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Mice21 [21]
3 years ago
13

What is the force of an object that has a mass of 42 kg and an acceleration of 3 m/s22?

Physics
1 answer:
Ivanshal [37]3 years ago
5 0

Answer:

\boxed {\boxed {\sf 126 \ Newtons}}

Explanation:

According to Newton's 2nd Law of Motion, force is the product of mass and acceleration.

F=ma

The mass is 42 kilograms and the acceleration is 3 meters per square second.

m= 42 \ kg \\a= 3 \ m/s^2

Substitute the values into the formula.

F= 42 \ kg * 3 \ m.s^2

Multiply.

F= 126 \ kg*m/s^2

  • 1 kilogram meter per square second is equal to 1 Newton.
  • The answer we calculated is equal to 126 Newtons.

F= 126 \ N

The object's force is <u>126 Newtons.</u>

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A potter’s wheel moves from rest to an angular speed of 0.20 rev/s in 23.8 s. Assuming constant angular acceleration, what is it
ser-zykov [4K]

Answer:

0.053 rad/s^2

Explanation:

0.2 rev/s = 0.2 rev/s * 2π rad/rev = 0.4π rad/s

Since the angular acceleration is assumed to be constant, and the wheel's angular speed is increasing from rest (0 rad/s) to 0.4π rad/s within 23.8s. Then the angular acceleration must be

\alpha = \Delta \omega / \Delta t = \frac{0.4 \pi - 0}{23.8} = 0.053 rad/s^2

8 0
3 years ago
Forest protect the soil from erosion ? True or False
Ivenika [448]

Answer:

<h2><em>True</em></h2>

Forests helps in binding the soil together. Thus, prevent soil from erosion.

4 0
2 years ago
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egoroff_w [7]
B is the correct answer since acid corrode on metals such as carbon, steel, zinc and such.
6 0
3 years ago
A 15 m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60° angle with horizontal. (a) Find t
scoray [572]

Answer:

a)    F₁ = 267.3 N,   N₁ = 1300 N,  b)    μ = 0.324

Explanation:

For this exercise we use the rotational equilibrium condition, we have a reference system is the floor and the anticlockwise rotations as positive, in the adjoint we can see a diagram of the forces

           

let's use subscript 1 for the ladder and 2 for the firefighter

            ∑ τ = 0

          -W₁ x₁ - W₂ x₂ + N₁ y = 0

           N₁ = \frac{W_1 x_1 + W_2 x_2}{y}          (1)

the center of mass of the ladder is at its geometric center,

d = L / 2 = 15/2 = 7.5 m

         cos 60 = x₁ / d₁

         x₁ = d₁ cos 60

         x₁ = 7.5 cos 60

         x₁ = 3.75 m

for the firefighter d₂ = 4 m

         cos 60 = x₂ / d₂

         x₂ = d₂ cos 60

          x₂ = 4 cos 60 = 2 m

for the fulcrum d₃ = 15 m

         sin 60 = y / d₃

         y = d₃ sin 60

         y = 15 sin 60

         y = 13 m

we look for the Normal by substituting in equation 1

         N₂ = \frac{500 \ 3.75 \ + 800 \ 2}{13}

         N₂ = 267.3 N

now let's use the translational equilibrium relations

 X axis

           F₁ - N₂ = 0

           F₁ = N₂

           F₁ = 267.3 N

Axis y

          N₁ - W₁ -W₂ = 0

          N₁ = W₁ + W₂

          N₁ = 500 + 800

          N₁ = 1300 N

b) for this case change the firefighter's distance d₂ = 9 m

          x₂ = 9 cos 60

          x₂ = 4.5 m

we substitute in 1

          N₂ = \frac{500 \ 3.75 \ + 800 \ 4.5}{13}  

          N₂ = 421.15 N

of the translational equilibrium equation on the x-axis

          fr = F₁ = N₂

          fr = 421.15 N

friction force has the expression

          fr = μ N

in this case the reaction of the Earth to the support of the ladder is N1 = 1300N

          μ = fr / N₁

          μ = 421.15 / 1300

          μ = 0.324

8 0
3 years ago
Using Excel, or some other graphing software, plot the values of y as a function of x. (You will not submit this spreadsheet. Ho
Evgesh-ka [11]

Answer:

a) > x<-c(1,2,3,4,5)

> y<-c(1.9,3.5,3.7,5.1,6)

> linearmodel<-lm(y~x)

And the output is given by:

> linearmodel

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept)            x  

      1.10         0.98  

b) y = 0.98 x +1.10

And if we compare this with the general model y = mx +b

We see that the slope is m= 0.98 and the intercept b = 1.10

Explanation:

Part a

For this case we have the following data:

x: 1,2,3,4,5

y: 1.9,3.5,3.7,5.1, 6

For this case we can use the following R code:

> x<-c(1,2,3,4,5)

> y<-c(1.9,3.5,3.7,5.1,6)

> linearmodel<-lm(y~x)

And the output is given by:

> linearmodel

Call:

lm(formula = y ~ x)

Coefficients:

(Intercept)            x  

      1.10         0.98  

Part b

For this case we have the following trend equation given:

y = 0.98 x +1.10

And if we compare this with the general model y = mx +b

We see that the slope is m= 0.98 and the intercept b = 1.10

7 0
3 years ago
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