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2 years ago

If the mass is 2.0 kg and the acceleration is 5 m/s/s, then what is the force?

Physics

1 answer:

pshichka [43]2 years ago

5 0

Answer:

Explanation:

F = ma = 2.0(5) = 10 N

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3 years ago

Suppose that the radius of the circular path is r when the speed of the rocket is v and the acceleration of the rocket has magni

SCORPION-xisa [38]

Answer:

Explanation:

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3 years ago

Be sure to answer all parts. Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 k

Harrizon [31]

Explanation:

Given that,

(a) Speed, $v=6.66\times 10^6\ m/s$

Mass of the electron, $m_e=9.11\times 10^{-31}\ kg$

Mass of the proton, $m_p=1.67\times 10^{-27}\ kg$

The wavelength of the electron is given by :

$\lambda_e=\dfrac{h}{m_ev}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{9.11\times 10^{-31}\times 6.66\times 10^6}$

$\lambda_e=1.09\times 10^{-10}\ m$

The wavelength of the proton is given by :

$\lambda_p=\dfrac{h}{m_p v}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{1.67\times 10^{-27}\times 6.66\times 10^6}$

$\lambda_p=5.96\times 10^{-14}\ m$

(b) Kinetic energy, $K=1.71\times 10^{-15}\ J$

The relation between the kinetic energy and the wavelength is given by :

$\lambda_e=\dfrac{h}{\sqrt{2m_eK}}$

$\lambda_e=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.11\times 10^{-31}\times 1.71\times 10^{-15}}}$

$\lambda_e=1.18\times 10^{-11}\ m$

$\lambda_p=\dfrac{h}{\sqrt{2m_pK}}$

$\lambda_p=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 1.67\times 10^{-27}\times 1.71\times 10^{-15}}}$

$\lambda_p=2.77\times 10^{-13}\ m$

Hence, this is the required solution.

6 0

3 years ago

A force of 30.0 N is applied to a 3.00 kg object for 3.00 seconds. Calculate the velocity experienced by the object.

olganol [36]

Answer:

Explanation:

F = ma and

$a=\frac{v}{t}$

We have F, we have m, but in order to solve for v, we need a.

30.0 = 3.00a so

a = 10.0 m/s/s. Plug that in for a in the second equation and solve for v:

$10.0=\frac{v}{3.00}$ so

v = 10.0(3.00) so

v = 30.0 m/s

6 0

2 years ago

Suppose the value of one division of vernier scale is 0.5mm and the value of one main scale

gregori [183]

Answer:

-0.01 mm

Explanation:

We are given that

The value of one division of vernier scale =0.5 mm

The value of one main scale division=0.49 mm

We have to find the value of least count of the instrument in mm.

We know that

Leas count of vernier caliper=1 main scale division-1 vernier scale division

Least count of vernier caliper=0.49-0.50=-0.01 mm

Hence, the least count of the instrument=-0.01 mm

Answer: -0.01 mm

8 0

3 years ago