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3 years ago

Pls answer quick & an accurate answer will get brainliest

Physics

1 answer:

Oliga [24]3 years ago

6 0

Answer:

3.766 kg or 3.7 kg

Explanation:

F= M*A equation

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Challenge: Determine the speed of the waves at each tension setting (high, medium and low). Explain what measurements you made t

Svetradugi [14.3K]

Answer:

high tension: 4.2 × 1.5 = 6.3 cm/s

medium tension: 2.8 ×1.5 = 4.2 cm/s

low tension: 0.8 × 1.5 = 1.2 cm/s

Explanation: Given Settings:

amplitude: 0.75 cm

damping: zero

Using

Speed = frequency ×wavelength

Where

Wavelength = 0.75 × 2 = 1.5 cm

Therefore:

high tension: 4.2 × 1.5 = 6.3 cm/s

medium tension: 2.8 ×1.5 = 4.2 cm/s

low tension: 0.8 × 1.5 = 1.2 cm/s

4 0

3 years ago

Three crates with various contents are pulled by a force F pull = 3615 N across a horizontal, frictionless roller‑conveyor syste

Dmitry_Shevchenko [17]

Answer:

m1=914.9kg

m2=604.9kg

m3=864.75kg

Explanation

I think we are suppose to find the mass of the crate.

The effective force that moves the body in positive x direction is 3615N

ΣFx = Σma

Then Fx=3615N

Then the masses be m1, m2 and m3

Then,

ΣF = Σ(ma)

3615=(m1+m2+m3)a

Given that a=1.516

The masses are

m1+m2+m3=, 2384.56. Equation 1

Between mass 1 and mass 2 is, F12=1387.

The effective force that pull mass 1 is 1387.

F12=m1 ×a

Therefore,

m1=F12/a

m1=1387/1.516

m1=914.9kg.

The effective force that pulls crate 1 and crate 2 is F23

F23=(m1+m2)a

Therefore

2304=(m1+m2)a

Therefore, since a=1.516

m1+m2=2304/1.516

m1+m2=1519.8kg

Since m1=914.9kg

So, m2=1519.8-m1

m2=1519.8-914.9

m2=604.9kg

Also from equation 1

m1+m2+m3=2384.56

Since m1=914.9kg and m2=604.9kg

Then, m3=2384.56-604.9-914.9

m3=864.75kg

3 0

3 years ago

when a 4.25 kg object is placed on top of a vertical spring the spring compress a distance of 2.62 cm what is the spring force c

Alina [70]

Answer:

425n

Explanation:

4 0

3 years ago

34)You find it takes 200 N of horizontal force tomove an unloaded pickup truck along a level road at a speed of2.4 m/s. You then

velikii [3]

Answer:

$F_H_n=230.04 N$

The Required horizontal force is 230.04N

Explanation:

Since the velocity is constant so acceleration is zero; a=0

Now the horizontal force required to move the pickup is equal to the frictional force.

$F_H_n=F_f\\F_h=mg*u$

where:

F_{Hn} is the required Force

u is the friction coefficient

m is the mass

g is gravitational acceleration=9.8m/s^2

$200=mg*u$ Eq (1)

Now, weight increases by 42% and friction coefficient decreases by 19%

New weight=(1.42*m*g) and new friction coefficient=0.81u

$F_H=(1.42m*g*.81u)$ Eq (2)

Divide Eq(2) and Eq (1)

$\frac{F_H_n}{200}=\frac{1.42m*g*0.81u}{m*g*u}\\F_H_n=1.42*0.81*200\\F_H_n=230.04 N$

The Required horizontal force is 230.04N

4 0

4 years ago

How many significant digits are in 0.3500

const2013 [10]

4 significant figures (the 0’s are considered significant because they follow a non-zero number)

7 0

4 years ago