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Sergio [31]
2 years ago
7

An eastbound car travels a displacement of 80.0 Km in 45 minutes. What is the velocity

Physics
1 answer:
alexgriva [62]2 years ago
4 0

Explanation:

d= 80km = 8000m

t = 45 min = 45/60 h

= 0.75 h

V= ?

we know that,

V = d /t

or,V= 80 km / 0.75 h

  1. or, V= 106.67 km/hr

or,V= 106.67×1000m / 3600 s

2. or, V= 29.63 m/s

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A 0.500 H inductor is connected in series with a 93 Ω resistor and an ac source. The voltage across the inductor is V = −(11.0V)
bezimeni [28]

Answer:

205 V

V_{R} = 2.05 V

Explanation:

L = Inductance in Henries, (H)  = 0.500 H

resistor is of 93 Ω so R = 93 Ω

The voltage across the inductor is

V_{L} = - IwLsin(wt)

w = 500 rad/s

IwL = 11.0 V

Current:

I = 11.0 V / wL

 = 11.0 V / 500 rad/s (0.500 H)

 = 11.0 / 250

I = 0.044 A

Now

V_{R} = IR

    = (0.044 A) (93 Ω)

V_{R} = 4.092 V

Deriving formula for voltage across the resistor

The derivative of sin is cos

V_{R} = V_{R} cos (wt)

Putting V_{R} = 4.092 V and w = 500 rad/s

V_{R} = V_{R} cos (wt)

    = (4.092 V) (cos(500 rad/s )t)

So the voltage across the resistor at 2.09 x 10-3 s is which means

t = 2.09 x 10⁻³

V_{R} = (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))

    =  (4.092 V) (cos (500 rads/s)(0.00209))

    = (4.092 V) (cos(1.045))

    = (4.092 V)(0.501902)

    = 2.053783

V_{R} = 2.05 V

8 0
3 years ago
What is the acceleration of a 59 kg object pushed with a force of 500 Newton's ?
Igoryamba

Answer:

a=500/50=10 m/s^2

Explanation:

f=m•a     a=f/m

a=500/50= 10 m/s^2

4 0
2 years ago
In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400
liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, f_{ra} = \mu mg\\............(1)

Since frictional force is a type of force, we are safe to say f_{ra} = ma.......(2)

Equating (1) and (2)

ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}

a) Speed of A at the start of the sliding

d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s

b) Speed of B at the start of the sliding

d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s

Let the speed of car B before collision = v_{B1}

Momentum of car B before collision = m_{B} v_{B1}

Momentum after collision = m_{A} v_{A} + m_{B} v_{B2}

Applying the law of conservation of momentum:

m_{B} v_{B1}  = m_{A} v_{A} +m_{B} v_{B2}

m_{A} = 1100 kg\\m_{B} = 1400 kg

(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s

3 0
3 years ago
Read 2 more answers
What two factor affect the speed of wave as it travels through a medium
Aleksandr [31]
I believe it’s density and temperature
5 0
1 year ago
classic physics problem states that if a projectile is shot vertically up into the air with an initial velocity of 128 feet per
ddd [48]

Answer:

t_1 = 0.28 s

t_2 = 7.72 s

Explanation:

Given that height of the projectile as a function of time is

h = -16 t^2 + 128 t + 112

here we know that

h = 147 ft

so from above equation

147 = -16 t^2 + 128 t + 112

16 t^2 - 128 t + 35 = 0

now by solving above quadratic equation we know that

t_1 = 0.28 s

t_2 = 7.72 s

6 0
2 years ago
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