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pav-90 [236]
3 years ago
15

My stoked.elle its a new acc

Physics
1 answer:
klasskru [66]3 years ago
3 0

Answer:

cool

Explanation:

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PLEASE HELP ME WITH A PHYSICS QUESTION!!!!!!!!!!
stira [4]

Answer:

C. 3.00 s

Explanation:

Given:

Δy = 1.80 m − 46.0 m = -44.2 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

-44.2 m = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 3.00 s

6 0
3 years ago
Read 2 more answers
Help please!
Talja [164]

The correct expression for the maximum speed of the object during its motion is \frac{FT}{m}.

<h3>Maximum speed of the object</h3>

The maximum speed of the object is determined using the following formulas;

v(max) = Aω

where;

  • A is the amplitude of the motion
  • ω is angular speed

The maximum speed of the object can also be obtained from the maximum net force on the object,

F = ma

where;

  • F is the maximum net force
  • a is the acceleration
  • m is mass of the object

F = m(v/t)

mv = Ft

v = Ft/m

Thus, the correct expression for the maximum speed of the object during its motion is \frac{FT}{m}.

Learn more about maximum speed here: brainly.com/question/4931057

3 0
2 years ago
The Venn diagram compares protons with neutrons. Which shared property belongs in the region marked "B"?
telo118 [61]

is there any choices?

4 0
3 years ago
Read 2 more answers
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
3 years ago
A uniform 1500-kg beam, 20.0 m long, supports a 15,000-kg printing press
Ghella [55]

Answer:

\mathbf{F_1=4.41*10^4\ N}

\mathbf{F_2 = 1.176*10^5 \ N}

Explanation:

The missing image of the figure slide is attached in below.

However, from the model, it is obvious that it is in equilibrium.

As a result, the relation of the force and the torque is said to be zero.

i.e.

\sum F = 0 and \sum \tau = 0

From the image, expressing the forces through the y-axis, we have:

F_1+F_2 = W_B + W_P \\ \\ \implies 9.8(1500+15000) \\ \\ \implies  \mathtt{1.617\times 10^5 \ N}

Also, let the force F_1 be the pivot and computing the torque to determine F_2:

Then:

F_1(0)+F_2(20.0) = 10.0W_B + 15.0W_P

F_2 = \dfrac{((10*1500)+(15*15000))*9.8}{20.0}

F_2 = 117600 \ N

\mathbf{F_2 = 1.176*10^5 \ N}

For the force equation:

F_1+F_2=1.617*10^5 \ N;

where:

F_2 = 1.176*10^5 \ N

Then:

F_1+1.176*10^5 \ N=1.617*10^5 \ N

F_1=1.617*10^5 \ N-1.176*10^5 \ N

F_1=44100\ N

\mathbf{F_1=4.41*10^4\ N}

8 0
3 years ago
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