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3 years ago

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a 3,000kg truck runs into the rear of a 1000kg car that was stationary. The truck and car are locked together after the collisio

n and move with speed 9m/s what was the speed of the truck before the collision?

1 answer:

joja [24]3 years ago

7 0

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What is the magnitude of δv12 if the bulb is removed from the socket (i.e. the circuit is not closed)?

andreev551 [17]

Since the circuit is incomplete or not closed, no current flows in the circuit. as per ohm's law , Voltage is directly proportional to current and is given as

V = Voltage = i R where i = current , R = resistance

as no current flows in the circuit, i = 0

the resistance R can not be zero. hence

V = 0 (R)

V = 0 Volts

so the magnitude of the Voltage is zero Volts

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4 years ago

Read 2 more answers

A train is accelerating at a rate of 2 km/hr/s. If its initial velocity is 20 km/hr, what is its velocity after 30 seconds?

Mademuasel [1]

Here it is. *WARNING* VERY LONG ANSWER

________________________________________...
<span>11) If Galileo had dropped a 5.0 kg cannon ball to the ground from a height of 12 m, the change in PE of the cannon ball would have been product of mass(m),acceleration(g)and height(h) </span>

<span>The change in PE =mgh=5*9.8*12=588 J </span>
<span>______________________________________... </span>
<span>12.) The 2000 Belmont Stakes winner, Commendable, ran the horse race at an average speed = v = 15.98 m/s. </span>

<span>Commendable and jockey Pat Day had a combined mass =M= 550.0 kg, </span>

<span>Their KE as they crossed the line=(1/2)Mv^2 </span>

<span>Their KE as they crossed the line=0.5*550*(15.98)^2 </span>

<span>Their KE as they crossed the line is 70224.11 J </span>

<span>______________________________________... </span>
<span>13)Brittany is changing the tire of her car on a steep hill of height =H= 20.0 m </span>

<span>She trips and drops the spare tire of mass = m = 10.0 kg, </span>

<span>The tire rolls down the hill with an intial speed = u = 2.00 m/s. </span>

<span>The height of top of the next hill = h = 5.00 m </span>

<span>Initial total mechanical energy =PE+KE=mgH+(1/2)mu^2 </span>

<span>Initial total mechanical energy =mgH+(1/2)mu^2 </span>

<span>Suppose the final speed at the top of second hill is v </span>

<span>Final total mechanical energy =PE+KE=mgh+(1/2)mv^2 </span>

<span>As mechanical energy is conserved, </span>

<span>Final total mechanical energy =Initial total mechanical energy </span>

<span>mgh+(1/2)mv^2=mgH+(1/2)mu^2 </span>

<span>v = sq rt [u^2+2g(H-h)] </span>

<span>v = sq rt [4+2*9.8(20-5)] </span>

<span>v = sq rt 298 </span>

<span>v =17.2627 m/s </span>

<span>The speed of the tire at the top of the next hill is 17.2627 m/s </span>
<span>______________________________________... </span>
<span>14.) A Mexican jumping bean jumps with the aid of a small worm that lives inside the bean. </span>

<span>a.)The mass of bean = m = 2.0 g </span>

<span>Height up to which the been jumps = h = 1.0 cm from hand </span>

<span>Potential energy gained in reaching its highest point= mgh=1.96*10^-4 J or 1960 erg </span>

<span>b.) The speed as the bean lands back in the palm of your hand =v=sq rt2gh =sqrt 0.196 =0.4427 m/s or 44.27 cm/s </span>
<span>_____________________________ </span>
<span>15.) A 500.-kg horse is standing at the top of a muddy hill on a rainy day. The hill is 100.0 m long with a vertical drop of 30.0 m. The pig slips and begins to slide down the hill. </span>

<span>The pig's speed a the bottom of the hill = sq rt 2gh = sq rt 2*9.8*30 =sq rt 588 =24.249 m/s </span>
<span>__________________________________ </span>
<span>16.) While on the moon, the Apollo astronauts Neil Armstrong jumped up with an intitial speed 'u'of 1.51 m/s to a height 'h' of 0.700 m, </span>

<span>The gravitational acceleration he experienced = u^2/2h = 2.2801 /(2*0.7) = 1.629 m/s^2 </span>
<span>______________________________________... </span>

<span>EDIT </span>
<span>1.) A train is accelerating at a rate = a = 2.0 km/hr/s. </span>

<span>Acceleration </span>

<span>Initial velocity = u = 20 km/hr, </span>

<span>Velocity after 30 seconds = v = u + at </span>

<span>Velocity after 30 seconds = v = 20 km/hr + 2 (km/hr/s)*30s = </span>

<span>Velocity after 30 seconds = v = 20 km/hr + 60 km/hr = 80 km/ hr </span>

<span>Velocity after 30 seconds = v = 80 km/hr=22.22 m/s </span>
<span>_______________________________- </span>
<span>2.) A runner achieves a velocity of 11.1 m/s 9 s after he begins. </span>

<span>His acceleration = a =11.1/9=1.233 m/s^2 </span>

<span>Distance he covered = s = (1/2)at^2=49.95 m</span>

7 0

3 years ago

A 10 m long clothesline is strung so that it is perfectly horizontal. when a shirt is hung in the exact center the line sags to

Leto [7]

Answer:

x = 0.0873 m

Explanation:

given,

length of clothesline = 10 m

the line sags to create an angle that is 1 degree below the horizontal on each end.

As mass is hang at the center the angle made let the deflection be 'x'

as the shirt is hang at the center distance will be equal to 5 m.

now

$tan \theta = \dfrac{d_e}{d}$

$tan \theta = \dfrac{x}{5}$

$tan 1^0 = \dfrac{W}{5}$

$W = 5 \times 0.0175$

x = 0.0873 m

Hence, the cloth line will be x = 0.0873 m from the original length of the clothesline.

3 0

3 years ago

A ball is thrown straight up with enough speed so that it is in the air for several seconds. Assume the positive direction is up

Andrew [12]

Answer:

a) v= 0 m/s b) v= 6.86 m/s

Explanation:

a) When the ball reaches to its highest point, under the influence of gravity, before starting to fall down, it momentarily comes to an stop (this is needed prior to change direction in any movement), so, applying the definition of acceleration, and replacing the acceleration a by g, we have:

vf = v₀ - g*t (1)

The minus sign means that the acceleration due to gravity is always downward, so if we assume that the positive direction is upwards it must be negative.

At the highest point, vf= 0.

b) Prior to solve this point, we need to know which is the time when the ball reaches to its highest point.

As we know vf=0, we can solve (1) for t, as follows:

th = v₀/g

Now, for a time that is 0.7 s before this time, applying the acceleration definition and solving for v again, we have:

v = v₀ -(g *(th-0.7 s)), but th= v₀/g, so we get:

v= v₀ -g((v₀/g)-0.7 s) = v₀ - v₀ + g*0.7 s

⇒ v=g*0.7 s = 9.8 m/s²*0.7 s

⇒ v = 6.86 m/s

6 0

3 years ago

A 4-kg object falls vertically a distance of 5 m. its potential energy has changed by approximately how much?

Brums [2.3K]

Given:
m = 4 kg, the mass of the object
h = 5 m, distance fallen

Neglect air resistance.

The PE (potential energy) is
PE = mgh = (4 kg)*(9.8 m/s²)*(5 m) = 196 J

The PE is converted into KE (kinetic energy) after the fall.
Therefore the PE decreased by 196 J ≈ 200 J

Answer: d. It has decreased by 200 J

7 0

4 years ago