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3 years ago

What is a benefit of having a standard measurement system that can be used by scientists worldwide

1 answer:

4 0

The current international standard metric system, it is mks system based on metre, kilogram and second as well as the kelvin ampere candela and mole

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How much work is done when a 214 newton force pushes a sleeping cow 37m across a field.

nata0808 [166]

Hello!

Answer:
7918 J

Explanation:

We are assuming that the floor (field) is completely horizontal since there's no information about that in the statement.

We are going to use the following formula:

$W= F . Cos \alpha . D$

Where:

$F=214 N$
$\alpha =0$ º
$D= 37m$

Then, by substituting we have:

$W=214N . Cos (0).37m= 7918 N.m=7918 J$

8 0

3 years ago

If a cup of coffee is at 90°C and a person with a body temperature of 36'C touches it,how will heat flow between them

Arlecino [84]

Answer:

the heat always transfers from high temperature to low temperature body without aid of any external energy to this law the heat transfers from cup of coffee to the person body until both bodies reaches to the equilibrium state

Explanation:

5 0

3 years ago

Look at the equation. What detail is missing? 3 m/s2= (33 m/s - X)/30 S <br>

Tems11 [23]

Answer:

The starting velocity.

Explanation:

We must understand that this equation comes from the following equation of kinematics.

$v_{f}=v_{o}+a*t$

where:

Vf = final velocity = 33 [m/s]

Vo = starting velocity [m/s]

a = acceleration = 3 [m/s²]

t = time = 30 [s]

So, these values can be assembly in the following way:

$v_{f}=v_{o}+a*t\\a*t=v_{f}-v_{o}\\3=\frac{33-v_{o}}{30}$

6 0

3 years ago

Give two reasons why circular water waves decrease in amplitude

Ronch [10]

1. friction between water molecules
2. the wave spreads out onto a larger and larger area, so per unit area, the energy of the wave goes down

3 0

3 years ago

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

Lesechka [4]

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

Mass of inertia is given by

$I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2$

Angular acceleration is given by

$\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}$

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454$

The coefficient of friction is 0.54454

At r = 0.25 m

$\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N$

The force needed to stop the wheel is 104.00902 N

5 0

3 years ago