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guapka [62]
2 years ago
12

a student measures the mass of a metal weight using three separate balances and collects the following data: 38.7g, 38.9g and 38

.6g. The actual mass of the weight is 34.2 g. Evaluate the accuracy and precision of this student's measurements.
Physics
1 answer:
Lesechka [4]2 years ago
6 0

The values are precise because the values are close together but they are not accurate because they are all far from the true value.

<h3>What is the accuracy and precision?</h3>

The term accuracy refers to how much close to the true or the actual value that a measurement is. The precision refer to how close together the values that are obtained in a series of measurements are.

Now, we have the values that were obtained as; 38.7g, 38.9g and 38.6g. We can see that values are precise because the values are close together but they are not accurate because they are all far from the true value.

Learn more about precision:brainly.com/question/28336863

#SPJ1

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Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force
Margaret [11]

Answer:

Speed of the spacecraft right before the collision: \displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}.

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

  • the kinetic energy of this spacecraft, and
  • the (gravitational) potential energy of this spacecraft.

Let m denote the mass of this spacecraft. At a distance of R from the center of the earth (with mass M_\text{e}), the gravitational potential energy (\mathrm{GPE}) of this spacecraft would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}.

Initially, R (the denominator of this fraction) is infinitely large. Therefore, the initial value of \mathrm{GPE} will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy (\rm KE) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance R between the spacecraft and the center of the earth would be approximately equal to R_\text{e}, the radius of the earth.

The \mathrm{GPE} of the spacecraft at that moment would be:

\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}.

Subtract this value from zero to find the loss in the \rm GPE of this spacecraft:

\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the \rm GPE of this spacecraft would be equal to the size of the gain in its \rm KE.

Therefore, right before collision, the \rm KE of this spacecraft would be:

\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}.

On the other hand, let v denote the speed of this spacecraft. The following equation that relates v\! and m to \rm KE:

\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2.

Rearrange this equation to find an equation for v:

\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}.

It is already found that right before the collision, \displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}. Make use of this equation to find v at that moment:

\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}.

6 0
3 years ago
What is the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b?
xeze [42]

The energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

<h3>Change in energy level of the electron</h3>

When photons jump from a higher energy level to a lower level, they emit or radiate energy.

The change in energy level of the electrons is calculated as follows;

ΔE = Eb - Ef

ΔE = -2.68 eV - (-5.74 eV)

ΔE = 3.06 eV

Thus, the energy of the photon emitted when an electron in a mercury atom drops from energy level f to energy level b is 3.06 eV.

Learn more about energy level here: brainly.com/question/14287666

#SPJ1

7 0
2 years ago
What will be the ratio between the initial and final kinetic energy of a car, if its velocity is tripled?
Katarina [22]

Answer:b

Explanation:

7 0
2 years ago
Find the velocity of the student using KE= 1/2 mv^2. A 50-kilogram student is running and has 225 joules of kinetic energy. The
defon
225 = 1/2 (50) (v2)
225 = 25 (v2)
225/25 = v2
9 = v2
√9 = v
v = 3 m/s
3 0
3 years ago
Who was albert einstien describe his contributions???
Nimfa-mama [501]

Answer:

Albert Einstein was from Germany he was born theoritical physicist, from childhood only he loved mechanical toys, he was highly gifted in Mathematics, He was a world citizen and a scientific genius too. His contribution were:

1) he developed the theory of relativity

2) he also discovered the process of nuclear fission

3)he developed the quantum theory of specific heat

4)theory of stimulated emission, on which laser device technology is based

5) law of photoelectric effect

hope it helped you :)

8 0
3 years ago
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