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3 years ago

First ANSWER = BRAINLIESTquestion 3 onwards

Physics

1 answer:

jok3333 [9.3K]3 years ago

3 0

3. Conservation of momentum for inelastic collisions (the trolleys collide and stick together):

v = (m₁v₁+m₂v₂)/(m₁+m₂)

Given values:

m₁ = m₂ = 1kg

v₁ = 5m/s

v₂ = 0m/s

Substitute the terms in the equation with the given values and solve for v:

v = (1×5+1×0)/(1+1)

4. Apply the same equation for inelastic collisions (the skaters hold on together after the collision):

v = (m₁v₁+m₂v₂)/(m₁+m₂)

m₁ = 60kg (moving skater's mass)

m₂ = 30kg (stationary skater's mass)

v₁ = 6m/s (moving skater's speed)

v₂ = 0m/s (stationary skater is at rest)

Substitute the values and solve for v:

v = (60×6+30×0)/(60+30)

a) Since the car and lorry are moving as one object after the collision, multiply the sum of their masses by their velocity to get their momentum:

p = v×∑m

Given values:

m₁ = 1000kg (sports car)

m₂ = 3000kg (lorry)

v = 25 m/s

Plug in the values and solve for p:

p = 25×(1000+3000)

b) The conservation of momentum guarantees that the total momentum of the objects before and after the collision will be equal. Therefore the total momentum before the collision is the same as the value obtained in the previous question:

c) Assume the lorry was at rest before the collision.

Recall the general equation for conservation of momentum:

m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'

Given values:

m₁ = 1000kg (sports car)

m₂ = 3000kg (lorry)

v₂ = 0m/s (lorry is initially at rest)

v₁' = v₂' = 25m/s (car and lorry stick together after collision, moving at the same speed)

Plug in the given values and solve for v₁ (car's speed before the collision):

1000v₁ + 3000×0 = 1000×25 + 3000×25

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When operated on a household 110.0-V line, typical hair dryers draw about 1650 W of power. We can model the current as a long st

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Answer:

Current in the hair dryer will be equal to 15 A

Explanation:

We have given that household is operated at 110 volt

So potential difference V =110 volt

Power drawn by hairdryer is P = 1650 watt

We have to find the current in the hair dryer

We know that power is given as P = VI, here V is potential difference and I is current

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4 years ago

Question 10 of 34

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Julia walks from the park, which is six blocks east of her house, to the store, which is three blocks east of her house. Julia walks for 5 minutes. This walk's average speed will be 1.2 blocks per minute. Option B is correct.

<h3>What is displacement?</h3>

Displacement is defined as the shortest distance between the two points. Distance is the horizontal length covered by the body. While displacement is the shortest distance between the two points.

Displacement is a vector quantity .its unit is m.

The average velocity on this walk will be;

$\rm v_{avg}= \frac{d}{t} \\\\ \rm v_{avg}= \frac{6 \ block+ 3 \ block }{5 \ minute } \\\\ v_{avg}=1.4 \ block /min$

Hence option B is correct.

To learn more about displacement refer to the link; brainly.com/question/10919017

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Humans affect the eath by burning fossil fuels

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Compare each pair of numbers using <, > or =.

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3 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

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Answer:

a) $F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

b) $\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

Explanation:

In order to solve this problem we must first do a drawing of the situation and a free body diagram. (Check attached picture).

After a close look at the diagram and the problem we can see that the crate will have a constant velocity. This means there will be no acceleration to the crate so the sum of the forces must be equal to zero according to Newton's third law. So we can build a sum of forces in both x and y-direction. Let's start with the analysis of the forces in the y-direction:

$\Sigma F_{y}=0$

We can see there are three forces acting in the y-direction, the weight of the crate, the normal force and the force in the y-direction, so our sum of forces is:

$-F_{y}-W+N=0$

When solving for the normal force we get:

$N=F_{y}+W$

we know that

W=mg

and

$F_{y}=Fsin \theta$

so after substituting we get that

N=F sin θ +mg

We also know that the kinetic friction is defined to be:

$f_{k}=\mu_{k}N$

so we can find the kinetic friction by substituting for N, so we get:

$f_{k}=\mu_{k}(F sin \theta +mg)$

Now we can find the sum of forces in x:

$\Sigma F_{x}=0$

so after analyzing the diagram we can build our sum of forces to be:

$-f+F_{x}=0$

we know that:

$F_{x}=Fcos \theta$

so we can substitute the equations we already have in the sum of forces on x so we get:

$-\mu_{k}(F sin \theta +mg)+Fcos \theta=0$

so now we can solve for the force, we start by distributing $\mu_{k}$ so we get:

$-\mu_{k}F sin \theta -\mu_{k}mg)+Fcos \theta=0$

we add $\mu_{k}mg$ to both sides so we get:

$-\mu_{k}F sin \theta +Fcos \theta=\mu_{k}mg$

Nos we factor F so we get:

$F(cos \theta-\mu_{k} sin \theta)=\mu_{k}mg$

and now we divide both sides of the equation into $(cos \theta-\mu_{k} sin \theta)$ so we get:

$F=\frac{\mu_{k}mg}{cos \theta-\mu_{k}sin \theta}$

which is our answer to part a.

Now, for part b, we will have the exact same free body diagram, with the difference that the friction coefficient we will use for this part will be the static friction coefficient, so by following the same procedure we followed on the previous problem we get the equations:

$f_{s}=\mu_{s}(F sin \theta +mg)$

and

F cos θ = f

when substituting one into the other we get:

$F cos \theta=\mu_{s}(F sin \theta +mg)$

which can be solved for the static friction coefficient so we get:

$\mu_{s}=\frac{Fcos \theta}{Fsin \theta +mg}$

which is the answer to part b.

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4 years ago

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