The mass of an atom is found by adding it protons and

A 75 kg bobsled is pushed along a horizontal surface by two athletes. After the bobsled is pushed a distance of 4.5m starting fr

sergejj [24]

Answer:

The net force acting on the bobsled is 300 N.

Explanation:

Given:

Mass of the bobsled is, $m=75\ kg$

Displacement is, $d=4.5\ m$

Initial speed is, $u=0$ m/s

Final speed is, $v=6.0$ m/s

Net acceleration acting on the bobsled can be determined using the following Newton's equation of motion:

$v^2=u^2+2ad$

Plug in all the given values and solve for acceleration, $a$ .

$(6.0)^2=0+2a(4.5)\\36=9a\\a=\frac{36}{9}=4\ m/s^2$

Now, as per Newton's second law, net force is the product of mass and acceleration. So,

$F_{net}=ma\\F_{net}=75\times 4=300\ N$

Therefore, the net force acting on the bobsled is 300 N.

7 0

3 years ago

The box resting on the inclined plane above has a mass of 20kg. The incline sits at a 30o angle. Find the friction force between

tekilochka [14]

The friction force between the box and the incline if the box does not slide down the incline will be 0.577

The force preventing sliding against one another of solid surfaces, fluid layers, and material components is known as friction. There are several kinds of friction: Two solid surfaces in touch are opposed to one another's relative lateral motion by dry friction.

Given the box resting on the inclined plane above has a mass of 20kg and the The incline sits at a 30 degree angle

We have to find the friction force between the box and the incline if the box does not slide down the incline

Since the frictional force F₁ must equal or exceed gravitational force F₂ down the incline:

F₁ = F₂

μmgcosΘ = mgsinΘ

μ = (mgsinΘ)/(mgcosΘ)

μ = tanΘ

μ = 0.577

Hence the friction force between the box and the incline if the box does not slide down the incline will be 0.577

Learn more about friction force here:

brainly.com/question/24386803

#SPJ4

3 0

2 years ago

Read 2 more answers

A spring loaded toy shoots straight upward with a velocity of 4.5 m/s. Determine the maximum height it reaches. Determine the ti

Angelina_Jolie [31]

Recall that

${v_y}^2-{v_{0y}}^2=2a_y(y-y_0)$

At its maximum height $y_{\mathrm{max}}$ , the toy will have 0 vertical velocity, so that

$-\left(4.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)y_{\mathrm{max}}$

$\implies y_{\mathrm{max}}=1.0\,\mathrm m$

For the toy to reach this maximum height, it takes time $t$ such that

$\dfrac{0+v_0}2t=y_{\mathrm{max}}\implies t=0.46\,\mathrm s$

which means it takes twice this time, i.e. $t=0.92\,\mathrm s$ , for the toy to reach its original position.

The velocity of the toy when it falls 1.0 m below its starting point is

${v_y}^2-\left(4.5\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)(0-1.0\,\mathrm m)$

$\implies{v_y}^2=39.85\,\dfrac{\mathrm m^2}{\mathrm s^2}$

$\implies v_y=-6.4\,\dfrac{\mathrm m}{\mathrm s}$

where we took the negative square root because we expect the toy to be moving in the downward direction.

6 0

3 years ago

A ball thrown at 4 meters/second exerts 200 Newtons of force. The mass of the ball is:

Harlamova29_29 [7]

Answer:

50

Explanation:

200 divided by 4 gives you the mass of 50

5 0

3 years ago

One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren

katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

$\sigma = \frac{F}{A}$

Where,

$\sigma =$ Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

$\sigma_y = \frac{F_y}{A}$

where,

$\sigma_y =$ Shear strength

$F_y =$ Shear Force

$A_0 =$ Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

$311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}$

Resolving for F,

$F= 1399.5N$

PART B) We need here to apply the shear strength equation, then

$\sigma_y = \frac{F_y}{A}$

$210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}$

$F_y = 945N$

In such a way that the material is more resistant to tensile strength than shear force.

6 0

4 years ago