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anygoal [31]
4 years ago
6

The average force applied to move an initially stationary golf cart over a period of 25 s is 500 N. What is the change in moment

um of the cart?
Physics
1 answer:
Yuki888 [10]4 years ago
6 0

Answer:

12500(kg*m/s)

Explanation:

F=ma=mv/t=p/t

p=F*t=500N*25 s=12500(kg*m/s)

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A car slams on the brakes to stop. The road pushes back up on the car with a normal force of 12,750 N and friction from the brak
ivann1987 [24]

Answer:

\displaystyle \mu_d=0.75

Explanation:

Coefficients of Friction

Objects in physical contact produce friction which usually manifests as thermal energy being dissipated in the surface where the objects are interacting. It's usually harder to start to move an object from rest, that keeps moving it at a constant speed on the same surface. That is why there are two different coefficients of friction: the static and the dynamic. As mentioned, the static coefficient \mu_s is greater than the dynamic coefficient \mu_d. The car is already moving and is attempting to stop. The coefficient of friction is defined as

\displaystyle \mu_d=\frac{F_r}{N}

Where Fr is the force of friction and N is the normal or the force the road pushes back up on the car. With the given data, we have

\displaystyle \mu_d=\frac{9,560\ N}{12,750\ N}

\displaystyle \boxed{\mu_d=0.75}

The coefficient of friction is dimensionless (doesn't have any units)

6 0
3 years ago
Kiara is walking her dog in the neighborhood. She begins by walking 3
Gre4nikov [31]

Answer:

2 blocks north 1 block east

Explanation   math lol

you go 3 up, 2 left then one back down which puts you at 2,2 then she goes 3 east so its x = 1         y = 2

4 0
3 years ago
Why is genetic stability important?​
kap26 [50]

Answer: DNA polymerase and DNA ligase also are important in DNA replication. In addition to maintaining the integrity of DNA sequences by DNA repair, an accurate duplication of DNA is a prerequisite for all cell divisions.

7 0
3 years ago
What are the three categories of adult development?
andreev551 [17]
Physical development , Cognitive Development , psychosocial development I think!
5 0
4 years ago
Suppose a 350-g kookaburra (a large kingfisher bird) picks up a 75-g snake and raises it 2.5 m from the ground to a branch. (a)
mafiozo [28]

(a) The work done by the bird to raise the snake on the branch is equal to the product between the weight of the snake and the height of the branch above the ground:

W_1 = mg h

where m=75 g=0.075 kg is the mass of the snake and h=2.5 m is the height of the branch with respect to the ground. Substituting numbers into the equation, we find

W_1 = (0.075 kg)(9.81 m/s^2)(2.5 m)=1.84 J


(b) The work the bird did to raise its own centre of mass from the ground to the branch is equal to the product between the bird's weight and the height of the branch:

W_2 = mgh

where m=350 g=0.35 kg is the mass of the bird. Substituting numbers into the equation, we find

W_2 = (0.35 kg)(9.81 m/s^2)(2.5 m)=8.58 J

6 0
3 years ago
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