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3 years ago

What is the acceleration of a 10kg pushed by a 5n force

Physics

1 answer:

Semenov [28]3 years ago

8 0

Force equals mass times acceleration. Or:
F=ma
Plug it in:
5=10a
5/10=(10a)/10
.5m/s²=a

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Can someone help me? Idk why I can't understand this

olya-2409 [2.1K]

It's been a while since I've studied this, but my answers would be:

13. 5730 years. The half-life of a substance is the amount of time it takes for half of it to decay, and, according to the graph, half of the substance remained at 5730 years.

14. 10740 years. According to the graph, only 25% of the carbon remained after 10740 years.

15. 15 atoms. According to the graph, only 12.5% of the carbon remained after 16110 years. 12.5% of 120 atoms is 15 atoms.

16. 1600 atoms. According to the graph, if a sample of carbon is 10740 years old, only 25% of it remains. To find the original amount, multiply the current amount by (100% / 25%), which equals 4. So, 4. 400 atoms * 4 = 1600 atoms is the original amount.

5 0

3 years ago

One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhau

suter [353]

Answer:

actual efficiency is 47.78 %

Carnot efficiency is 67.65 %

power output is 5.20 × 10^3 hp

Explanation:

given data

temperature = 1900°C = 1900+ 273 K = 2173 K

exhaust temperature = 430°C = 430 + 273 K = 703 K

fuel = 7.0 × 10^9 cal

work = 1.4 × 10^10 J

to find out

actual efficiency and Carnot efficiency and power output of engine

solution

first we find actual efficiency that is = work / heat input

put the value and

input energy = 7.0 × 10^9 cal (4.184 J/1 cal) = 29.29 × 10^9 J

actual efficiency = 1.4 × 10^10 / ( 29.29 × 10^9 )

actual efficiency = 0.4778

actual efficiency is 47.78 %

and

Carnot efficiency is = 1 - ( 703 / 2173 )

so Carnot efficiency is = 0.67648

Carnot efficiency is 67.65 %

and

power output = work / time

power output = 1.4 × 10^10 / 3600 sec

power output = 3.88 × 10^6 W

power output = 3.88 × 10^6 W / 746 hp

so power output is 5.20 × 10^3 hp

5 0

3 years ago

A low-pass first-order instrument has a time constant of 20 ms. find the frequency, in hertz, of the input at which the output w

Vladimir [108]

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8 0

3 years ago

in baseball, a pitcher can accelerate 0.15kg ball from rest to 98 mi/hr in a distance of 1.7m. (a) What is the average force exe

Step2247 [10]

Answer:

F=84.68 N

Required force increases

Explanation:

By the equation

$v^{2}=u^{2} +2as\\ 98^{2}*1609.344^{2}/(60*60)^{2}=0+2*a*1.7\\a = 98^{2}*1609.344^{2}/((60*60)^{2}*2*1.7)\\a = 564.50m/s^{2} \\$

By newton's second law :

$F=ma\\F=0.15*564.50 \\F=84.68 N$

When the mass increased the force increases as the equation.

F=ma

3 0

3 years ago

Read 2 more answers

A point charge of -0.70 μC is fixed to one corner of a square. An identical charge is fixed to the diagonally opposite corner. A

MakcuM [25]

Answer:

The magnitude and algebraic sign of q is $14\sqrt{2}\ \mu C$

Explanation:

Given that,

Point charge = -0.70 μC[/tex]

We need to calculate the force for all charges

The electric force at first corner

$F_{1}=\dfrac{-k0.70\times10^{-6}q}{r^2}$

The electric force at opposite corner

$F_{3}=\dfrac{-k0.70\times10^{-6}q}{r^2}$

The net force is

$F=\sqrt{F_{1}^2+F_{2}^2}$

Put the value into the formula

$F=\sqrt{(\dfrac{-k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}$

The electric force at second corner

$F_{2}=\dfrac{-kq^2}{2r^2}$

The net force acting on either of the charges is zero.

So, $F=F'$

$\sqrt{(\dfrac{k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}=\dfrac{kq^2}{2r^2}$

$\sqrt{2}\times\dfrac{0.70\times10^{-6}kq}{r^2}=\dfrac{kq^2}{2r^2}$

$q=14\sqrt{2}\ \mu C$

Hence, The magnitude and algebraic sign of q is $14\sqrt{2}\ \mu C$

8 0

3 years ago