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andrew11 [14]
3 years ago
13

A car speeds upu as it rolls down a hill. Which is this an example of?

Physics
2 answers:
olga2289 [7]3 years ago
8 0
<span>positive acceleration</span>
anygoal [31]3 years ago
4 0
Positive acceleration
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"A parcel moving in a horizontal direction with speed v0 = 13 m/s breaks into two fragments of weights 1.4 N and 1.9 N, respecti
Lady bird [3.3K]

Answer:

<em>the smaller particle moves with speed of 8.706 m/s in the opposite direction to the bigger particle.</em>

<em></em>

Explanation:

Speed of the original particle = 13 m/s

We designate particles as A and B

The final weights of the component particles are

Particle A = 1.4 N

particle B = 1.9 N

The speed of the larger piece (particle B) = 29 m/s

We know that weight is the product of a body's mass and acceleration due to gravity g which is equal to 9.81 m/s^2, therefore, masses of the particles are

particle A = 1.4/9.81 = 0.143 kg

Particle B = 1.9/9.81 = 0.194 kg

The momentum of a body is the product of its mass and its velocity i.e

P = mv

This means that the mass of the particle before splitting is  

0.143 kg + 0.194 kg = 0.337 kg

Momentum of the initial whole particle = mv

==> 0.337 x 13 = 4.381 kg-m/s

The bigger particle B remains horizontal, and has a momentum of

mv = 0.194 x 29 = 5.626 kg-m/s

<em>According to the conservation of momentum, the total initial momentum of a system must be equal tot the total final momentum of the system.</em>

Initial total momentum of the system = 4.381 kg-m/s (momentum of original particle before splitting)

Final total momentum of the system = Total momentum of the particles after splitting = 5.626 kg-m/s + ( 0.143 kg x V_{B})

where  V_{B}  is the velocity of smaller particle A

final total momentum of the system = 5.626 + 0.143V_{B}

Equating the two momenta of the system, we'll have

4.381 = 5.626 + 0.143V_{B}

4.381 - 5.626 = 0.143V_{B}

-1.245 = 0.143V_{B}

V_{B}  = -1.245/0.143 =<em> -8.706 m/s</em>

<em>The negative sign indicates that the smaller particle moves in the opposite direction to the bigger particle</em>

5 0
4 years ago
The driver of a 1750 kg car traveling on a horizontal road at 110 km/h suddenly applies the brakes. Due to a slippery pavement,
Alexeev081 [22]

Answer: a=-2.4525 m/s^2

d=s=190.3 m

Explanation:The only force that is stopping the car and causing deceleration is the frictional force Fr

Fr = 25% of weight

W=mg

W=1750*9.81

W=17167.5

Hence

Fr=\frac{25}{100} * -17167.5\\\\Fr=-4291.875 N

Frictional force is negative as it acts in opposite direction

According to newton second law of motion

F=ma

hence

a=Fr/m

a=-4291.875/1750\\a=-2.4525

given

u= 110 km/h

u=110*1000/3600

u=30.55 m/s

to get t we know that final velocity v=0

v^2=u^2+2as\\0=30.55^2-2*2.4525*s\\s=190.34m

3 0
3 years ago
Sam blew up a balloon and rubbed it on his head. Then he stuck the balloon on the wall. This was all the result of electrostatic
yan [13]

Answer: If the balloon acquires charge, Sam's hair loses charge.

Explanation:

I JUST GOT THIS QUESTION RIGHT ON THE IA4 <3

3 0
3 years ago
Read 2 more answers
True or false? The plates keep moving and are still moving today.
wariber [46]

Answer:

true

Explanation:

takes time for most plates to move but throughout time they keep moving just at a slow pace

6 0
3 years ago
Two round rods, one steel andthe other copper, are joined end to end. Each rod is 0.750 mlong and 1.50 cmin diameter. The combin
nikitadnepr [17]

Answer: a) Strain on Steel rod = 0.0001078

b) elongation on the steel rod = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain on Copper Rod = 0.000189

Explanation: a) To obtain the strain of the steel rod, we invoke Hooke's law which states that, provided the elastic limit of A material isn't exceeded, the stress it undergoes is directly proportional to its strain.

(Stress, σ) ∝ (Strain, ε)

The constant of proportionality is called Young's modulus, E.

σ = Eε

For steel, Younger Modulus as obtained from literature = 210GPa.

Strain = Stress/Young's Modulus

Stress = (Force or Load applied)/ Cross sectional Area.

Force applied For the steel = 4000N

Cross sectional Area = (π(D^2))/4

D = 1.50cm = 0.015m

A = 0.0001767 m2

σ = 4000/0.0001767 = 22637238.257 N/m2

ε = σ/E = 22637238.257/(210 × (10^9)) = 0.0001078.

b) To solve for elongation.

Strain, ε = (elongation, dl)/(original length, lo)

Elongation, dl = strain × original length

dl = 0.0001078 × 0.75 = 0.00008085m = 0.008085cm = 0.0081mm.

c) strain in Copper

ε = σ/E; σ = 22637238.257 N/m2

Young's modulus of Copper, from literature, = 120GPa

ε = 22637238.257/(120 × (10^9)) = 0.000189

4 0
3 years ago
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