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3 years ago

If a car changes its velocity from 32 km/hr to 54 km/hr in 8.0 seconds, what is its acceleration?

Physics

2 answers:

igomit [66]3 years ago

5 0

Answer:

a = 2.75 m/s^2

Explanation:

Using the equation for the acceleration = (final velocity - initial velocity)/ time

= 54 - 32 / 8

= 22 / 8

= 2.75 m/s^2

Stella [2.4K]3 years ago

3 0

Use the equation for the acceleration
A = final velocity - initial velocity divided by time final - time initial
A= 54 - 32 / 8 - 0
A= 22 / 8
A= 2.75 m/s^2
Hope this helps!

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Which object has the most kinetic energy? a bus a car a plane a bicycle

loris [4]

Answer:I would guess a plane

Assuming they all Thad the same velocity....

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3 years ago

Un atleta de 70 kg de masa que ha efectuado un salto de altura cae una vez que ha

Allushta [10]

Answer:

a) the elastic force of the pole directed upwards and the force of gravity with dissects downwards

Explanation:

The forces on the athlete are

a) at this moment the athlete presses the garrolla against the floor, therefore it acquires a lot of elastic energy, which is absorbed by the athlete to rise and gain potential energy,

therefore the forces are the elastic force of the pole directed upwards and the force of gravity with dissects downwards

b) when it falls, in this case the only force to act is batrachium by the planet, this is a projectile movement for very high angles

c) When it reaches the floor, it receives an impulse that opposes the movement created by the mat. The attractive force is the attraction of gravity.

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2 years ago

Help pls i need this right now

pantera1 [17]

Answer:

The x-component of $F_{3}$ is 56.148 newtons.

Explanation:

From 1st and 2nd Newton's Law we know that a system is at rest when net acceleration is zero. Then, the vectorial sum of the three forces must be equal to zero. That is:

$\vec F_{1} + \vec F_{2} + \vec F_{3} = \vec O$ (1)

Where:

$\vec F_{1}$ , $\vec F_{2}$ , $\vec F_{3}$ - External forces exerted on the ring, measured in newtons.

$\vec O$ - Vector zero, measured in newtons.

If we know that $\vec F_{1} = (70.711,70.711)\,[N]$ , $\vec F_{2} = (-126.859, 46.173)\,[N]$ , $F_{3} = (F_{3,x},F_{3,y})$ and $\vec O = (0,0)\,[N]$ , then we construct the following system of linear equations:

$\Sigma F_{x} = 70.711\,N - 126.859\,N +F_{3,x} = 0\,N$ (2)

$\Sigma F_{y} = 70.711\,N + 46.173\,N+F_{3,y} = 0\,N$ (3)

The solution of this system is:

$F_{3,x} = 56.148\,N$ , $F_{3,y} = -116.884\,N$

The x-component of $F_{3}$ is 56.148 newtons.

5 0

2 years ago

A 4.40-kilogram hoop starts from rest at a height 1.70 m above the base of an inclined plane and rolls down under the influence

Anestetic [448]

Answer:

The linear velocity is $v=4.08m/s$

Explanation:

According to the law of conservation of energy

The potential energy possessed by the hoop at the top of the inclined plane is converted to the kinetic energy at the foot of the inclined plane

The kinetic energy can be mathematically represented as

$KE = \frac{mv^2}{2} + \frac{Iw}{2}$

Where $I$ is the moment of inertia possessed by the hoop which is mathematically represented as

$I = mr^2$

Here R is the radius of the hoop

$w$ is the angular velocity which the hoop has at the bottom of the lower part of the inclined plane which is mathematically represented as

$w = \frac{v}{r}$

Where v linear speed of the hoop's center of mass just as the hoop leaves the incline and rolls onto a horizontal surface

Now expressing the above statement mathematically

$potential \ energy = \frac{mv^2}{y} + \frac{Iw^2}{2}$

$mgh = \frac{mv^2}{y} + \frac{Iw^2}{2}$

=> $mgh =\frac{mv^2}{2} + \frac{(mr^2)(\frac{v}{r})^2 }{2}$

=> $mgh = \frac{mv^2}{2} + \frac{mv^2}{2}$

=> $mgh = mv^2$

=> $v = \sqrt{gh}$

Substituting values

$v = \sqrt{9.81 * 1.7}$

$v=4.08m/s$

4 0

3 years ago

Read 2 more answers

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butalik [34]

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3 years ago