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Paul [167]
2 years ago
8

What is the acceleration of a 100 kg object pushed by Beau with a force of 500 N? Write out your equation.

Physics
1 answer:
Aneli [31]2 years ago
4 0

Answer:

ProvideD :

  • Mass ( m ) = 100 kg , Force ( F ) = 500 N

To FinD :

  • Acceleration ( a )

SolutioN :

  • We know , F = ma so put up the values and then solve for a

\longrightarrow \tt \: 500 = 100a

\longrightarrow \tt \:  \frac{500}{100}  = a

\longrightarrow  \tt \: a =  \: \boxed { \tt{5 \: ms ^{ - 2}  }}

  • The acceleration produced is 5 ms ^ -2 .

--------------- HappY LearninG <3 --------------

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Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen
kakasveta [241]

Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

\lambda = 436.1 \ nm

or,

  =436.1\times 10^{-9} \ m

Distance,

d = 0.31 \ mm

or,

  =0.31\times 10^{-3} \ m

Distance between the 1st and 2nd dark fringes,

(y_2-y_1) = 6\times 10^{-3} \ m

As we know,

⇒ \frac{d}{L} (y_2-y_1) = \lambda

or,

⇒ L=\frac{d(y_2-y_1)}{\lambda}

By substituting the values, we get

       =\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}

       =\frac{0.31\times 6\times 10^3}{436.1}

       =\frac{1860}{436.1}

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3 0
3 years ago
g calculate the effectiveness radiation dosage in sieverts for a 79 kg person who is exposed 6.8x10^9
Elza [17]

Answer:

The answer is "\bold{dosage = 0.031 rem}"

Explanation:

please find the complete question in the attached file.

Given value:

m = 79\  kg  \\\\n = 3.4 \times  10^9 \\\\E = 5.5  \times  10^{-13} \\\\ RBE = 15

\to E = n E\\

        = 3.4  \times  10^9  \times  5.5  \times  10^{-13} \\\\      = 1.87 \times  10^{-3}

\to E(absorbed) = 1.87  \times 10^{-3}  \times  0.87 = 1.63  \times  10^{-3}

calculating the radiation absorbed per kg:

= \frac{1.63  \times  10^{-3}}{79}  \\\\ = 2.06  \times  10^{-5} \\\\ = 0.00206 \  rad

\to Dosage = 0.00206  \times  15 \\

                 = 0.031 \ \ rem

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