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3 years ago

7

What are the difference between clinical and laboratory thermometer?

1 answer:

Mamont248 [21]3 years ago

6 0

Answer:

Explanation:

Clinical Thermometer is meant for clinical purposes. It is developed for measuring the human body temperature. A laboratory thermometer, which is colloquially known as the lab thermometer, is used for measuring temperatures other than the human body temperature.

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Acceleration can occur when a car is moving at a constant speed. what must cause this acceleration?

igomit [66]

The gasoline is charging the the car to move faster

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2 years ago

Assume that a person bouncing a ball represents a closed system. Which statement best describes how the amounts of the ball's po

Zinaida [17]

Answer:

C.

Explanation:

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3 years ago

A 600-kg car traveling at 30.0 m/s is going around a curve having a radius of 120 m that is banked at an angle of 25.0°. The coe

ikadub [295]

Answer:

$f_{fr}=1590.85 N$

Explanation:

Here the total force at the horizontal components will be equal to the centripetal force on the car. So we will have:

$f_{fr}cos(25)+Nsin(25)=m\frac{v^{2}}{r}$ (1)

f(fr) is the friction force
N is the normal force

Now, the sum of forces at the vertical direction is equal to 0.

$Ncos(25)-mg-f_{fr}sin(25)=0$ (2)

Let's combine (1) and (2) to find f(fr)

$f_{fr}=\frac{(mv^{2}/r)-mgtan(25)}{cos(25)+tan(25)sin(25)}$

$f_{fr}=\frac{(600*30^{2}/120)-600*9.81*tan(25)}{cos(25)+tan(25)sin(25)}$

$f_{fr}=1590.85 N$

I hope it helps you!

5 0

3 years ago

Physicist ____ was found to have completely made up some of the data on which his astounding discoveries were made.

GaryK [48]

Answer:

Jan Hendrik Schön I believe

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3 years ago

A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k

Luba_88 [7]

Answer:

The frictional force $F_{fri} =$ 6.446 N

The acceleration of the block a = 6.04 $\frac{m}{s^{2} }$

Explanation:

Mass of the block = 3.9 kg

$\theta = 40$ °

$\mu$ = 0.22

(a). The frictional force is given by

$F_{fri} = \mu R_{N}$

$R_{N} = mg \cos \theta$

$R_{N} =$ 3.9 × 9.81 × $\cos 40$

$R_{N} =$ 29.3 N

Therefore the frictional force

$F_{fri} =$ 0.22 × 29.3

$F_{fri} =$ 6.446 N

(b). Block acceleration is given by

$F_{net} = F - F_{fri}$

F = 30 N

$F_{fri}$ = 6.446 N

$F_{net}$ = 30 - 6.446

$F_{net}$ = 23.554 N

The net force acting on the block is given by

$F_{net} = ma$

23.554 = 3.9 × a

a = 6.04 $\frac{m}{s^{2} }$

This is the acceleration of the block.

8 0

2 years ago