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Kazeer [188]
2 years ago
7

A car has a velocity of 10m/s.it now accelerates for 1m/s for 1/4 minutes. Find the distance travelled in this time and final sp

eed of the car​
Physics
1 answer:
Sonja [21]2 years ago
6 0

¡Hellow!

For this problem, lets recabe information:

v (Velocity) = 10 m/s

a (Aceleration) = 1 m/s²

t (Time) = 1/4 min = 25 s

d (Distance) = ?

v' (Final velocity) = ?

First, for calculate distance, lets applicate formula:

                                        \boxed{\boxed{\text{d = Vo * t + (a * t}^{2})\text{ * 0,5}   } }

Lets replace according we information and let's resolve it:

d = 10 m/s * 25 s + (1 m/s² * (25 s)²) * 0,5

d = 250 m + (625 m) * 0,5

d = 2,5 m + 312,5 m

d = 314 meters.

Now, for calculate final speed, lets applicate formula:

                                                 \boxed{\boxed{\text{v' = v + a * t}   } }

Lets replace according we information and let's resolve it:

v' = 10 m/s + 1 m/s² * 25 s

v' = 10 m/s + 25 m/s

v' = 35 m/s

¿Good Luck?

Att: That guy who use the "ñ".

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An electric car is being designed to have an average power output of 4,600 W for 2 h before needing to be recharged. (Assume the
MAVERICK [17]

Answer:

a)3312 x 10⁴ J

b)I = 57.5 A

c)9200 W

Explanation:

Given that

P =4600 W

Time t= 2 h = 2 x 3600 s= 7200 s

We know that

1 W = 1 J/s

a)

Energy stored in the battery = P .t

                                              =4600 x 7200 J

                                            =3312 x 10⁴ J

b)

We know that power P given as

P = V .I

V=Voltage  ,I =Current

4600 = 80 x I

I = 57.5 A

c)

The energy supplied = 4600 x 2 = 9200 W

7 0
3 years ago
A 27 kg bear slides, from rest, 14 m down a lodgepole pine tree, moving with a speed of 6.1 m/s just before hitting the ground.
Nadusha1986 [10]
<h2>Thus the force of friction is 235 N</h2>

Explanation:

When the bear was at the height of 14 m . Its potential energy = m g h

here m is the mass of bear , g is acceleration due to gravity and h is the height .

Thus P.E =  27 x 10 x 14 = 3780 J

The K.E of the bear just before hitting = \frac{1}{2} m v²

=   \frac{1}{2} x 27 x ( 6.1 )²  = 490 J

The force of friction f = P.E - K.E = 3290 J

Because the work done = Force x Distance

Thus frictional force = \frac{3290}{14} = 235 N

3 0
3 years ago
A 120-V rms voltage at 1000 Hz is applied to an inductor, a 2.00-μF capacitor and a 100-Ω resistor, all in series. If the rms va
natima [27]

Answer:

The inductance of the inductor is 35.8 mH

Explanation:

Given that,

Voltage = 120-V

Frequency = 1000 Hz

Capacitor C= 2.00\mu F

Current = 0.680 A

We need to calculate the inductance of the inductor

Using formula of current

I = \dfrac{V}{Z}

Z=\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}

Put the value of Z into the formula

I=\dfrac{V}{\sqrt{R^2+(L\omega-\dfrac{1}{C\omega})^2}}

Put the value into the formula

0.680=\dfrac{120}{\sqrt{(100)^2+(L\times2\pi\times1000-\dfrac{1}{2\times10^{-6}\times2\pi\times1000})^2}}

L=35.8\ mH

Hence, The inductance of the inductor is 35.8 mH

4 0
3 years ago
Read 2 more answers
What is the symbol of calcium in it's most stable form?
Licemer1 [7]

Answer:Ca

Explanation:fact according to the periodic table

6 0
2 years ago
A grocery cart with a mass of 15 kg is pushed at constant speed along an aisle by a force fp = 12 n which acts at an angle of 17
Korolek [52]

<span>Given:

Mass of cart: 15kg</span>

Aisle length = 14m

Angle = 17° below the horizontal

Force fp = 12 N

 

So, the solution would be like this for this specific problem:

 

<span><span>1)    </span>W(by applied force) = F(applied) x s x cosθ <span>
=>W(a) = 12 x 14 x cos17* = 160.66 J </span></span>
<span><span>
2)    </span>By F(net) = F(applied) - F(friction) <span>
=>As v = constant => a = 0 => F(net) = 0 
=>F(applied) = F(friction) 
<span>=>W(friction) = -F(friction) x s
=>W(friction) = -F(applied) x s 
=>W(f) = -12 x 14 x cos17* = -160.56 J </span></span></span>
<span><span>
3)    </span>0, as the displacement is perpendicular to Force </span>
<span><span>
4)    </span>0, as the displacement is perpendicular to Force</span>  
<span>
To add, the force that is applied<span> to an object by a person or another object is called the applied force.</span></span>
8 0
3 years ago
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