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3 years ago

What is the ACCELERATION of a 50 kg object pushed with a force of 500 newtons *

Physics

1 answer:

Anon25 [30]3 years ago

5 0

Answer:

The answer is

Explanation:

To find the acceleration of an object given the force and mass we use the formula

<h3> $acceleration = \frac{force}{mass}$ </h3>

From the question

mass of object = 50 kg

force = 500 N

So the acceleration is

<h3> $acceleration = \frac{500}{50} \\ = \frac{50}{5}$ </h3>

We have the final answer as

Hope this helps you

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An insulated rigid tank initially contains 1.4-kg saturated liquid water and water vapor at 200°C. At this state, 25 percent of

Elan Coil [88]

Solution:

Mass of liquid water and water vapor in the insulated tank initially = 1.4 kg

Temperature = 200 °C

And 25% of the volume by liquid water is steam.

State 1

$m=\frac{V}{v}$

$m=m_f+m_g$

$1.4=\frac{0.25V}{v_f}+\frac{0.75V}{v_g}$

$1.4=\frac{0.25V}{1.1565 \times 10^{-3}}+\frac{0.75V}{0.1274}$ (taking the value of $v_g$ and $v_g$ at 200°C )

$V=6.304 \times 10^{-3}$

Now quality of vapor

$x=\frac{m_g}{m}$

$=3.377 \times 10^{-3}$

Internal energy at state 1 can be found out by

$u_1=u_f+xu_{fg}$

$=850.65+3.377\times10^{-3}\times 1744.65$

= 856.54 kJ/kg

After heating with the resistor for 20 minutes, at state 2, the tank contains saturated water vapor $v_2=v_g \text { and }\ x=1$

Tank is rigid, so volume of tank is constant.

$v_g=v_2=\frac{V}{m}$

$v_g=\frac{6.304\times 10^{-3}}{1.4}$

$v_g=4.502 \times 10^{-3} \ m^3 /kg$

Now interpolate the value to get temperature at state 2 with specific volume value to get final temperature

$T_2=360+(374.14-360)\left(\frac{0.004502-0.006945}{0.003155-0.006945}\right)$

= 369.11° C

Internal energy at state 2

$u_2=2154.9 \ kJ/kg$

Now power rating of the resistor

$P=\frac{m(u_2-u_1)}{t}$

$P=\frac{1.4(2154.9-856.54)}{20 \times 60}$

= 1.51 kW

6 0

3 years ago

With a frequency of 500 hz, what is the period of a wave

Pavlova-9 [17]

<span>500 hz means 500 times in a second therefore its 2.</span>

5 0

3 years ago

A force of 3600 N is exerted on a piston that has an area of 0.030 m2. What force is exerted on a second piston that has an area

Airida [17]

For Pascal's law, the pressure is transmitted with equal intensity to every part of the fluid:
$p_1 = p_2$
which becomes
$\frac{F_1}{A_1}= \frac{F_2}{A_2}$
where
$F_1=3600 N$ is the force on the first piston
$A_1=0.030 m^2$ is the area of the first piston
$F_2$ is the force on the second piston
$A_2=0.015 m^2$ is the area of the second piston

If we rearrange the equation and we use these data, we can find the intensity of the force on the second piston:
$F_2=F_1 \frac{A_2}{A_1}=(3600 N) \frac{0.015 m^2}{0.030 m^2}= 1800 N$

7 0

3 years ago

Read 2 more answers

46 points :)

IgorC [24]

Answer:

Its is dividing by 2

Explanation:

It starts with 100 them it goes to 50, 25, 12.5 so its a cycle of dividing by 2

3 0

3 years ago

Read 2 more answers