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krok68 [10]
4 years ago
13

After exercise, your heart rate isand you need to slowly bring it to normal.

Physics
2 answers:
tino4ka555 [31]4 years ago
8 0

You will have to do a cool of and rest to bring it down....hope this help plz mark brainliest !!:)

DedPeter [7]4 years ago
3 0

So we have to rest instead for some minute

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Why is foam a good insulator?
r-ruslan [8.4K]
Air or any sort of gas is not that good in conducting heat in relation to other materials. 
<span>since foam consists of many bubbles with trapped gas it both stops convection by simply being in position, and prevents conduction of heat through it, due to the gas inside</span>
7 0
3 years ago
The left end of a long glass rod 8.00 cm in diameter and with an index of refraction of 1.60 is ground and polished to a convex
lesya692 [45]

Answer:

A) 0.1477

B) 0.65388 mm

C) object is inverted

Explanation:

The formula for object - image relationships for spherical reflecting surface is given as;

n1/s + n2/s' = = (n2 - n1)/R

Where;

n1 & n2 are the Refractive index of both surfaces

s is the object distance from the vertex of the spherical surface

s' is the image distance from the vertex of the spherical surface

R is the radius of the spherical surface

We are given;

index of refraction of glass; n2 = 1.60

s = 24 cm = 0.24 m

R = 4 cm = 0.04 m

index of refraction of air has a standard value of 1. Thus; n1 = 1

a) So, making s' the subject from the initial equation, we have;

s' = n2/[((n2 - n1)/R) - n1/s]

Plugging in the relevant values, we have;

s' = 1.6/[((1.6 - 1)/0.04) - 1/0.24]

s' = 0.1477

b) The formula for lateral magnification of spherical reflecting surfaces is;

m = -(n1 × s')/(n2 × s) = y'/y

Where;

m is the magnification

n1, n2, s & s' remain as earlier explained

y is the height of the object

y' is the height of the image

Making y' the subject, we have;

y' = -(n1 × s' × y)/(n2 × s)

We are given y = 1.7 mm = 0.0017 m and all the other terms remain as before.

Thus;

y' = -(1 × 0.1477 × 0.0017)/(1.6 × 0.24)

y' = - 0.00065388021 m = -0.65388 mm

C) since y' is negative and y is positive therefore, m = y'/y would result in a negative value.

Now, in object - image relationships for spherical reflecting surface, when magnification is positive, it means the object is erect and when magnification is negative, it means the object is inverted.

Thus, the object is inverted since m is negative.

6 0
3 years ago
A girl is whirling a ball on a string around her head in a horizontal plane. She wants to let go at precisely the right time so
hodyreva [135]

Answer:

<u>The girl should leave the string a time when the ball is at a position tangent at that point of the circle passes through the location of target.</u>

Explanation:

The direction of an object moving in circular path changes endlessly with time. The direction of the object at some extent are within the direction of tangential drawn to the circle at that point,  

The  centripetal force on the ball becomes zero once the girl leaves the string. hence, the ball can go into the direction tangential to the circle at the point whenever she left the string.

5 0
4 years ago
Energy is transferred through a solid
Alex Ar [27]

Answer:

Conduction is the transfer of thermal energy through direct contact between particles of a substance, without moving the particles to a new location. Usually occurs in solids. When heat is supplied to one end, molecules at that end start to move more quickly.

Explanation:

5 0
3 years ago
After a 0.320-kg rubber ball is dropped from a height of 19.0 m, it bounces off a concrete floor and rebounds to a height of 15.
GarryVolchara [31]

Answer:

Imp = 11.666\,\frac{kg\cdot m}{s}

Explanation:

Speed experimented by the ball before and after collision are determined by using Principle of Energy Conservation:

Before collision:

(0.32\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (19\,m) = \frac{1}{2}\cdot (0.320\,kg)\cdot v_{A}^{2}

v_{A} \approx 19.304\,\frac{m}{s}

After collision:

\frac{1}{2}\cdot (0.320\,kg)\cdot v_{B}^{2} = (0.32\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot (15\,m)

v_{B} \approx 17.153\,\frac{m}{s}

The magnitude of the impulse delivered to the ball by the floor is calculated by the Impulse Theorem:

Imp = (0.32\,kg)\cdot [(17.153\,\frac{m}{s} )-(-19.304\,\frac{m}{s} )]

Imp = 11.666\,\frac{kg\cdot m}{s}

5 0
3 years ago
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