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1 year ago

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a car initially at rest can accelerate at 7 m/s^2 how long will it take the car to reach 60 m/s and how far will it travel durin

g this time

1 answer:

Alex787 [66]1 year ago

7 0

1. The time taken for the car to reach a velocity of 60 m/s is 8.57 s

2. The distance travelled during the time is 257.14 m

<h3>What is acceleration? </h3>

The acceleration of an object is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

a is the acceleration
v is the final velocity
u is the initial velocity
t is the time

1. How to determine the time

Initial velocity (u) = 0 m/s
Acceleration (a) = 7 m/s²
Final velocity (v) = 60 m/s
Time (t) =?

a = (v – u) / t

Thus,

t = (v – u) / a

t = (60 – 0) / 7

t = 8.57 s

2. How to determine the distance

Initial velocity (u) = 0 m/s
Acceleration (a) = 7 m/s²
Final velocity (v) = 60 m/s
Distance (s) = ?

v² = u² + 2as

60² = 0² + (2 × 7 × s)

3600 = 0 + 14s

3600 = 14s

Divide both sides by 14

s = 3600 / 14

s = 257.14 m

Learn more about acceleration and velocity:

brainly.com/question/491732

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A mass of 267 g is attached to a spring and set into simple harmonic motion with a period of 0.176 s. If the total energy of the

Gnoma [55]

Answer:

(a) 7.1 m /sec

(b) 339.9 N/m

(c) 19.91 cm

Explanation:

We have given mass m = 267 gram = 0.267 kg

Time period T = 0.176 sec

Total energy of the oscillating system = 6.74 J

We know that energy is given by

(a) $Ke=\frac{1}{2}mv_{max}^2$

$6.74=\frac{1}{2}\times 0.267\times v_{max}^2$

$v_{max}=7.1m/sec$

(b) Now $\omega =\frac{2\pi }{T}=\frac{2\times 3.14}{0.176}=35.681rad/sec$

We know that $\omega =\sqrt{\frac{k}{m}}$

$35.68=\sqrt{\frac{k}{0.267}}$

$k=339.9N/m$

(c) We know that energy is given by

$E=\frac{1}{2}KA^2$

$6.74=\frac{1}{2}\times 339.9\times A^2$

$A=19.91cm$

4 0

3 years ago

A projectile is shot horizontally at 23.4 m/s from the roof of a building 55.0 m tall. What is the time necessary for the projec

baherus [9]

The horizontal speed has no effect on the answer.

It doesn't matter whether you flick a marble horizontally from the roof,
fire a high-power rifle horizontally from the roof, drive a school bus straight
off the roof, or drop a bowling ball from the roof with zero horizontal speed.
Their vertical speed is completely determined by gravity, (and it happens to
be the same for all of them).

Handy dandy formula for the distance covered by anything that starts out
with zero speed and accelerates to the end:

      Distance = (1/2) (acceleration) x (time)²

If the beginning of the journey is on Earth, then the acceleration is
9.8 m/s² ... the acceleration of gravity on Earth. We'll assume that
the 55-meter rooftop in the question is part of a building on Earth.

   55 meters = (1/2) (9.8 m/s²) x (time)²

Divide each side
by 4.9 m/s² : 55 m / 4.9 m/s² = (time)²

   (time)² = (55/4.9) sec²

Square-root
each side: time = √(55/4.9 sec²)

   =    3.35 sec .

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3 years ago

Two objects with masses m and 3m are connected by a compressed spring so that the energy stored in the spring is 375 joules. If

cluponka [151]

Answer:

281.25 J

Explanation:

We are told that the two objects with masses m and 3m.

Also that energy stored in the spring is 375 joules.

Now, initially the centre of mass of the system took place at rest, it means v1 = v and v2 = v/3

Thus, from principle of conservation of energy, we have;

½mv² + ½(3m)(v/3)² = 375J

(m + 3m/9)½v² = 375

(4/3)m × ½v² = 375

Multiply both sides by ¾ to get;

½mv² = 375 × ¾

½mv² = 281.25 J

Therefore, energy of lighter body is 281.25 J

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2 years ago

To practice Problem-Solving Strategy 30.1: Inductors in Circuits. A circuit has a 1 V battery connected in series with a switch.

ki77a [65]

Answer:

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Explanation:

Since we have two sections, for the Inductor region there would be a current $i_1$ . In the case of resistance 2, it will cross a current $i_2$

Defined this we proceed to obtain our equations,

For $i_1$ ,

$\frac{di_1}{dt}+i_1R_1 = V$

$I_1 = \frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})$

For $i_2$ ,

$I_2R_2 =V$

$I_2 = \frac{V}{R_2}$

The current in the entire battery is equivalent to,

$i_t = I_1+I_2$

$i_t = \frac{V}{R_2}+\frac{V}{R_1} (1-e^{-\frac{R_1t}{L}})$

Our values are,

$V=1V$

$R_1 = 95\Omega$

$L= 1.5*10^{-2}H$

$R_2 =360\Omega$

Replacing in the current for t= 0.4m/s

$i=\frac{1}{360}+\frac{1}{95}(1-e^{-\frac{95*0.4}{1.5*10^{-2}}})$

$i= 0.0133A$

$i_1 = 0.01052A$

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3 years ago

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I found this!! maybe this will help :)

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2 years ago