All categories

3 years ago

A 2.0-kg object moving at 5.0 m/s encounters a 30-Newton restive force

Physics

1 answer:

sveta [45]3 years ago

6 0

The impulse experienced by the object is 3 N s.

<u>Explanation:</u>

Impulse is also termed as change in the momentum of the object. So, it is directly proportional to the force acting on the object and the time for which the force is acting on that object.

Thus, impulse experienced by an object is the product of force acting on the object for a given time period. So, it is the sudden influence of force on the given volume.

As the force is given as 30 N and the duration or the time is given as 0.1 seconds. Then, the impulse will be product of force with duration.

Impulse = Force × ΔTime = Force × Duration

Impulse = 30 × 0.1 = 3 N s.

Thus, the impulse experienced by the object is 3 N s.

You might be interested in

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$

When the string length is reduced by one-third, the fundamental frequency increases to one-half or 50% to 840 Hz.

6 0

3 years ago

Physical science semester B exam 66 problems , please don’t delete I’m on my last leg I need this please god help me

krek1111 [17]

HI!! If you need answers to your exam, then go to https://quizlet.com/ then search for what you need!! I hope this helps I am not sure what your exam is for, and I don't have enough info to tell you all the answers, but hopefully, this will help you!!

7 0

3 years ago

Can you tell from the coefficient of restitution whether a collision has added kinetic energy to a system, taken some away, or l

Maurinko [17]

For an inelastic collision where coefficient of restitution,e, is equal to 0, the momentum is conserved but not the kinetic energy. So, there is addition or elimination of kinetic energy.

On the otherhand, when e = 1, like for an elastic collision, kinetic energy and momentum is conserved. Thus, the system's kinetic energy is unchanged.

6 0

3 years ago

According to the big bang theory, the universe is continually getting

oee [108]

The answer on Edge would be (A.)= Larger and Cooler ! I'm doing the same thing as y'all. Good luck everyone.

7 0

3 years ago

Read 2 more answers

A water line enters a house 2.0 m below ground. A smaller diameter pipe carries water to a faucet 5.0 m above ground, on the sec

nirvana33 [79]

Answer:

P₁- P₂ = 91.1 10³ Pa

Explanation:

For this exercise we will use Bernoulli's equation, where point 1 is at the bottom of the house and point 2 on the second floor

P₁ + ½ ρ v₁² + ρ g y₁ = P₂ + ½ ρ v₂² + ρ g y₂

P1-P2 = ½ ρ (v₂² - v₁²) + ρ g (y₂-y₁)

In the exercise they give us the speeds and the height of the turbid, so we can calculate the pressure difference

For heights let's set a reference system on the ground floor of the house, so we have 5m for the second floor and an entrance at -2m

P₁-P₂ = ½ 1.0 10³ (7² - 2²) + 1.0 10³ 9.8 (5 + 2)

P₁-P₂ = 22.5 10³ + 68.6 10³

P₁- P₂ = 91.1 10³ Pa

3 0

3 years ago