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hram777 [196]
1 year ago
12

this is a 3 part questionOn vacation, your 1400-kg car pulls a 560-kg trailer away from a stoplight with an acceleration of 1.85

m>s2. (a) What is the net force exerted on the trailer? (b) What force does the trailer exert on the car? (c) What is the net force acting on the car?
Physics
1 answer:
mamaluj [8]1 year ago
3 0

ANSWER:

(a) 1036 N

(b) -1036 N

(c) 2590 N

STEP-BY-STEP EXPLANATION:

Given:

Mc = 1400 kg

Mt = 560 kg

a = 1.85 m/s^2

(a)

Force by car on trailer:

\begin{gathered} F_c=m\cdot a \\ F_c=560\cdot1.85 \\ F_c=1036\text{ N} \end{gathered}

(b)

\begin{gathered} F_t=-F_c \\ F_t=-1036\text{ N} \end{gathered}

(c)

\begin{gathered} F_n=1400\cdot1.85 \\ F_n=2590\text{ N} \end{gathered}

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7 0
3 years ago
A disk of radius 10 cm speeds up from rest. it turns 60 radians reaching an angular velocity of 15 rad/s. what was the angular a
stepan [7]

Answer:

a) α = 1.875 \frac{rad}{s^{2} }

b) t = 8 s

Explanation:

Given:

ω1 = 0 \frac{rad}{s}

ω2 = 15 \frac{rad}{s}

theta (angular displacement) = 60 rad

*side note: you can replace regular, linear variables in kinematic equations with angular variables (must entirely replace equations with angular variables)*

a) α = ?

(ω2)^2 = (ω1)^2 + 2α(theta)

15^{2} = 0^{2} + 2(α)(60)

225 = 120α

α = 1.875 \frac{rad}{s^{2} }

b)

α = (ω2-ω1)/t

t = (ω2-ω1)/α = (15-0)/1.875 = 8

t = 8 s

4 0
3 years ago
A -turn rectangular coil with length and width is in a region with its axis initially aligned to a horizontally directed uniform
madam [21]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The maximum emf is \epsilon_{max}= 26.8 V

The emf induced at t = 1.00 s is \epsilon = 24.1V

The maximum rate of change of magnetic flux is   \frac{d \o}{dt}|_{max}  =26.8V

Explanation:

    From the question we are told that

        The number of turns is N = 44 turns

          The length of the coil is  l = 15.0 cm = \frac{15}{100} = 0.15m

          The width of the coil is  w = 8.50 cm =\frac{8.50}{100} =0.085 m

          The magnetic field is  B = 745 \ mT

          The angular speed is w = 64.0 rad/s

Generally the induced emf is mathematically represented as

        \epsilon = \epsilon_{max} sin (wt)

 Where \epsilon_{max} is the maximum induced emf and this is mathematically represented as

            \epsilon_{max} = N\ B\ A\ w

Where \o is the magnetic flux

            N is the number of turns

             A is the area of the coil which is mathematically evaluated as

             A = l *w

        Substituting values

           A = 0.15 * 0.085

               = 0.01275m^2

substituting values into the equation for  maximum induced emf

         \epsilon_{max} = 44* 745 *10^{-3} * 0.01275 * 64.0

                 \epsilon_{max}= 26.8 V

 given that the time t = 1.0sec

substituting values into the equation for induced emf  \epsilon = \epsilon_{max} sin (wt)

      \epsilon = 26.8 sin (64 * 1)

        \epsilon = 24.1V

   The maximum induced emf can also be represented mathematically as

              \epsilon_{max} = \frac{d \o}{dt}|_{max}

  Where  \o is the magnetic flux and \frac{d \o}{dt}|_{max} is the maximum rate at which magnetic flux changes the value of the maximum rate of change of magnetic flux is

         \frac{d \o}{dt}|_{max}  =26.8V

8 0
3 years ago
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at 12.8 m/s2 . At t1 the rocket e
Simora [160]

Answer:4.39 s

Explanation:

Given

initial velocity u=0

acceleration a=12.8 m/s^2

velocity acquired by sled in t_1 time

v=0+at

v=12.8t_1

distance traveled by sled in t_1 s

v^2-u^2=2as

(12.8t_1)^2-0=2\times 12.8\times s_1

s_1=6.4\cdot t_1^2

distance traveled in t_2 time with velocity v=12.8t_1

s_2=v\times t_2

s_2=12.8\times t_1\times t_2

s_2=12.8\cdot t_1\cdot t_2

s_1+s_2=5.37\times 10^3

6.4t_1^2+12.8t_1t_2=5370----1

t_1+t_2=97.7 s

t_2=97.7-t_1

substitute the value of t_2 in 1

we get

6.4t_1^2-1250.56t_1+5370=0

thus t_1=\frac{1250.56-1194.33}{12.8}=4.39 s

t_1=4.39 s

5 0
3 years ago
A thin hoop is hung on a wall, supported by a horizontal nail. The hoop's mass is M=2.0 kg and its radius is R=0.6 m. What is th
boyakko [2]

Answer:

Explanation:

Given that,

Mass of the thin hoop

M = 2kg

Radius of the hoop

R = 0.6m

Moment of inertial of a hoop is

I = MR²

I = 2 × 0.6²

I = 0.72 kgm²

Period of a physical pendulum of small amplitude is given by

T = 2π √(I / Mgd)

Where,

T is the period in seconds

I is the moment of inertia in kgm²

I = 0.72 kgm²

M is the mass of the hoop

M = 2kg

g is the acceleration due to gravity

g = 9.8m/s²

d is the distance from rotational axis to center of of gravity

Therefore, d = r = 0.6m

Then, applying the formula

T = 2π √ (I / MgR)

T = 2π √ (0.72 / (2 × 9.8× 0.6)

T = 2π √ ( 0.72 / 11.76)

T = 2π √0.06122

T = 2π × 0.2474

T = 1.5547 seconds

T ≈ 1.55 seconds to 2d•p

Then, the period of oscillation is 1.55seconds

6 0
3 years ago
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