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3 years ago

Which of the following is an example of Class 3 lever system

2 answers:

6 0

If the fulcrum is closer to the effort, then the load will move a greater distance. A pair of tweezers, swinging a baseball bat or using your arm to lift something are examples of third class levers.

tankabanditka [31]3 years ago

6 0

Answer:

d. bicep curl

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A projectile is fired with a velocity of 45 m/s at an angle of 32°. What is the

Dennis_Churaev [7]

Answer:

Explanation:

To calculate adjacent of triangle use

$\cos(32) = \frac{a}{45}$

where 45 is the hypotenuse, and a is the adjacent side (horizontal component)

$45 \cos(32) = 38.16$

then round to 38.2

8 0

3 years ago

A 30-g bullet is fired with a horizontal velocity of 450 m/s and becomes embedded in block B, which has a mass of 3 kg. After th

Zolol [24]

Answer:

(a) The velocity of the bullet and B after the first impact is 4.4554 m/s.

(b) The velocity of the carrier is 0.40872 m/s.

Explanation:

(a) To solve the question, we apply the principle of conservation of linear momentum as follows.

we note that the distance between B and C is 0.5 m

Then we have

Sum of initial momentum = Sum of final momentum

0.03 kg × 450 m/s = (0.03 kg + 3 kg) × v₂

Therefore v₂ = (13.5 kg·m/s)÷(3.03 kg) = 4.4554 m/s

The velocity of the bullet and B after the first impact = 4.4554 m/s

(b) The velocity of the carrier is given as follows

Therefore from the conservation of linear momentum we also have

(m₁ + m₂)×v₂ = (m₁ + m₂ + m₃)×v₃

Where:

m₃ = Mass of the carrier = 30 kg

Therefore

(3.03 kg)×(4.4554 m/s) = (3.03 kg+30 kg) × v₃

v₃ = (13.5 kg·m/s)÷ (33.03 kg) = 0.40872 m/s

The velocity of the carrier = 0.40872 m/s.

3 0

3 years ago

How much force is required to accelerate a 5 kg object at 5 m/s^2?

kramer

Answer:25n

Explanation:

3 0

3 years ago

calculate the acceleration of a 41-kg crate of baseball gear when pulled sideways with a net force of 1255 N. neglect frictional

lara31 [8.8K]

In his celebrated Second Law of Motion, Newton wrote:

Net force = (mass) x (acceleration).

By the process of "plugging in numbers", we can write

1255 N = (41 kg) x (acceleration)

Now, after dividing each side by (41 kg), we have

(1255 N) / (41 kg) = acceleration.

But (1255N)/(41kg) = 30.61 m/s² .

So unless we have carelessly blooped the calculations somewhere,
that 30.61 m/s² is the answer we're looking for.

7 0

4 years ago

What is the mass of the other block?

xeze [42]

Let us consider the tension produced on both the sides of the rope is T.

We have been given that the rope is mass less and the rope is passing through a pulley which is stationary.

$Let\ m_{1} and\ m_{2}\ are\ the\ masses\ of\ two\ blocks$

$Let\ m_{1}\ is\ moving\ in\ vertical\ upward\ direction\ while\ m_{2}\ is\ in\ downward\ \ direction$

$Hence\ m_{2} =4.5\ kg$

The body is moving downward with an acceleration of $\frac{3}{4} g$

As the rope is the same one which is passing over a mass less pulley and connected by two masses,hence, acceleration of each block will be the same in magnitude.

$For\ body\ m_{1}$

Here the tension is acting in vertical upward direction and the weight is acting in vertical downward direction. Here,the body is moving in vertical upward direction. Hence, the net force acting on it is-

$T-m_{1} g=m_{1}a$ [1]

$For\ body\ m_{2}$

Here the tension is acting in vertical upward direction while weight is in vertical downward direction. The body is moving in downward direction. Hence the net force acting on it will be-

$m_{2} g-T=m_{2} a$ [2]