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2 years ago

The change in momentum that occurs when a 1. 0 kg ball traveling at 4. 0 m/s strikes a wall and bounces back at 2. 0 m/s is.

Physics

1 answer:

Doss [256]2 years ago

7 0

Answer:

The change is momentum is given by ∆p=p(inital) - p(final) =4-2=2 kg.m/s

Explanation:

momentum is the product of mass and velocity (speed)

So it's initial momentum would be:

p=mv=(1)(4)=4 kg.m/s

It's final momentum is given by:

p=mv=(1)(2)=2 kg.m/s

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A light bulb provides a resistance of 20 Ω to the 30 A current that runs through it. Determine the voltage of the battery in the

likoan [24]

Explanation:

Ohm's law:

V = IR

V = (30 A) (20 Ω)

V = 600 V

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3 years ago

Read 2 more answers

When an object moves in a circular path, it accelerates toward the center of the circle as a result of ______.

kolbaska11 [484]

Answer:

Centripetal acceleration

Explanation:

An object moving around a xirxular path maintains its route as a result of centripetal force. However, its acceleration is caused by centripetal acceleration. Despite centripetal acceleration not being among the choices, it is the right answer.

Centripetal acceleration helps an object that navigates around a circular path to accelerate while centripetal force enables the movement of an object around a circular path to move inwards. Momentum, given as one of the choices is product of mass and velocity while friction is the force opposing movement of an object around a surface.

3 0

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

rjkz [21]

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:

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