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2 years ago

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

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

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