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Bess [88]
3 years ago
7

A 1500 kg car originally moving at 10 m/s crashes into a wall and comes to rest in 0.25 s. What is the magnitude of the impulse

given to the car?
a. 37.5 kg m/s
b. 150 kg m/s
c. 15,000 kg m/s
d. 60,000 kg m/s
Physics
1 answer:
kirill [66]3 years ago
6 0

Answer:

Option C

Explanation:

Impulse is the change in momentum in an object hence given by

Impulse=F∆t

Impulse=∆p=m(v-u)

Here, F is force applied, t is rime, m is mass, v is final velocity and u is initial velocity

Given that the question has time but no magnitude of the force, we use the sexond equation.

Taking forward velocity as negative then u=-10 m/s, also since the car comes to rest, final velocity is 0 m/s. Substititing mass with 1500 kgs then

Impulse=1500(0--10)=15000 kg.m/s

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A satellite is circling the moon (radius 1700 km) close to the surface at a speed v. A projectile is launched vertically up from
Elina [12.6K]

Answer:

 Rmax = 3.4 10⁶ m

Explanation:

For this exercise we will use the concept of energy

Initial. On the surface of the luma

    Em₀ = K + U

    Em₀ = ½ m v² - G m M / R_moon

Final. At the furthest point

    Emf = U

    Emf = - g m M / R_max

    Em₀ = Emf

    ½ m v² - G m M / R_moon = - G m M / R_max

    ½ v² + G M (-1 / R_moon + 1 / R_max) = 0           (1)

Let's use the fact that tells us that the speed of the rocket is equal to the speed of a satellite that rotates around the moon near the surface, let's use Newton's second law

        F = m a

Acceleration is centripetal

       a = v² / r

        r= R_moon

       G m M / R_moon² = m v² / R_mon

        G M / R_moon = v²

We substitute in 1

     ½ G M / R_moon + G M (1 / R_max - 1 / R_moon) = 0

    1 / R_max = 1 / R_moon (1- ½)

     R_max = R_moon 2

     Rmax = 2 1700 103

     Rmax = 3.4 10⁶ m

5 0
3 years ago
a 2000 kg elevator with broken cables in a test rig is falling at 4.00 m/s when it contacts a cushioning spring at the bottom of
Ratling [72]

A. The speed of the elevator after it has moved downward 1.00 m from the point where it first contacts a spring is 3.65m/s

B. The acceleration when the elevator is 1.00 {\rm m} below point where it first contacts a spring is 4m/s²

In calculating the speed of the elevator and acceleration, first we have to find the force of gravity F on the elevator, which is the force pulling the elevator in downward direction. Using the equation for force of gravity which is:

F = mg

Where:

Mass of the elevator; m= 2000kg

Acceleration due to gravity; g = 9.8m/s

2000kg × 9.8m/s²= 19600N

F = 19600

Force of opposing friction clamp of gravity = 17000N

Net force on the elevator = force of gravity - Force of opposing friction clamp

Net force on the elevator = 19600 - 17000

Net force on the elevator = 2600 N

We will also find the kinetic energy K.E; of the elevator at the point of contact with the spring using:

K.E = 1/2 mv²

Where

Mass of the elevator; m = 2000kg

Velocity of the elevator = 4.00m/s

K.E = (1/2)*2000kg*(4m/s)²

K.E = 16000J

The kinetic energy and energy gained will be absorbed by the spring across the next 2m

Therefore,

Energy; E = K.E + P.E

Where:

Kinetic energy K.E = 16000J

Potential Energy P.E = ?

P.E of spring = net force absorbed × distance at compression

Where:

Net force absorbed = 2600N

Distance at compression = 2.0m

P.E = 2600*2

P.E = 5200J

E = 16000J + 5200J

E = 21200J

Spring constant = k

To find k

Using:

E = (1/2)*k*(x)²

Where:

E = 21200J

k = ?

x = 2m

21200J = (1/2)*k*(2m)²

21200J*2 = (4m)k

K = 42400J/4m

K = 10600N/m

Therefore,

Acceleration at 1m compression = ?

Using:

F = K*X

Where

F is force provided by the spring = 10600N/m,

K = 10600 N/m

X = 1m

F = 10600N/m * 1m

F = 10600N (upward)

A. The speed of the elevator after it has moved downward 1.00 {\rm m} from the point where it first contacts a spring?

Using:

Original Kinetic energy + net force on the elevator = final kinetic energy + spring energy

16000N + 2600N = (1/2)mv² + (1/2)k x²

18600 = (1/2)(2000)(v²) + (1/2)(10600N)(1²)

18600 = 1000(v²) + 5300

18600 - 5300 = 1000(v²)

13300 = 1000(v²)

V² = 13.300

V =3.65m/s

B. The acceleration of the elevator is 1.00m below point where it first contacts a spring

Spring constant = net force on the elevator + resultant force

Where:

Spring constant = 10600N

Net force on the elevator = 2600N

Resultant force = ?

10600N = 2600N + resultant force

Resultant force = 10600N - 2600N

Resultant force = 8000N

Using the equation for Newton's 2nd law where F = ma,

a = F/m

Where:

Resultant force; F =8000N

Mass of the elevator; m =2000kg)

a = 8000 / 2000

a = 4m/s²

Here's the complete question:

In a "worst-case" design scenario, a 2000kg elevator with broken cables is falling at 4.00m/s when it first contacts a cushioning spring at thebottom of the shaft. The spring is supposed to stop the elevator,compressing 2.00m as it does so. During the motion a safety clampapplies a constant 17000N frictional force to the elevator.

1. What is the speed of the elevator after it has moved downward 1.00m from the point where it first contacts aspring?

2. When the elevator is 1.00m below point where it first contacts a spring, what is its acceleration?

Learn more about calculating speed of an elevator from:

brainly.com/question/3850823?referrer=searchResults

#SPJ4

6 0
1 year ago
True or False. If statement is false, change the word in parentheses to make it true.
xz_007 [3.2K]

False

Explanation:

Loudness is the human perception of sound intensity and not pitch. Pitch is a property of waves.

Loudness measures the intensity of sound to the hear and the frequency is often expressed in decibels.

  • Pitch is how high or low a note is being played.
  • It is a property that makes it to judge whether sounds are high or low.
  • Loudness on the other hand is depends on the human perception of how high or low sound is to a person.

learn more:

Amplitude of sound waves brainly.com/question/2845448

#learnwithBrainly

6 0
3 years ago
Two 4.952 cm by 4.952 cm plates that form a parallel-plate capacitor are charged to +/- 0.576 nC. What is the electric field str
horrorfan [7]

In order to solve this question, we will need to know the E-field of a capacitor

E = sigma/epsilon naught

To find sigma (charge density), we will need to divide the total charge by the area

sigma = (.576 x 10^-9)/(4.952x10^-2)^2 = 2.34 x 10^-7

E = 2.34 x 10^-7/ (8.85x10^-12) = 26541.04 V/m

7 0
1 year ago
Help please 10 pts and quick!
wariber [46]

Answer:

B

Explanation:

6 0
3 years ago
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