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3 years ago

• How much work isrequired to lift a 2kgobject 2m high?

1 answer:

7 0

Answer You need to consider that the gravity on earth is 9.8 m/s/s. This means any object you let go on the earths surface will gain 9.8 m/s of speed every second. You need to apply a force on the object in the opposite direction to avoid this acceleration. If you are pushing something up at a constant speed, you are just resisting earths acceleration. The more massive and object is, the greater force is needed to accelerate it. The equation is Force = mass*acceleration. So for a 2kg object in a 9.8 m/s/s gravity you need 2kg*9.8m/s/s = 19.6 Newtons to counteract gravity. Work or energy = force * distance. So to push with 19.6 N over a distance of 2 meters = 19.6 N*2 m = 39.2 Joules of energy. There is an equation that puts together those two equations I just used and it is E = mgh

The amount of Energy to lift an object is (mass) * (acceleration due to gravity) * (height)

:Hence, the Work done to life the mass of 2 kg to a height of 10 m is 196 J. Hope it helps❤️❤️❤️

Explanation:

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A rope swing is hung from a tree right at the edge of a small creek. The rope is 5.0 m long; the creek is 3.0 m wide.

irina [24]

Answer:

1) T = 4.5 s

2) T = 4.5 s

3) v = 9.9 m/s

Explanation:

We can use the equation

T = 2π√(L/g)

1) T = 2π√(5m/9.81 m/s²) = 4.5 s

2) T = 2π√(L/g)

T = 2π√(5m/9.81 m/s²) = 4.5 s

3) v = √(2gR)

v = √(2(9.81 m/s²)(5m))

v = 9.9 m/s

8 0

4 years ago

Read 2 more answers

V,? = V2 + 2a Ax

Ivan

Answer:

Explanation:

$V^2=V^2_i+2a(x_f-x_i)$

$0.76^2=0^2+2a(15.2)$

$0.5776=30.4a$

$a=0.019m/s^2$

3 0

3 years ago

PLSSS HELPP

photoshop1234 [79]

Answer:

Well, if you look up countries with the highest birth rates, also those countries poverty rates and what class the majority of the people are in then you should have a better understanding. Also, education is not passed through genes it is learned through expirience and being taught. Also being rich vs being poor probably doesn't have that big of an effect either. It is the living conditions.

Explanation:

5 0

3 years ago

Read 2 more answers

A heavy flywheel is accelerated (rotationally) by a motor that provides constant torque and therefore a constant angular acceler

ki77a [65]

Answer:

a) $t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

b) $\theta_1=\frac{w_1^2}{2\alpha}rad$

c) $t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

Explanation:

1) Basic concepts

Angular displacement is defined as the angle changed by an object. The units are rad/s.

Angular velocity is defined as the rate of change of angular displacement respect to the change of time, given by this formula:

$w=\frac{\Delat \theta}{\Delta t}$

Angular acceleration is the rate of change of the angular velocity respect to the time

$\alpha=\frac{dw}{dt}$

2) Part a

We can define some notation

$w_o=0\frac{rad}{s}$ ,represent the initial angular velocity of the wheel

$w_1=?\frac{rad}{s}$ , represent the final angular velocity of the wheel

$\alpha$ , represent the angular acceleration of the flywheel

$t_1$ time taken in order to reach the final angular velocity

So we can apply this formula from kinematics:

$w_1=w_o +\alpha t_1$

And solving for t1 we got:

$t_1=\frac{w_1-w_o}{\alpha}=\frac{w_1}{\alpha}sec$

3) Part b

We can use other formula from kinematics in order to find the angular displacement, on this case the following:

$\Delta \theta=wt+\frac{1}{2}\alpha t^2$

Replacing the values for our case we got:

$\Delta \theta=w_o t+\frac{1}{2}\alpha t_1^2$

And we can replace $t_1$ from the result for part a, like this:

$\theta_1-\theta_o=w_o t+\frac{1}{2}\alpha (\frac{w_1}{\alpha})^2$

Since $\theta_o=0$ and $w_o=0$ then we have:

$\theta_1=\frac{1}{2}\alpha \frac{w_1^2}{\alpha^2}$

And simplifying:

$\theta_1=\frac{w_1^2}{2\alpha}rad$

4) Part c

For this case we can assume that the angular acceleration in order to stop applied on the wheel is $\alpha_1 =-5\alpha \frac{rad}{s}$

We have an initial angular velocity $w_1$ , and since at the end stops we have that $w_2 =0$

Assuming that $t_2$ represent the time in order to stop the wheel, we cna use the following formula

$w_2 =w_1 +\alpha_1 t_2$

Since $w_2=0$ if we solve for $t_2$ we got

$t_2=\frac{0-w_1}{\alpha_1}=\frac{-w_1}{-5\alpha}$

And from part a) we can see that $w_1=\alpha t_1$ , and replacing into the last equation we got:

$t_2=\frac{\alpha t_1}{5\alpha}=\frac{t_1}{5}sec$

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3 years ago

4. An object has a mass in air of 0.0832 kg, apparent mass in water of 0.0673 kg, and apparent mass in another liquid of 0.0718

Dmitry_Shevchenko [17]

Question

4. An object has a mass in air of 0.0832 kg, apparent mass in water of 0.0673 kg, and apparent mass in another liquid of 0.0718 kg. What is the specific gravity of the other liquid

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Answer:

Explanation:4. An object has a mass in air of 0.0832 kg, apparent mass in water of 0.0673 kg, and apparent mass in another liquid of 0.0718 kg. What is the specific gravity of the other liquid

8 0

2 years ago

• How much work isrequired to lift a 2kgobject 2m high?​

• How much work isrequired to lift a 2kgobject 2m high?