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3 years ago

3. Thekinetic energy of an object of

Physics

1 answer:

lawyer [7]3 years ago

7 0

Answer:

1. 100 J

2. 225 J

Explanation:

We'll begin by calculating the mass of the object. This can be obtained as follow:

Velocity (v) = 5 ms¯¹

Kinetic energy (KE) = 25 J

Mass (m) =?

KE = ½mv²

25 = ½ × m × 5²

25 = ½ × m × 25

25 = 25m / 2

Cross multiply

25m = 25 × 2

25m = 50

Divide both side by 25

m = 50 / 25

m = 2 Kg

1. Determination of the kinetic energy when the velocity is doubled.

Mass (m) = 2 Kg

Velocity (v) = double the initial velocity

= 2 × 5 ms¯¹

= 10 ms¯¹

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 2 × 10²

KE = ½ × 2 × 100

KE = 100 J

2. Determination of the kinetic energy when the velocity increased three times.

Mass (m) = 2 Kg

Velocity (v) = three times the initial velocity

= 3 × 5 ms¯¹

= 15 ms¯¹

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 2 × 15²

KE = ½ × 2 × 225

KE = 225 J

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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:

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

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.

3 0

3 years ago

Read 2 more answers

The diagram shows what happens to a system undergoing an adiabatic process.

posledela

The answer is:
B. X: Work is done to the system and temperature increases.
Y: Work is done by the system and temperature decreases.

5 0

3 years ago

Read 2 more answers

to measure the static friction coefficient between a block and a vertical wall, a spring is attached to the block, is pushed on

Stolb23 [73]

Answer:

μ = mg/kx

Explanation:

Since the bock does not slip, the frictional force equals the weight of the block. So, F = mg. Now, the frictional force, F = μN where μ = coefficient of static friction and N = Normal force.

Now, the normal force equals the spring force F' = kx where k = spring constant and x = compression of spring.

N = F' = kx

So, F = μN = μkx

μkx = mg

So, μ = mg/kx

8 0

3 years ago

How does the elbow medical and lateral epicondylitis?

dlinn [17]

Lateral epicondylitis, or “tennis elbow,” is an inflammation of the tendons that join the forearm muscles on the outside of the elbow. The bony bump on the outside (lateral side) of the elbow is called the lateral epicondyle. The ECRB muscle and tendon is usually involved in tennis elbow. 
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4 0

3 years ago

A permanent magnet is pushed into a wire, left there for a while, and then pulled out. During which time does a current run thou

lakkis [162]

Answer:

C. while the magnet is moving

Explanation:

Electromagnetic induction implies the production of electric current by mere movement of a magnet with respect to a coil or wire.

In the given question, current would be induced in the wire only when the magnet moves. That is either when the magnet is pushed into a wire, or when pulled out. But no current would flow through the wire when the magnet is left there for a while.

The current is induced because of the motion involved. Thus, the appropriate option is C.

8 0

3 years ago