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Keith_Richards [23]
3 years ago
9

A driver uses his/her _____ vision to detect the motion from the sides

Physics
1 answer:
labwork [276]3 years ago
8 0
A driver would use peripheral vision
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A 28 g bullet pierces a sand bag 29 cm thick. If the initial bullet velocity was 55 m/s and it emerged from the sandbag with 18
Bumek [7]

The correct answer to the question is 130.4 N.

CALCULATION;

The mass of the bullet is given as m = 28 gram = 0. 028 kg.

The initial velocity of the bullet u = 55 m/s

The final velocity of the bullet  v = 18 m/s.

The distance covered by the bullet through the sand bag s = 29 cm.

                                                                                                   = 0.29 m

Let the acceleration of the bullet is a .

From equation of kinematics, we know that-

                                         v^2-u^2=\ 2as

                                        ⇒ a=\ \frac{v^2-u^2}{2s}

                                               =\ \frac{(18)^2-(55)^2}{2\times 0.29}\ m/s^2

                                              =\ -4656.897\ m/s^2

The negative sign is used due to the fact that force is opposing in nature. Its velocity is decreasing with time.

From Newton's second law of motion, we know that net force on a body is equal to the product of mass with acceleration.

Mathematically F = ma.

Hence, the frictional force exerted on the bullet is calculated as -

                                              F = m × a

                                                 = 0.028 × (-4656.897) N

                                                 = -130.4 N                     [ANS]

Here, N ( newton) stands for the unit of force.

8 0
3 years ago
Is mixture an example of water
Evgesh-ka [11]
No I don't believe so
7 0
3 years ago
Read 2 more answers
An object has a mass of 785 g and a volume of 15 cm³. What is its density? (Give your answer in g/cm³ to 2 decimal places).
Anvisha [2.4K]

Answer:

denisity = 52.33 g/cm^{3}

Explanation:

Density:

d = \frac{m}{v}

We have that m = 785 and that v = 15 cm^{3}.

d = \frac{785}{15}

d = 52.33 m^{3}

4 0
3 years ago
A certain field line diagram illustrates the electric field due to three particles that carry charges 5.0 μC, -3.0 μC, and -2.0
4vir4ik [10]

Answer:

6

Explanation:

Number of lines emanate from + 5 micro coulomb is 15 .

They terminates at negative charges that means at - 3 micro coulomb and - 2 micro Coulomb.

the electric field lines terminates at - 3 micro Coulomb and - 2 micro Coulomb is in the ratio of 3 : 2.

So the lines terminating at - 3 micro coulomb

                                    = \frac{3}{5}\times 15 = 9

So the lines terminating at - 2 micro coulomb

                                    = \frac{2}{5}\times 15 = 6

So, the number of filed lines terminates at - 2 micro Coulomb are 6.

4 0
3 years ago
How much would a person whose mass is 60kg weight on the moon​
Valentin [98]
Approximately 10kg

Given that gravity on the Moon has approximately 1/6th of the strength of gravity on Earth, a man who weighs 60kg on Earth would weigh approximately 10kg on the Moon.
4 0
3 years ago
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