What is the correct answer?

Please help! the mass of an object is measured on a pan balance with a precision of 0.005 g and the recorded value of 128.01 g.

inysia [295]

Yes, these two objects have different masses.

<h3>How can we calculate that this statement is right ?</h3>

To calculate the precision of mass we are using the formula,

Precision(P) = $\frac{m_1}{m_1+\triangle m}$

Or, $\triangle$ m= $\frac{m_1}{P}$ -m₁

For the first case we are given,

m₁= The recorded value of mass

= 128.01 g

P= Precision of the mass

=0.005g

So, according to the formula, $\triangle$ m will be,

$\triangle$ m= $\frac{128.01}{0.005}$ -128.01

Or, $\triangle$ m=25,473.99 g

Or, $\triangle$ m=25.47 Kg

For the first case $\triangle$ m is 25.47 Kg..

For the second case we are given,

m₁= The recorded value of mass

= 0.13 Kg

P= Precision of the mass

=0.005 Kg

So, according to the formula, $\triangle$ m will be,

$\triangle$ m= $\frac{0.13}{0.005}$ -0.13

Or, $\triangle$ m= 25.87 kg

For the second case $\triangle$ m is 25.87 Kg.

For the two cases $\triangle$ m has different values, 25.47 Kg≠25.87 Kg.

Therefore we can conclude that, these two objects have different masses.

Learn more about Mass:

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7 0

2 years ago

Detlev walks 1000 meters north to the store and walks back 900 meters south to his friends house. The friends house is 100 meter

lilavasa [31]

1000 is the correct answer

3 0

3 years ago

A solenoid has a radius Rs = 14.0 cm, length L = 3.50 m, and Ns = 6500 turns. The current in the solenoid decreases at the rate

babymother [125]

Answer:

E = 58.7 V/m

Explanation:

As we know that flux linked with the coil is given as

$\phi = NBA$

here we have

$A = \pi R_s^2$

$B = \mu_o N i/L$

now we have

$\phi = N(\mu_o N i/L)(\pi R_s^2)$

now the induced EMF is rate of change in magnetic flux

$EMF = \frac{d\phi}{dt} = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L$

now for induced electric field in the coil is linked with the EMF as

$\int E. dL = EMF$

$E(2\pi r_c) = \mu_o N^2 \pi R_s^2 \frac{di}{dt}/L$

$E = \frac{\mu_o N^2 R_s^2 \frac{di}{dt}}{2 r_c L}$

$E = \frac{(4\pi \times 10^{-7})(6500^2)(0.14^2)(79)}{2(0.20)(3.50)}$

$E = 58.7 V/m$

3 0

3 years ago

A 60-watt light bulb has a voltage of 120 bolts applied across it and a current of 0.5 amperes flows through the bulb. What is t

Andre45 [30]

Answer:

240 ohms

Explanation:

From Ohms law we deduce that V=IR and making R the subject of the formula then R=V/I where R is resistance, I is current and V is coltage across. Substituting 120 V for V and 0.5 A for A then

R=120/0.5=240 Ohms

Alternatively, resistance is equal to voltage squared divided by watts hence $\frac {120^{2}}{60}=240$

7 0

3 years ago

A circular loop of flexible iron wire has an initial circumference of 165.0cm, but its circumference is decreasing at a constant

vovangra [49]

Answer:

emf induced in the loop, at the instant when 9.0s have passed = 1.576 * 10 ⁻² V.

Direction is counter clockwise.

Explanation:

See attached pictures.

6 0

3 years ago