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ziro4ka [17]
2 years ago
12

A toy car is given an initial velocity of 5.0 m/s and experiences a constant acceleration of 2.0 m/s656-03-02-00-00_files/i02900

00.jpg. What is the final velocity after 6.0 s?
Physics
2 answers:
Anvisha [2.4K]2 years ago
7 0

Answer:

Final velocity of the car, v = 17 m/s

Explanation:

It is given that,

Initial velocity of the car, u = 5 m/s

Acceleration of the car, a=2\ m/s^2

Time, t = 6 s

We need to find the final velocity of the car after 6 seconds. It can be calculated using first equation of motion as :

v=u+at, v is the final velocity of the car

v=5+2\times 6

v = 17 m/s

So, the final velocity of the car is 17 m/s. Hence, this is the required solution.

Whitepunk [10]2 years ago
5 0
It would be 17 m/s

If we use

V2 = V1 + a*t
Sub in 5 for v1
2m/s*2 for a
And
6 for t
That should give you the answer.
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You run due east at a constant speed of 3.00 m/s for a distance of 120.0 m and then continue running east at a constant speed of
Leni [432]

Answer:

Explanation:

Given

Speed while running towards east is v_1=3\ m/s

Distance traveled in east direction x_1=120\ m

For Another interval you  run with velocity

v_2=5\ m/s

x_2=240\ m

Total displacement=x_1+x_2

=120+120=240\ m

Time for first interval

t_1=\frac{x_1}{v_1}=\frac{120}{3}

t_1=\frac{120}{3}=40\ s

Time for second interval

t_2=\frac{x_2}{v_2}=\frac{120}{5}=24\ s

total time t=t_1+t_2

t=40+24=64\ s

average velocity v_{avg}=\frac{x_1+x_2}{t}

v_{avg}=\frac{240}{64}=3.75\ m/s

Therefore average velocity is less than 4 m/s  

7 0
2 years ago
A straight, horizontal length of copper wire has a current i=28A through it. What are the magnitude and direction of the minimum
seropon [69]

Answer:

0.01631 T

Explanation:

current, i = 28 A

mass per unit length, m/l = 46.6 g/m = 0.0466 kg/m

Let the magnetic field is B.

the weight of the wire is balanced by the magnetic force .

mg = i l B

B = mg / i l

B = (m/l) x g/i

B = 0.0466 x 9.8 / 28

B = 0.01631 T

Thus, the magnetic field is 0.01631 T.

6 0
2 years ago
A crane uses a block and tackle to lift a 2200N flagstone to a height of 25m
Cloud [144]

Remember the headline:  ENERGY IS NEVER CREATED OR DESTROYED

The amount of energy before and after are always equal.  All we ever do with energy is move it around from one place to another.

a). A crane can't create energy.  Lifting the same rock in 20 different ways always takes the <u><em>same amount of work</em></u>.  It doesn't matter whether one person picks the rock straight up, or 50 people get around it and lift it, or roll it up a ramp, or lift it with 16 pulleys and a mile of rope, or use a giant steam crane.

You want to lift a 2200N weight up 25m, you're going to have to supply

(2200N) x (25m) = <em>55,000 Joules</em> of work.

c). YOU put out 55,000 Joules of energy.  It had to GO someplace. Where is it now ? ===>  It's the potential energy the rock has now, from being 25m higher than it was before.  That <em>55,000 Joules</em> is NOW the potential energy  of the rock.

No energy was created or destroyed.  It just got moved around.  

55,000 Joules of energy began as nuclear energy in the core of the sun. Solar radiation carried it to the Earth. Plants absorbed it, and stored it as chemical energy.  You ... or a cow that you ate later ... ate the plants and took the chemical energy.  One way or the other, the chemical energy got stored in your blood and fat.  When you needed to put it out somewhere, you moved it into your muscles, and they converted it into mechanical energy.  Then you used the mechanical energy to exert forces.  Today, you used the original 55,000 joules to lift the flagstone, and NOW that energy is in the flagstone, 25 meters up off the ground !

6 0
2 years ago
Two objects are maintained at constant temperatures, one hot and one cold. Two identical bars can be attached end to end, as in
irina1246 [14]

Answer:

Qa/Qb = k/2×2k = 1/4

Explanation:

a situation is series and b situation is parallel.

let conductance of each plate = k

so net conductance in series = k/2

net conductance in parallel = 2k

so Qa/Qb = k/2×2k = 1/4

7 0
2 years ago
A person is standing on a scale placed on the floor of an elevator. At time t1, the elevator is at rest and the reading on the s
sergiy2304 [10]

the following statements about the motion of the elevator could be true at time t2 is

c. The elevator is moving downward at constant speed.

Explanation:

  • Under given situation, the elevator is at rest and the reading on the scale is 500N.
  • After some time t2, the person is still standing on the scale and the reading on the scale is 400N .
  • It is because if you stand on a scale in an elevator which is accelerating upward, your body will feel heavier because the elevator's floor pressure which presses harder on your feet, and this is why the scale will show a higher reading than the time when the elevator is at rest.
  • Similarly on the other hand, when the elevator accelerates downward, your body will feel lighter. The force which is exerted by the scale is called as the apparent weight; which means it does not change with constant speed.
  • Applying Newton's second law, which concludes about this particular statement about force exerted on a body when at rest and when in motion.

7 0
2 years ago
Read 2 more answers
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