All categories

3 years ago

What is The distance traveled by a vehicle in 12 minutes,if it’s speed is 35km/h

Physics

1 answer:

Delvig [45]3 years ago

3 0

Answer: 7 km

Explanation: 12 minutes is 1/5 of an hour. 35/5=7

You might be interested in

Pls help me its due today

poizon [28]

Answer:

pls give me brainliest

Explanation:

3 0

3 years ago

A helicopter is ascending vertically with a speed of 5.10m/s. At a height of 105m above the Earth, a package is dropped from a w

Luden [163]

Here it is given that initial speed of the package will be same as speed of the helicopter

$v_i = 5.10 m/s$

displacement of the package as it is dropped on ground

$d = -105 m$

acceleration is due to gravity

$a = -9.8 m/s^2$

now by kinematics

$y = v* t + \frac{1}{2}at^2$

$-105 = 5.1 * t - \frac{1}{2}*9.8*t^2$

$4.9 t^2 - 5.1 t - 105 = 0$

by solving above equation we have

$t = 5.2 s$

so it will take 5.2 s to reach the ground

7 0

3 years ago

People often use simple machines like pulleys, levers, and ramps because they say the machine “makes the work easier.” Which of

dimulka [17.4K]

I think the answer is D.
Hope this help

3 0

2 years ago

in a 4 kilometer race, a runner completes the first kilometer in 5.9 minutes, the second kilometer in 6.2 minutes, the third kil

ohaa [14]

To find the average of data collected, add all of the measurements: 5.9+6.2+6.3+6= 12.2

Then, divide the total amount by the number of data collected which is 4: 12.2/4= 3.05

The average speed of the runner of the race is approximately 3.05 km/min

Feel free to ask me any other questions you might have :)

5 0

3 years ago

Read 2 more answers

A catapult used by medieval armies hurls a stone of mass 32.0 kg with a velocity of 50.0 m/s at a 30.0 degree angle above the ho

NISA [10]

Answer:

The horizontal distance traveled when the stone returns to its original height = 220.81 m

Explanation:

Considering vertical motion of catapult:-

At maximum height,

Initial velocity, u = 50 sin30 = 25 m/s

Acceleration , a = -9.81 m/s²

Final velocity, v = 0 m/s

We have equation of motion v = u + at

Substituting

v = u + at

0 = 25 - 9.81 x t

t = 2.55 s

Time of flight = 2 x Time to reach maximum height = 2 x 2.55 = 3.1 s

Considering horizontal motion of catapult:-

Initial velocity, u = 50 cos30 = 43.30 m/s

Acceleration , a = 0 m/s²

Time, t = 5.10 s

We have equation of motion s= ut + 0.5 at²

Substituting

s= ut + 0.5 at²

s = 43.30 x 5.10 + 0.5 x 0 x 5.10²

s = 220.81 m

The horizontal distance traveled when the stone returns to its original height = 220.81 m

5 0

3 years ago