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8_murik_8 [283]
3 years ago
8

I'm not quite sure how the formula works. I thought it was

Physics
1 answer:
juin [17]3 years ago
4 0

Answer:

See the explanation below.

Explanation:

Of course yes, after my explanation you will understand very well how to analyze this type of problems.

First, we must differentiate the weight of the mass, the mass has units of kilograms or pounds, while the weight is the product of the mass by gravitational acceleration. In this way, we have the following formula for the weight of anybody.

w = m*g

where:

w = weight [N] (units of Newtons]

m = mass [kg]

g = gravity acceleration = 9.81 [m/s²]

When we have the product of these units [kg]*[m/s²], We are always going to get as a result the units of Newtons [N], The Newton is a unit of force.

Now, to solve this problem we must use Newton's second law, which tells us that the sum of forces on a body must be equal to the product of mass by acceleration.

ΣF = m*a

where:

ΣF  = forces acting over the gymnast [N]

m = mass = 55 [kg]

a = desacceleration = 6*9.81 = 58.86 [m/²]

Now we have two forces acting over the gymnast the first one is his or her weight (the mass by the gravity acceleration) and the second force is the one made it by his or her legs upwards in order to be standing. Let's take as positive signs the movements and forces upwards and negative downwards

- (m*g) + Fl = m*a

where:

Fl = force made it by the legs [N]

As the deceleration is upward because it prevents the person from continuing to go down, it is taken as a positive sign.

- (55*9.81) + Fl = (55*58.86)

Fl =  (55*58.86) + (55*9.81)

Fl = 3776.85 [N]

I hope you can understand :)

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A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th
shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

F = ma

a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}

Now, we can calculate the final speed of the crate at the end of 10.0 m:

v_{f}^{2} = v_{0}^{2} + 2ad_{1}                  

v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s    

For the next 10.5 meters we have frictional force:

F - F_{\mu} = ma

F - \mu mg = ma

So, the acceleration is:

a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

v_{f}^{2} = v_{0}^{2} + 2ad_{2}  

v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s  

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.  

I hope it helps you!                              

7 0
3 years ago
A meter stick is balanced at the 50 cm mark. You tie a 20 N weight at the 20 cm mark. Where should a 30 N weight be placed so th
vivado [14]

Answer:

20 cm to the right of the center or 20+50 = 70 cm from the left side.

Explanation:

The length of meter stick is 1 m = 100 cm

Balance point on 50 cm

From the center the 20 N weight is 50-20 = 30 cm

Torque is obtained when force is multiplied with the distance

As the force is conserved we have

\dfrac{20}{30}\times 30=20\ cm

The distance will be 20 cm to the right of the center or 20+50 = 70 cm from the left side.

7 0
3 years ago
a linear function has the same y-intercept as x + 4y equals 16 and it's graph contains the point (4,5). Find the slope of the li
navik [9.2K]

Answer:  \bold{\text{Slope (m)}=\dfrac{1}{4}}

<u>Explanation:</u>

A linear equation is of the form: y = mx + b   where

  • m is the slope
  • b is the y-intercept (where it crosses the y-axis)

x + 4y = 16

     4y = -x + 16

       y = -\dfrac{1}{4}x+\dfrac{16}{4}

       y=-\dfrac{1}{4}x+4

The y-intercept (b) = 4

Next, find the slope given point (4, 5) and b = 4

y=mx+b\\\\5=m(4)+4\\\\1=4m\\\\\dfrac{1}{4}=m\\\\\\\\\large\boxed{Slope (m)=\dfrac{1}{4}}

6 0
3 years ago
Definition of distance travelled?​
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Answer:

distance traveled is a total length of the path traveled between two positions.

6 0
3 years ago
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2.4 physics assignment.
kvv77 [185]

Answer:The distance o the ramp that the car traveled is given by d=(1/2)at^2=(0.5)(3.96)(5.76)^2=65.69 meters. The horizontal component of this travel is 65.69*

Explanation:

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2 years ago
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