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4 years ago

How to solve 5,897,159 milligrams into kilograms?

Physics

2 answers:

OlgaM077 [116]4 years ago

8 0

Answer:

5.897159 kilograms

Afina-wow [57]4 years ago

5 0

Answer:

5.897159 kilograms

Explanation:

5,897,159÷1000000=5.897159

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In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-tempera

Talja [164]

Answer:

η = 40 %

Explanation:

Given that

Qa ,Heat addition= 1000 J

Qr,Heat rejection= 600 J

Work done ,W= 400 J

We know that ,efficiency of a engine given as

$\eta=\dfrac{W(net)}{Q(heat\ addition)}$

Now by putting the values in the above equation ,then we get

$\eta=\dfrac{400}{1000}$

η = 0.4

The efficiency in percentage is given as

η = 0.4 x 100 %

η = 40 %

Therefore the answer will be 40%.

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3 years ago

A horizontal force of 92.7 N is applied to a 40.5 kg crate on a rough, level surface. If the crate accelerates at 1.13 m/s2, wha

lord [1]

Answer:

The value is $F_f = 46.935 \ N$

Explanation:

From the question we are told that

The magnitude of the horizontal force is $F = 92.7 \ N$

The mass of the crate is $m = 40.5 \ kg$

The acceleration of the crate is $a = 1.13 \ m/s$

Generally the net force acting on the crate is mathematically represented as

$F_{net} = F - F_f = ma$

Here $F_f$ is force of kinetic friction (in N) acting on the crate

$92.7 - F_f = 40.5 * 1.13$

=> $F_f = 46.935 \ N$

5 0

3 years ago

What is the most important safety rule in cycling?

Mekhanik [1.2K]

Answer:

Wear a helmet. Stay visible; use bike lights and/or wear bright clothes. Look and Signal; use hand signals to let drivers know where you're going, try to make eye contact with them and look before you go.

Explanation:

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3 years ago

Read 2 more answers

A child's top is held in place upright on a frictionless surface. The axle has a radius of ????=3.21 mm . Two strings are wrappe

tekilochka [14]

Answer:

Angular momentum, $L=6.47\times 10^{-3}\ m$

Explanation:

It is given that,

Radius of the axle, $r=3.21\ mm=3.21\times 10^{-3}\ m$

Tension acting on the top, T = 3.15 N

Time taken by the string to unwind, t = 0.32 s

We know that the rate of change of angular momentum is equal to the torque acting on the torque. The relation is given by :

$\tau=\dfrac{dL}{dt}$

Torque acting on the top is given by :

$\tau=F\times r$

Here, F is the tension acting on it. Torque acting on the top is given by :

$\tau=2F\times r$

$2T\times r=\dfrac{L}{t}$

$L=2T\times r \times t$

$L=2\times 3.15\times 3.21\times 10^{-3}\times 0.32$

$L=6.47\times 10^{-3}\ m$

So, the angular momentum acquired by the top is $6.47\times 10^{-3}\ m$ . Hence, this is the required solution.

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4 years ago

A ball is thrown straight up with a velocity of 50 m/s.(use g = -10 m/s^2) a) What is the velocity in m/s after 2 seconds?

Andrei [34K]

Answer:

(a) 30 m/sec

(b) -50 m/sec

Explanation:

We have given initial velocity of ball u = 50 m/sec

Acceleration due to gravity $g=-10m/sec^2$

(a) Time t = 2 sec

Now according to first equation of v = u-gt

So v=50-10×2=30 m/sec

(b) Time t = 10 sec

Now according to first equation of motion

So final velocity v = u-gt = 50-10×10 =-50 m/sec

3 0

4 years ago