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2 years ago

Naruto is running towards Zetsu. Starting from rest, he accelerates to a velocity of 15

Physics

1 answer:

Angelina_Jolie [31]2 years ago

8 0

Answer:

1.2 m/s^2

Explanation:

average acceleration = vf-vi/t = (15 m/s-0 m/s)/10 s = 1.2 m/s^2

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A 0.050 kg toy truck moving right at 0.20 m/s collided with a toy car weighing 0.015 kg initially at rest, on a frictionless tra

Kamila [148]

Answer:

0.1667 m/s

Explanation:

m1V1 + m2V2 = m1V3 + m2V4

0.01 = ( 0.0075) + (0.015 * V4)

V4 = (0.01 - 0.0075) / (0.015)

V4= 0.1667

7 0

2 years ago

Read 2 more answers

A train travels 77 kilometers in 1 hour, and the 66 kilometers in 1 hour. What is the average speed?

Kipish [7]

Average speed = (total distance covered) / (time to cover the distance)

Total distance = (77km + 66km) = 143 kilometers

Time to cover the distance = 2 hours

Average speed = (143 km) / (2 hours) = 71.5 km per hour

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2 years ago

In what way would a digital thermometer be preferable to those from a liquid-based thermometer?

garik1379 [7]

Reading a digital thermometer is more preferable because you can just easily read the measurement from a screen, a number will just show in a screen as compared to a liquid-based thermometer where you still have to read the measurement by counting the markers in the thermometer. Hope this helps! Mark brainly please!

5 0

2 years ago

A helicopter starting on the ground is rising directly into the air at a rate of 25 ft/s. You are running on the ground starting

rusak2 [61]

Answer:

The rate of change of the distance between the helicopter and yourself (in ft/s) after 5 s is $\sqrt{725}$ ft/ sec

Explanation:

Given:

h(t) = 25 ft/sec

x(t) = 10 ft/ sec

h(5) = 25 ft/sec . 5 = 125 ft

x(5) = 10 ft/sec . 5 = 50 ft

Now we can calculate the distance between the person and the helicopter by using the Pythagorean theorem

$D(t) = \sqrt{h^2 + x^2}$

Lets find the derivative of distance with respect to time

$\frac{dD}{dt} (t) = \frac{2h \cdot \frac{dh}{dt} +2x \cdot\frac{dx}{dt}} {2\sqrt{h^2 + x^2}}$

Substituting the values of h(t) and x(t) and simplifying we get,

$\frac{dD}{dt}(t) = \frac{50t \cdot \frac{dh}{dt} + 20 \cdot \frac{dx}dt}{2\sqrt{625\cdot t^2 + 100 \cdot t^2}}$

$\frac{dh}{dt} = 25ft/sec$

$\frac{dx}{dt} = 10 ft/sec$

$\frac{Dd}{dt} (t) = \frac{1250t +200t}{2\sqrt{725}t}$ = $\frac{725}{\sqrt{725}}$ = $\sqrt{725}$ ft / sec

5 0

2 years ago

Ross training for a half marathon which is 13.1 miles long. Within four months he has progressed to an 11 mile practice run if R

damaskus [11]

If Ross takes two months off from training, his fitness level will reduce in comparison to what it was two months ago.

In as little as 3–4 weeks after beginning strength training, Ross will probably experience weight increase, energy loss, diminished balance, diminished strength (making it tougher to carry out daily tasks), and overall fewer fitness levels.

Many people mistakenly believe they lose muscle mass far more quickly than they actually do because their muscles' ability to store water and glycogen is declining.

A decrease in strength and muscle mass, with beginners experiencing a smaller decline in strength than experienced lifters.

Ross will experience Increased VO2 Max from exercise. VO2 Max is almost completely lost in people who train at lower intensities.

learn more about fitness here: brainly.com/question/13490156

#SPJ10

4 0

1 year ago