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Leona [35]
3 years ago
14

If constants aren't given in an experiment, what can be constant in the experiment?

Physics
1 answer:
mario62 [17]3 years ago
6 0

Answer:

Anything in an experiment that remains unchanged.

Explanation:

An example could be the temperature of the laboratory room. If there is something that has an effect on an experiment that is not variable, it is a constant. Another constant could be, say, if you were doing calculations with the same amount and kind of fluid throughout the experiment, then that fluid would also be a constant.

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what causes a star to shine brightly:

by squeezing atoms together in its core

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The length of a rectangle is 2 cm more than the width. If the area of the rectangle is
damaskus [11]
I would think 10 but I would have to see the picture
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A fan blade, initially at rest, rotates with a constant acceleration of 0.029 rad/s2. What is the time interval required for it
LenKa [72]

Answer:

The time interval is  t = 21.30 \ s

Explanation:

From the question we are told that

    The constant acceleration is \alpha  = 0.029 \ rad / s^2

    The displacement is  \theta  =  6.58 \ rad

     

According to the second equation of motion we have that

    \theta  =  w_i* t  +  \frac{1}{2} *  \alpha  t^2

given that the blade started from rest

     w_i which is the initial angular velocity is 0

 So  

       \theta  = 0 +  \frac{1}{2} *  \alpha  t^2

 =>  t = \sqrt{ \frac{2 * \theta }{\alpha } }

substituting values  

=>    t = \sqrt{ \frac{2 * 6.58 }{0.029 } }

=>    t = 21.30 \ s

6 0
3 years ago
Fan can accelerate from the starting blocks to running 11m/s down the track in 5 seconds. What is her acceleration?
Firdavs [7]
To answer this question we subtract the initial velocity from her final velocity. On the starting blocks the initial velocity was 0 m/s. Her final velocity was 11 m/s.t = 11m/s - 0 m/s, so the change in velocity is 11m/s. Time was 5 seconds.Plug it into the formula: 11 m/s ÷ 5 s= 2.2 m/s<span>2</span>
4 0
3 years ago
A cylindrical tank of methanol has a mass of 70 kg and a volume of 75 L. Determine the methanol’s weight, density, and specific
dem82 [27]

Answer:

Weight=686.7N, \rho=933kg/m^{3}, S.G.=0.933, F=17.5N

Explanation:

So, the first value the problem is asking us for is the weight of methanol. (This is supposing there is a mass of methanol of 70kg inside the tank). We can find this by using the formula:

W=mg

so we can substitute the data the problem provided us with to get:

W=70kg(9.81m/s^{2})

which yields:

W=686.7N

Next, we need to find the density of methanol, which can be found by using the following formula:

\rho=\frac{m}{V}

we know the volume of methanol is 75L, so we can convert that to m^{3} like this:

75L*\frac{0.001m^{3}}{1L}=0.075m^{3}

so we can now use the density formula to find our the methanol's density, so we get:

\rho=\frac{m}{V}

\rho=\frac{70kg}{0.075m^{3}}

\rho=933.33kg/m^{3}

Next, we can us these values to find the specific gravity of methanol by using the formula:

S.G.=\frac{\rho_{sample}}{\rho_{H_{2}O}}

when substituting the known values we get:

S.G.=\frac{933.33kg/m^{3}}{1000kg/m^{3}}

so:

S.G.=0.933

We can now find the force it takes to accelerate this tank linearly at 0.25m/s^{2}

F=ma

F=(70kg)(0.25m/s^{2})

F=17.5N

6 0
3 years ago
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