iron bolt attracted to a magnet?

icang [17]

Answer:

A. When it is in a magnetic field, it becomes a temporary magnet.

Explanation:

An iron bolt is attracted to a magnet because when in a magnetic field, the iron becomes a temporary magnet.

This is because the iron aligns their electrons in the magnetic fields.

This causes that attraction between the magnet and the iron.
Metals like iron are said to be ferromagnetic
Unpaired electrons in iron spin in such a way that they align with the magnetic fields of the magnet.

8 0

3 years ago

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Crude oil coal and peat are examples of fossil fuels name another one<br><br>

Nikolay [14]

Answer:

petroleum and Natural gas are fossil fuels

Explanation:

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3 years ago

The peak intensity of radiation from a star named sigma is 2 x 10^6 mmkay. What is the average surface temperature of Sigma roun

Igoryamba

Answer:

1.45 K

I had the same question and i got it right.

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4 years ago

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A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500

jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5 ×10 ⁻³ kg

initial velocity of stone ( $u_{stone}$ ) = 0 m/s

Initial velocity of bullet ( $u_{bullet}$ ) = (500 m/s)i

Speed of the bullet after collision ( $v_{bullet}$ ) = (300 m/s) j

Suppose we represent $(v_{stone})$ to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

$m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}$

$(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}$

$(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j= (0.100 \ kg)v_{stone}$

$v_{stone}= (1.25\ kg.m/s)i-(0.75\ kg m/s)j$

$v_{stone}= (12.5\ m/s)i-(7.5\ m/s)j$

∴

The magnitude now is:

$v_{stone}=\sqrt{ (12.5\ m/s)^2-(7.5\ m/s)^2}$

$\mathbf{v_{stone}= 14.6 \ m/s}$

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

$\theta = tan^{-1} (\dfrac{-7.5}{12.5})$

$\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}$

5 0

3 years ago

A father fashions a swing for his children out of a long rope that he fastens to the limb of a tall tree. As one of the children

trasher [3.6K]

Answer:

The centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

Explanation:

The centripetal acceleration is given by:

$a_{c} = \frac{v^{2}}{r}$

Where:

$v^{2}$ : is the tangential speed = 9.50 m/s

r: is the distance = 6.00 m

Hence, the centripetal acceleration is:

$a_{c} = \frac{v^{2}}{r} = \frac{(9.50 m/s)^{2}}{6.00 m} = 15.04 m/s^{2}$

Therefore, the centripetal acceleration of the child at the bottom of the swing is 15.04 m/s².

I hope it helps you!

3 0

3 years ago

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