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2 years ago

17.a car is traveling at a constant velocity of 60 km/h for 4 hours means

Physics

1 answer:

Setler [38]2 years ago

7 0

<span>Assuming the car is travelling in the same direction for the entire hour, the acceleration is zero.</span>

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Does it take more force to slow something down than to speed it up.

ANTONII [103]

To bring something to a stop the same force that was applied to speed it up can be used to stop it. If a greater force is used it will stop quicker.

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2 years ago

A 1640 kg merry-go-round with a radius of 7.50 m accelerates from rest to a rate of 1.00 revolution per 8.00 s. Estimate the mer

son4ous [18]

Solution :

Given data :

Mass of the merry-go-round, m= 1640 kg

Radius of the merry-go-round, r = 7.50 m

Angular speed, $\omega = \frac{1}{8}$ rev/sec

$=\frac{2 \pi \times 7.5}{8}$ rad/sec

= 5.89 rad/sec

Therefore, force required,

$F=m.\omega^2.r$

$$$=1640 \times (5.89)^2 \times 7.5$

= 427126.9 N

Thus, the net work done for the acceleration is given by :

W = F x r

= 427126.9 x 7.5

= 3,203,451.75 J

6 0

2 years ago

What do submarines translate into pictures?

prohojiy [21]

It lets the viewer know it's something to do with underwater.

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3 years ago

You hold a metal block of mass 40 kg above your head at a height of 2 m.

Kitty [74]

Answer:

The work done by gravity is 784 J.

Explanation:

Given:

Mass of the block is, $m=40\ kg$

Height to which it is raised is, $h=2\ m$

Acceleration due to gravity is, $g=9.8\ m/s^2$

Now, work done by gravity is equal to the product of force of gravity and the distance moved in the direction of gravity. So,

$\textrm{Work by gravity}=F_g\times h$

Force of gravity is given as the product of mass and acceleration due to gravity.

$\therefore F_g=mg=40\times 9.8=392\ N$ . Now,

$\textrm{Work by gravity}=F_g\times h=392\times 2=784\ J$

Therefore, the work done by gravity is 784 J.

5 0

2 years ago

A yacht is pushed along at a constant velocity by the wind. The wind force is 180N at an angle of 25 degrees to the direction of

damaskus [11]

The parallel component is given by
F=180cos(25)=163.14N

4 0

3 years ago