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tensa zangetsu [6.8K]
2 years ago
9

A train moves with a uniform velocity of 36km/hr 10sec. calculate the distance travelled​

Physics
1 answer:
galben [10]2 years ago
5 0

<u>Given:-</u>

Speed = 36 km/hr

converting speed into m/s

Speed = 36*5/18

Speed = 10 m/s

t = 10 sec  

By using the Formula

Distance = Speed * time

D = 10*10

<u>D = 100 m</u>

<em>Hope it helps....</em>

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A car with a mass of 710 kg is traveling at 37 km/hr. It accelerates to a speed of 120 km/hr in 12.6 seconds. What is the net fo
guajiro [1.7K]

Answer: The net force acting on the car 1,299.3 N.

Explanation:

Mass of the car = 710 kg

Initial velocity of the car of the ,u= 37 km/h= 10.27 m/s (1km\h=\frac{5}{18} m/s)

Final velocity of the car,v = 120 km/h = 33.33 m/s

time taken b y car = 12.6 sec

v-u=at

33.33m/s-10.27m/s=23.06 m/s=a(12.6 sec)

a = 1.83 m/s^2

Force=mass\times acceleration

Force=710 kg\times 1.83 m/s^2

Force=1,299.3 N

The net force acting on the car 1,299.3 N.

8 0
3 years ago
The seafloor spreading process at ridges produces what kind of faults?
irga5000 [103]
Active transform faults are between two tectonic<span> structures or faults.</span>
4 0
3 years ago
20 characters or more
andreev551 [17]

Answer:

yes 20 characters or more

Explanation:

3 0
2 years ago
The intensity of the Sun's light in the vicinity of the Earth is about 1000W/m^2. Imagine a spacecraft with a mirrored square sa
NeX [460]

Explanation:

Formula to represent thrust is as follows.

             F = \frac{dP}{dt}

                = \frac{2p}{dt}

or,           p = \frac{E}{c}

         \frac{p}{dt} = \frac{W}{c}

                  F = \frac{2IA}{c}

                     = \frac{2 (1000 W/m^{2})(5.5 \times 10^{3} m)^{2}}{3 \times 10^{8} m/s}

                     = 201.67 N

Thus, we can conclude that the thrust is 201.67 N.

8 0
3 years ago
Using energy considerations, calculate the average force (in N) a 62.0 kg sprinter exerts backward on the track to accelerate fr
slava [35]

Answer:

69.68 N

Explanation:

Work done is equal to change in kinetic energy

W = ΔK = Kf - Ki = \frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

W = F_{total} .d

where m = mass of the sprinter

vf = final velocity

vi = initial velocity

W  = workdone

kf = final kinetic energy

ki = initial kinetic energy

d = distance traveled

Ftotal = total force

vf = 8m/s

vi= 2m/s

d = 25m

m = 60kg

inserting parameters to get:

W = ΔK = Kf - Ki = \frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

F_{total} .d =\frac{1}{2} mv^{2} _{f}  - \frac{1}{2} mv^{2} _{i}

F_{total} = \frac{\frac{1}{2} mv^{2} _{f} - \frac{1}{2} mv^{2} _{i}}{d}

F_{total=} \frac{\frac{1}{2} X 62 X6^{2} -\frac{1}{2} X 62 X2^{2} }{25}

= 39.7

we know that the force the sprinter exerted F sprinter, the force of the headwind Fwind = 30N

F_{sprinter} = F_{total} + F_{wind}  = 39.7 + 30 = 69.68 N

7 0
3 years ago
Read 2 more answers
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