The acceleration of an object is constant when its velocity is :

Please help me solve this and give an explanation

Arlecino [84]

Answer:

6.5

Explanation:

Because 1.5+5=6.5

7 0

1 year ago

An object of mass m = 4.0 kg, starting from rest, slides down an inclined plane of length l = 3.0 m. The plane is inclined by an

kirill [66]

Answer:

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane:

Wg= 58.8 J is positive

Explanation:

Nomenclature

vf: final velocity

v₀ :initial velocity

a: acceleleration

d: distance

Ff: Friction force

W: weight

m:mass

g: acceleration due to gravity

Graphic attached

The attached graph describes the variables related to the kinetics of the object (forces and accelerations)

Calculation de of the components of W in the inclined plane

W=m*g

Wx₁ = m*g*sin30°

Wy₁= m*g*cos30°

Object kinematics on the inclined plane

vf₁²=v₀₁²+2*a₁*d₁

v₀₁=0

vf₁²=2*a₁*d₁

$v_{f1} = \sqrt{2*a_{1}*d_{1} }$ Equation (1)

Object kinetics on the inclined plane (μ= 0.2)

∑Fx₁=ma₁ :Newton's second law

-Ff₁+Wx₁ = ma₁ , Ff₁=μN₁

-μ₁N₁+Wx₁ = ma₁ Equation (2)

∑Fy₁=0 : Newton's first law

N₁-Wy₁= 0

N₁- m*g*cos30°=0

N₁ = m*g*cos30°

We replace N₁ = m*g*cos30 and Wx₁ = m*g*sin30° in the equation (2)

-μ₁m*g*cos30₁+m*g*sin30° = ma₁ : We divide by m

-μ₁*g*cos30°+g*sin30° = a₁

g*(-μ₁*cos30°+sin30°) = a₁

a₁ =9.8(-0.2*cos30°+sin30°)=3.2 m/s²

We replace a₁ =3.2 m/s² and d₁= 3m in the equation (1)

$v_{f1} = \sqrt{2*3.2*3} }$

$v_{f1} =\sqrt{2*3.2*3}$

$v_{f1} = 4.38 m/s$

Rough surface kinematics

vf₂²=v₀₂²+2*a₂*d₂ v₀₂=vf₁=4.38 m/s

0 =4.38²+2*a₂*d₂ Equation (3)

Rough surface kinetics (μ= 0.3)

∑Fx₂=ma₂ :Newton's second law

-Ff₂=ma₂

--μ₂*N₂ = ma₂ Equation (4)

∑Fy₂= 0 :Newton's first law

N₂-W=0

N₂=W=m*g

We replace N₂=m*g inthe equation (4)

--μ₂*m*g = ma₂ We divide by m

--μ₂*g = a₂

a₂ =-0.2*9.8= -1.96m/s²

We replace a₂ = -1.96m/s² in the equation (3)

0 =4.38²+2*-1.96*d₂

3.92*d₂ = 4.38²

d₂=4.38²/3.92

(a-1) d₂=4.89 m: The object slides 4.89 m along the rough surface

(a-2) Work (Wf) done by the friction force while the mass is sliding down the in- clined plane:

Wf = - Ff₁*d₁

Ff₁= μ₁N₁= μ₁*m*g*cos30°= -0.2*4*9.8*cos30° = 6,79 N

Wf= - 6.79*3 = 20.4 N*m

Wf= -20.4 J is negative

(b) Work (Wg) done by the gravitational force while the mass is sliding down the inclined plane

Wg=W₁x*d= m*g*sin30*3=4*9.8*0.5*3= 58.8 N*m

Wg= 58.8 J is positive

6 0

3 years ago

Consider the equation . The dimensions of the variables v, x, and t are , , and , respectively. The numerical factor 3 is dimens

Vinvika [58]

Answer:

Part of the question is missing but here is the equation for the function;

Consider the equation v = (1/3)zxt2. The dimensions of the variables v, x, and t are [L/T], [L], and [T] respectively.

Answer = The dimension for z = 1/T3 i.e 1/ T - raised to power 3

Explanation:

What is applied is the principle of dimensional homogenuity

From the equation V = (1/3)zxt2.

V has a dimension of [L/T]

x has a dimension of [L]

t has a dimension of [T]
from the equation, make z the subject of the relation
z = v/xt2 where 1/3 is treated as a constant
Substituting into the equation for z
z = L/T / L x T2
the dimension for z = 1/T3 i.e 1/ T - raised to power 3

5 0

2 years ago

Label each formula in the chemical quation below as either a reactant of a product.

Karolina [17]

Fe and S are both reactants: they react with each other to give a different compound.

FeS is the product of the reaction: it was formed or produced as result of the reaction of Fe and S.

Answer:

Fe: reactant
S: reactant
FeS: product

3 0

3 years ago

Two streams merge to form a river. One stream has a width of 8.3 m, depth of 3.2 m, and current speed of 2.2 m/s. The other stre

katovenus [111]

Answer:

The depth of the resulting stream is 3.8 meters.

Explanation:

Under the assumption that streams are formed by incompressible fluids, so that volume flow can observed conservation:

$\dot V_{1} + \dot V_{2} = \dot V_{3}$ (1)

All volume flows are measured in cubic meters per second.

Dimensionally speaking, we can determine the depth of the resulting stream ( $h_{3}$ ), in meters, by expanding (1) in this manner:

$w_{1}\cdot h_{1}\cdot v_{1} + w_{2}\cdot h_{2}\cdot v_{2} = w_{3}\cdot h_{3}\cdot v_{3}$

$h_{3} = \frac{w_{1}\cdot h_{1}\cdot v_{1}+w_{2}\cdot h_{2}\cdot v_{2}}{w_{3}\cdot v_{3}}$ (2)

$v_{1}, v_{2}$ - Speed of the merging streams, in meters per second.

$h_{1}, h_{2}$ - Depth of the merging streams, in meters.

$w_{1}, w_{2}$ - Width of the merging streams, in meters.

$w_{3}$ - Width of the resulting stream, in meters.

$v_{3}$ - Speed of the resulting stream, in meters per second.

If we know that $w_{1} = 8.3\,m$ , $h_{1} = 3.2\,m$ , $v_{1} = 2.2\,\frac{m}{s}$ , $w_{2} = 6.8\,m$ , $h_{2} = 3.2\,m$ , $v_{2} = 2.4\,\frac{m}{s}$ , $w_{3} = 10.4\,m$ and $v_{3} = 2.8\,\frac{m}{s}$ , then the depth of the resulting stream is:

$h_{3} = \frac{(8.3\,m)\cdot (3.2\,m)\cdot \left(2.2\,\frac{m}{s} \right) + (6.8\,m)\cdot (3.2\,m)\cdot \left(2.4\,\frac{m}{s} \right)}{(10.4\,m)\cdot \left(2.8\,\frac{m}{s} \right)}$

$h_{3} = 3.8\,m$

The depth of the resulting stream is 3.8 meters.

3 0

2 years ago