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saul85 [17]
3 years ago
11

A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are

in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.00 s.
Physics
1 answer:
Hatshy [7]3 years ago
3 0

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

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Answer:

the answer is A because

from tate 4 dozen is 48 and from joe the sixth multiple of eight is 48

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Decreasing the temperature of a reaction decreases the reaction rate because
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D) decreasing the temperature lowers the average kinetic energy of the reactants.
8 0
3 years ago
For a specific volume of 0.2 m3/kg, find the quality of steam if the absolute pressure is (a) 40 kPa and (b) 630 kPa. What is th
ICE Princess25 [194]

Answer:

x=0.0498

x'=0.659

Explanation:

Specific Volume V=0.2m_3/kg

Absolute Pressure (a) P_a= 40kpa

Giving

T_a=75.87

v_f=1.265*10^{-3}m^3/kg

v_g=3.993m^3/kg

                               (b) P_a= 630kpa

Giving

T_b=160.13C

v_f'=1.10282*10^{-3} m^3/kg

v_g'=0.30286 m^3/kg

(a)

Generally the equation for quality of Steam X  is mathematically given by

x=\frac{v-v_f}{v_g-v_f}

x=\frac{0.2-1.0265*10^{-3}}{3.993-1.0265*10^{-3}}

x=0.0498

(b)

Generally the equation for quality of Steam X  is mathematically given by

x'=\frac{v-v_f'}{v_g'-v_f'}

x'=\frac{0.2-1.10*10^{-3}}{3.30-1.1*10^{-3}}

x'=0.659

4 0
3 years ago
An eagle is flying horizontally at a speed of 3.80 m/s when the fish in her talons wiggles loose and falls into the lake 3.90 m
Dovator [93]

Answer:

the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal

Explanation:

initial veetical speed V₀y=0

Horizontal speed Vx = Vx₀= 3.80m/s

Vertical drop height= 3.90m

Let Vy = vertical speed when it got to the water downward.

g= 9.81m/s² = acceleration due to gravity

From kinematics equation of motion for vertical drop

Vy²= V₀y² +2 gh

Vy²= 0 + ( 2× 9.8 × 3.90)

Vy= √76.518

Vy=8.747457

Then we can calculate the velocity of the fish relative to the water when it hits the water using Resultant speed formula below

V= √Vy² + Vx²

V=√3.80² + 8.747457²

V=9.537m/s

The angle can also be calculated as

θ=tan⁻¹(Vy/Vx)

tan⁻¹( 8.747457/3.80)

=66.52⁰

the velocity of the fish relative to the water when it hits the water is 9.537m/s and 66.52⁰ below horizontal

6 0
3 years ago
A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis
bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

B=\frac{\mu_o I_r}{2\pi r}     (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

I_r=JA_r

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by  A_r=\pi(r^2-R_1^2)

The current density is given by the total area and the total current:

J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current  = 4 A

Then, the current in the wire for a distance r is:

I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}  (2)

You replace the last result of equation (2) into the equation (1):

B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})

Finally. you replace the values of all parameters:

B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0
2 years ago
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