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2 years ago

I

Physics

1 answer:

valentinak56 [21]2 years ago

7 0

Answer:

mass is 56.25 N

Explanation:

F=ma

325=4m

m=325/4

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true or false/ force field that exists in the space around every massive body is called electric field

musickatia [10]

Answer:

True

Explanation:

An electric field is a region around a charged particle or object within which a force would be exerted on other charged particles or objects.

3 0

3 years ago

A car is traveling at the bottom of a 9.00-meter-radius circular hill with a constant speed v. The moment the car is at the bott

Rina8888 [55]

Answer:

Explanation:

reading of scale = reaction force of surface R

centripetal force = R - mg = m v² / R , m is mass , v is velocity and R is radius of the circular path .

R = mg + m v² / R

given ,

m v² / R = .80 mg

v² = .80 x g x R

= .8 x 9.8 x 9 = 70.56

v = 8.4 m /s

3 0

2 years 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

Which of the following is true regarding transverse waves?

DIA [1.3K]

Answer:

in a transverse wave, The directions that energy and matter travel in are perpendicular to one another.

these waves are types of mechanical waves that doesn't require any material medium to transfer.

4 0

3 years ago

A 1-kilogram mass is attached to a spring whose constant is 14 N/m, and the entire system is then submerged in a liquid that imp

ch4aika [34]

Answer:

Part(a): The equation of motion is $\bf{x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}}$ .

Part(b): The equation of motion is $\bf{x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}}$ .

Explanation:

If ' $m$ ' be the mass of the object, ' $k$ ' be the force constant and ' $\beta$ ' be the damping constant, then the equation of motion of the particle can be written as

$\dfrac{d^{2}x}{dt^{2}} + \dfrac{\beta}{m} \dfrac{dx}{dt} + \dfrac{k}{m}x= 0.........................................(I)$

Given $m$ = 1 Kg, $k$ = 14 $N~m^{-1}$ , $\beta$ = 9. Substituting these values in equation (I),

$\dfrac{d^{2}x}{dt^{2}} + 9~\dfrac{dx}{dt} + 14~x= 0$

Taking a trial solution $x(t) = e^{mt}$ , the auxiliary equation can be written as

$m^{2} + 9m + 14 = 0............................................................(II)$

and its solutions are $m_{1} = -2~and~m_{2} = -7$ , resulting the general solution

$x(t) = C_{1}~e^{-2t} + C_{2}~e^{-7t}....................................................................(III)$

The velocity at any instant of time of the mass is

$v(t) = -2C_{1}~e^{-2t} _7~C_{2}~e^{-7t}..............................................................(IV)$

Part(a):

Given $x(t=0) = 1 m,~and~v(t=0) = 0~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(V)\\&and,& 0 = -2C_{1} - 7C_{2}.......................(VI)$

Solving equations (V) and (VI), we have

$C_{1} = \dfrac{7}{5}~and~C_{2} = \dfrac{-2}{5}$

So the equation of motion is

$x(t) = \dfrac{7}{5}~e^{-2t} - \dfrac{2}{5}~e^{-7t}$

Part(b):

Given $x(t=0) = 1 m,~and~v(t=0) = - 12~m~s^{-1}$ . Substituting these values in equation (III) and (IV),

$&& 1 = C_{1} + C_{2}......................(VII)\\&and,& -12 = -2C_{1} -7C_{2}.......................(VIII)$

Solving equations (V) and (VI), we have

$C_{1} = -1~and~C_{2} = \dfrac{11}{5}$

So the equation of motion is

$x(t) = -e^{-2t} + \dfrac{11}{5}~e^{-7t}$

3 0

3 years ago