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svetoff [14.1K]
2 years ago
5

A speeding car is going 72 mi/hr. The driver hits the brakes to slow the car down to the 45 mi/hr speed limit. The car has a mas

s of 1.0 metric tons. What is the magnitude of the impulse on the car?
Physics
1 answer:
MatroZZZ [7]2 years ago
8 0

Answer:

I = -12070 N.s

Explanation:

Given that:

mass of 1.0 metric tons = 1000 kg

initial speed = 72mi/hr = 32.19 m/s

final speed = 45mi/hr = 20.12 m/s

Impulse = change in momentum

I = mv-mu

I = m(v-u)

I = 10³(20.12 - 32.19)

I = 10³(-12.07)

I = -12070 N.s

The negative  sign signifies that the impulse way is to the opposite direction.

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WILL MARK BRAINLIEST IF RIGHT & HAS EXPLINATION
Colt1911 [192]

Answer:

s= 64m I'm not 100% sure

Explanation: like 97.9%

7 0
3 years ago
A box falls off of a tailgate and slides along the street for a distance of 62.5 m. Friction slows the box at –5.0 m/s2. At what
Mila [183]

Answer:

25 m/s

Explanation:

This question can be solved using equation of motion

v^2 = u^2 + 2as

where

v is the final velocity

u is the initial velocity

s is the distance covered while moving from initial to final velocity

a is the acceleration

_____________________________________________

Given

box moved for distance of 62.5 m

Friction slows the box at –5.0 m/s2----> this statement means that there is deceleration , speed of truck decreases by 5 m/s in every second until the box comes to rest. Friction causes this deceleration.

thus in this problem

a = -5.0 m/s2

V = 0   as body came to rest due to friction deceleration

u the initial velocity we have to find

the initial velocity of box will be the same as speed of truck, as the box was in the truck and hence box will pick the speed of truck.

so if we find speed of box, we will be able get sped of truck as well.

using equation of motion

v^2 = u^2 + 2as\\0^2 = u^2 + 2*-5* 62.5\\0 = u^2 - 625\\u^2 = 625\\\sqrt{u^2} = \sqrt{625} \\u = 25

Thus, initial speed with the truck was travelling was 25 m/s.

3 0
2 years ago
A 48.0-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocit
Kipish [7]

Answer:

F = 47.6 N

Explanation:

  • Newton's 2nd law can be expressed as the rate of change of the total momentum, respect of time, as follows:

       F = \frac{\Delta p}{\Delta t}

  • So, in order to find the average force exerted by the skater on the wall, we can find the change in momentum due to the force exerted by the wall (which is equal and opposite to the one exerted by the skater), and  divide it by the time interval , as follows:

       F_{wall} = \frac{\Delta p}{\Delta t} =\frac{(48.0 kg*(-1.06m/s)}{1.07s} = -47.6 N

       ⇒ Fsk = 47.6 N (normal to the wall)

3 0
3 years ago
A 3.0-kg cart is rolling across a frictionless, horizontal track toward a 1.3-kg cart that is held initially at rest. The carts
Karolina [17]

Answer:

Explanation:

a ) Momentum of first cart = mass x velocity

= 3 x 4.6 =+13.8 kg m /s

Momentum of second cart = 1.3 x - 1.9 = - 2.47 kg m /s

Total momentum = 13.8 - 2.47

= +11.33 kg m /s

b )

Let the velocity of first cart be v at the moment when second cart was at rest

total momentum = 3 x v + 0 = 3 v

Applying conservation of momentum law

3 v  = +11.33

v = +3.77 m /s

6 0
2 years ago
Identify each part of this chemical equation that describes the burning of methane and oxygen. B (blue box): D (number): E (purp
Allushta [10]

Answer:

The correct answer is -

A (the entire green box): Chemical Equation

B (the blue box): Reactants

C (the arrow): Reacts to Form

D (the number): Coefficient

E (the purple box): Products

Explanation:

The chemical reaction of burning methane and oxygen is as follows;

Here, the green part A is the chemical equation that includes various parts that are reactants B, methane, and oxygen, C is an arrow that indicates the formation of products.

2 is here coefficient that indicates the moles of the oxygen which forms carbon dioxide and water in box E is products

5 0
2 years ago
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