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myrzilka [38]
2 years ago
8

Problem 6: A bullet in a gun is accelerated from rest from the firing chamber to the end of the barrel at an average rate of 6.3

× 105 m/s2 for 8.2 × 10-4 s.Ball,removedc795646bb4371e1754411a7dadf94458c503446af1b6450bb3269c1f97e8ef53removedremoved58b1e9a401041b69266daacea519e828d050d14013adc67f8c64697e40f2ef89removedtheexpertta - tracking id: 0W86-2A-6A-4E-962A-28979. In accordance with Expert TA's Terms of Service. copying this information to any solutions sharing website is strictly forbidden. Doing so may result in termination of your Expert TA Account.show answer No Attempt What is the gun’s muzzle velocity (that is, the bullet’s final speed), in meters per second
Physics
1 answer:
solong [7]2 years ago
3 0

Answer:

v = 5.166 10² m / s

Explanation:

We can solve this exercise using the kinematics equations

           v = v₀ + at

as the bullet starts from rest its initial velocity is zero

           v = a t

let's calculate

           v = 6.3 10⁵   8.2 10⁻⁴

           v = 5.166 10² m / s

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Suppose that you connect the terminals of two batteries of different emfs positive to positive and negative to negative (opposin
gulaghasi [49]

Answer:

Answer is explained in the explanation section below.

Explanation:

This question is very basic and easy. The answer to this question is:

Answer: If both batteries are connected we would get less amount of charge as compared to connected a single battery.

Reasoning:

If both batteries are connected in a manner of positive terminal to positive terminal and negative terminal to negative terminal then a capacitor is added to charge it from the batteries then, total electromotive force (emf) would decrease.

As a result, the capacitor added would get less amount of charge stored. But capacitor added will get more amount of charge stored when a single battery is connected.

7 0
2 years ago
A 100 kg bungee jumper leaps from a bridge. The bungee cord has an un-streched equilibrium length of 10 m, and a spring constant
nika2105 [10]

Answer:

11.78meters

Explanation:

Given data

Mass m = 100kg

Length of cord= 10m

Spring constant k= 35N/m

At the greatest vertical distance, the spring potential energy is equal to the gravitational potential energy

That is

Us=Ug

Us= 1/2kx^2

Ug= mgh

1/2kx^2= mgh

0.5*35*10^2= 100*9.81*h

0.5*35*100=981h

1750=981h

h= 1750/981

h= 1.78

Hence the bungee jumper will reach 1.78+10= 11.78meters below the surface of the bridge

6 0
3 years ago
Read 2 more answers
Calculate the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m
otez555 [7]

Answer: The period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

Explanation:

Given: Mass = 5 kg

Spring constant = 6 N/m

Formula used to calculate period is as follows.

T = 2 \pi \sqrt\frac{m}{k}

where,

T = period

m = mass

k = spring constant

Substitute the values into above formula as follows.

T = 2 \pi \sqrt\frac{m}{k}\\= 2 \times 3.14 \times \sqrt\frac{5}{6}\\= 5.73 s

Thus, we can conclude that the period of a spring if it has a mass of 5 kg and a spring constant of 6 N/m is 5.73 sec.

5 0
3 years ago
A weightlifter lifts a 1400 N barbell 2.5 meters. Calculate the work done during the lift
Andre45 [30]

Answer:

Force = 1400N

Displacement = 2.5 m

work done = F×D

=1400×2.5

= 3500 joule

8 0
3 years ago
Read 2 more answers
The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actu
Juliette [100K]

\orange{\underline{\huge{\bold{\textit{\green{\bf{QUESTION}}}}}}}

The apparent depth of an object at the bottom of a tank filled with a liquid of refractive index 1.3 is 7.7 cm. What is the actual depth of the liquid in the tank?

{\bold{\blue{GIVEN}}}

REFRACTIVE INDEX = 1.3

APPARENT DEPTH = 7.7 cm

{ \bold{\green{To  \: Find}}}

REAL DEPTH OF THE OBJECT.

{\red{FORMULA   \:  \: USED }}

Reflective  \:  Index  =  \frac{Real \:  Depth}{Apparent  \: Depth }

\huge\mathbb{\red A \pink{N}\purple{S} \blue{W} \orange{ER}}

Refractive Index = 1.3

Apparent Depth = 7.7 cm

Putting the values in the formula:-

Reflective  \:  Index  =  \frac{Real \:  Depth}{Apparent  \: Depth }  \\  \\ 1.3 =  \frac{Real \:  Depth}{7.7 \: cm}  \\ \\  1.3 \times 7.7 = Real \:  Depth \\  \\ 10.01 \:  \: cm = Real \:  Depth

3 0
2 years ago
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