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Norma-Jean [14]

3 years ago

15

Suppose that a constant force is applied to an object with a mass of 12kg, it’s creates an acceleration of 5m/s^2. The accelerat

ion of another object produced by the same force is 4m/s^2, what is the mass of this object?

1 answer:

Klio2033 [76]3 years ago

5 0

Answer:

Mass of second object will be 15 kg

Explanation:

We have given mass of first object = 12 kg

Acceleration $a=5m/sec^2$

According to second law of motion we know that force F = MA

So force $F=12\times 5=60N$

As the same force is applied to the second object of acceleration $a=4m/Sec^2$

So force = ma

$60=m\times 4$

m = 15 kg

So mass of second object will be 15 kg

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(6 m/s) / (1.9 m/s²) = 3.158 seconds (rounded)

to lose all of her initial speed, and stop.

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Help me please I can't get the final step

inna [77]

Answer:

$\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}$

Explanation:

<u>Dimensional Analysis</u>

It's given the relation between quantities A, B, and C as follows:

$\displaystyle A=\frac{3}{2}B^mC^n$

and the dimensions of each variable is:

$A=L^2T^2$

$B=LT^{-1}$

$C=LT^2$

Substituting the dimensions into the relation (the coefficient is not important in dimension analysis):

$\displaystyle L^2T^2=\left(LT^{-1}\right)^m\left(LT^2\right)^n$

Operating:

$L^2T^2=\left(L^mT^{-m}\right)\left(L^nT^{2n}\right)$

$L^2T^2=L^{m+m}T^{-m+2n}$

Equating the exponents:

$m+n=2$

$-m+2n=2$

Adding both equations:

$3n=4$

Solving:

$n=4/3$

$m=2-4/3=2/3$

Answer:

$\mathbf{\displaystyle m=\frac{2}{3},\ n=\frac{4}{3}}$

6 0

3 years ago

A long, straight, horizontal wire carries a left-to-right current of 40 A. If the wire is placed in a uniform magnetic field of

sladkih [1.3K]

Answer:

The magnitude of the resultant of the magnetic field is $4.11\times10^{-5}\ T$

Explanation:

Given that,

Current = 40 A

Magnetic field $B=3.7\times10^{-5}\ T$

Distance = 22 cm

We need to calculate the magnetic field

Using formula of magnetic field

$B'=\dfrac{\mu_{0}I}{2\pi r}$

Where, r = distance

I = current

Put the value into the formula

$B'=\dfrac{4\pi\times10^{-7}\times20}{2\pi\times0.22}$

$B'=1.8\times10^{-5}\ T$

We need to calculate the magnitude of the resultant of the magnetic field

Using formula of resultant

$B''=\sqrt{B^2+B'^2}$

Put the value into the formula

$B''=\sqrt{(3.7\times10^{-5})^2+(1.8\times10^{-5})^2}$

$B''=4.11\times10^{-5}\ T$

Hence, The magnitude of the resultant of the magnetic field is $4.11\times10^{-5}\ T$

6 0

3 years ago

The overall energy involved in the formation of CsCl from Cs(s) and Cl2(g) is −443 kJ/mol. Given the following information: heat

german

Answer : The magnitude of the lattice energy for CsCl is, 667 KJ/mole

Explanation :

The steps involved in the born-Haber cycle for the formation of $CsCl$ :

(1) Conversion of solid calcium into gaseous cesium atoms.

$Cs(s)\overset{\Delta H_s}\rightarrow Cs(g)$

$\Delta H_s$ = sublimation energy of calcium

(2) Conversion of gaseous cesium atoms into gaseous cesium ions.

$Ca(g)\overset{\Delta H_I}\rightarrow Ca^{+1}(g)$

$\Delta H_I$ = ionization energy of calcium

(3) Conversion of molecular gaseous chlorine into gaseous chlorine atoms.

$Cl_2(g)\overset{\frac{1}{2}\Delta H_D}\rightarrow Cl(g)$

$\Delta H_D$ = dissociation energy of chlorine

(4) Conversion of gaseous chlorine atoms into gaseous chlorine ions.

$Cl(g)\overset{\Delta H_E}\rightarrow Cl^-(g)$

$\Delta H_E$ = electron affinity energy of chlorine

(5) Conversion of gaseous cations and gaseous anion into solid cesium chloride.

$Cs^{1+}(g)+Cl^-(g)\overset{\Delta H_L}\rightarrow CsCl(s)$

$\Delta H_L$ = lattice energy of calcium chloride

To calculate the overall energy from the born-Haber cycle, the equation used will be:

$\Delta H_f^o=\Delta H_s+\Delta H_I+\Delta H_D+\Delta H_E+\Delta H_L$

Now put all the given values in this equation, we get:

$-443KJ/mole=76KJ/mole+376KJ/mole+121KJ/mole+(-349KJ/mole)+\Delta H_L$

$\Delta H_L=-667KJ/mole$

The negative sign indicates that for exothermic reaction, the lattice energy will be negative.

Therefore, the magnitude of the lattice energy for CsCl is, 667 KJ/mole

5 0

3 years ago