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bulgar [2K]
2 years ago
14

Two workers push on a wooden crate. One worker push with a force of 543 N and the other with a force of 333 N. The mass of the w

ooden crate is 206 kg. What is the acceleration of the crate?
Physics
1 answer:
Vinil7 [7]2 years ago
5 0

Answer:

a=4.2524\ m.s^{-2}

Explanation:

Given:

two workers push a crate.

  • Force by first worker, F_1=543\ N
  • force by second worker, F_2=333\ N
  • mass of crate, m=206\ kg

(Assuming that both the workers push in the same direction.)

We know that,

<u>Acceleration is given as:</u>

a=\frac{F}{m}

a=\frac{F_1+F_2}{m}

a=\frac{543+333}{206}

a=4.2524\ m.s^{-2}

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Beth, a construction worker, attempts to pull a stake out of the ground by pulling on a rope that is attached to the stake. The
AnnyKZ [126]

Answer:

Fy=107.2 N

Explanation:

Conceptual analysis

For a right triangle :

sinβ = y/h formula (1)

cosβ = x/h formula (2)

x: side adjacent to the β angle

y:  opposite side of the β angle

h: hypotenuse

Known data

h = T = 153.8 N : rope tension

β= 44.2°with the horizontal (x)

Problem development

We apply the formula (1) to calculate Ty : vertical component of the rope force.

sin44.2° = Ty/153.8 N

Ty = (153.8 N ) *(sen44.2°)= 107.2 N  directed down

for equilibrium system

Fy= Ty=107.2 N

Fy=107.2 N upward component of the force acting on the stake

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3 years ago
25.16 what is the reaction product of acetic acid and ethylamine at room temperature?
Ne4ueva [31]

The ammonium salt of acetic acid is the reaction product of acetic acid and ethylamine at room temperature

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Acetic acid is used in the production  of acetic anhydride, cellulose acetate, vinyl acetate monomer, acetic ester, chloroacetic acid, plastics, dyes, insecticides, photographic chemicals, and rubber. Other commercial uses include the production of vitamins, antibiotics, hormones,  organic chemicals, and as a food additive. Typical concentrations of acetic acid found naturally in foods are 700 to 1200 milligrams/kg (mg/kg) in wine, up to 860 mg/kg in aged cheeses, and 2.8 mg/kg in aged cheeses. fresh orange juice.

 

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Who is responsible for documenting the crime scene in detail and collecting physical
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7 0
3 years ago
Two particles with masses 2m and 9m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m i
s2008m [1.1K]

Answer:

The final speed for the mass 2m is v_{2y}=-1,51\ v_{i} and the final speed for the mass 9m is v_{1f} =0,85\ v_{i}.

The angle at which the particle 9m is scattered is \theta = -66,68^{o} with respect to the - y axis.

Explanation:

In an elastic collision the total linear momentum and the total kinetic energy is conserved.

<u>Conservation of linear momentum:</u>

Because the linear momentum is a vector quantity we consider the conservation of the components of momentum in the x and y axis.

The subindex 1 will refer to the particle 9m and the subindex 2 will refer to the particle 2m

\vec{p}=m\vec{v}

p_{xi} =p_{xf}

In the x axis before the collision we have

p_{xi}=9m\ v_{i} - 2m\ v_{i}

and after the collision we have that

p_{xf} =9m\ v_{1x}

In the y axis before the collision p_{yi} =0

after the collision we have that

p_{yf} =9m\ v_{1y} - 2m\ v_{2y}

so

p_{xi} =p_{xf} \\7m\ v_{i} =9m\ v_{1x}\Rightarrow v_{1x} =\frac{7}{9}\ v_{i}

then

p_{yi} =p_{yf} \\0=9m\ v_{1y} -2m\ v_{2y} \\v_{1y}=\frac{2}{9} \ v_{2y}

<u>Conservation of kinetic energy:</u>

\frac{1}{2}\ 9m\ v_{i} ^{2} +\frac{1}{2}\ 2m\ v_{i} ^{2}=\frac{1}{2}\ 9m\ v_{1f} ^{2} +\frac{1}{2}\ 2m\ v_{2f} ^{2}

so

\frac{11}{2}\ m\ v_{i} ^{2} =\frac{1}{2} \ 9m\ [(\frac{7}{9}) ^{2}\ v_{i} ^{2}+ (\frac{2}{9}) ^{2}\ v_{2y} ^{2}]+ m\ v_{2y} ^{2}

Putting in one side of the equation each speed we get

\frac{25}{9}\ m\ v_{i} ^{2} =\frac{11}{9}\ m\ v_{2y} ^{2}\\v_{2y} =-1,51\ v_{i}

We know that the particle 2m travels in the -y axis because it was stated in the question.

Now we can get the y component of the  speed of the 9m particle:

v_{1y} =\frac{2}{9}\ v_{2y} \\v_{1y} =-0,335\ v_{i}

the magnitude of the final speed of the particle 9m is

v_{1f} =\sqrt{v_{1x} ^{2}+v_{1y} ^{2} }

v_{1f} =\sqrt{(\frac{7}{9}) ^{2}\ v_{i} ^{2}+(-0,335)^{2}\ v_{i} ^{2} }\Rightarrow \ v_{1f} =0,85\ v_{i}

The tangent that the speed of the particle 9m makes with the -y axis is

tan(\theta)=\frac{v_{1x} }{v_{1y}} =-2,321 \Rightarrow\theta=-66,68^{o}

As a vector the speed of the particle 9m is:

\vec{v_{1f} }=\frac{7}{9} v_{i} \hat{x}-0,335\ v_{i}\ \hat{y}

As a vector the speed of the particle 2m is:

\vec{v_{2f} }=-1,51\ v_{i}\ \hat{y}

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