How much work is done in lifting a 6 kg object from the ground to a height of 4m?

The latent heat of fusion for Aluminium is 3.97 x 105. How much energy would be required to melt 0.75 kg of it?

RoseWind [281]

Answer:

$E = 2.9775\times10^5 J$

Explanation:

Given: The latent heat of fusion for Aluminum is $L = 3.97\times10^5 J/Kg$

mass to be malted m = 0.75 Kg

Energy require to melt E = mL

$E = 3.97\times10^5\times0.75 = 2.9775\times10^5 J$

Therefore, energy required to melt 0.75 Kg aluminum

$E = 2.9775\times10^5 J$

5 0

3 years ago

If your brother changes the channel, you will change it back, so then he will change it, then you will change it, etc, etc. This

KonstantinChe [14]

The answer is the law of inetia

7 0

3 years ago

A train starts at rest in a station and accelerates at a constant 0.987 m/s2 for 182 seconds. Then the train decelerates at a co

algol [13]

Answer:

$\displaystyle X_T=66.6\ km$

Explanation:

Accelerated Motion

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

$\displaystyle V_f=V_o+a\ t$

where a is the acceleration, and vo is the initial speed .

The train has two different types of motion. It first starts from rest and has a constant acceleration of $0.987 m/s^2$ for 182 seconds. Then it brakes with a constant acceleration of $-0.321 m/s^2$ until it comes to a stop. We need to find the total distance traveled.

The equation for the distance is

$\displaystyle X=V_o\ t+\frac{a\ t^2}{2}$

Our data is

$\displaystyle V_o=0,a=0.987m/s^2,\ t=182\ sec$

Let's compute the first distance X1

$\displaystyle X_1=0+\frac{0.987\times 182^2}{2}$

$\displaystyle X_1=16,346.7\ m$

Now, we find the speed at the end of the first period of time

$\displaystyle V_{f1}=0+0.987\times 182$

$\displaystyle V_{f1}=179.6\ m/s$

That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

$\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0$

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}$

$\displaystyle t=559.5\ sec$

Computing the second distance

$\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}$

$\displaystyle X_2=50,243.2\ m$

The total distance is

$\displaystyle X_t=x_1+x_2=16,346.7+50,243.2$

$\displaystyle X_t=66,589.9\ m$

$\displaystyle \boxed{X_T=66.6\ km}$

8 0

3 years ago

A 50 Kg box sits at rest on a 30 degree ramp where the coef of static friction is 0.5773. If your push was directed at an angle

emmasim [6.3K]

Given data:

m= 50 Kg,

W= m×g = 50 × 9.81 = 490.5 N

ramp angle (α) = 30 degrees,

coefficient of friction (μs) = 0.5773,

Push at an angle (Θ) = 40 degrees,

Determine: Push to get box move up (P)=?

From the figure,

Resolving the forces along the plane

W sinα + μs.R = P cos Θ --------------------- (i)

Resolving the forces perpendicular to the inclined plane

W cosα = R+Psin Θ => R= W cosα - Psin Θ -------------- (ii)

Solving (i) and (ii) and keeping μs = tan Φ, Φ = Θ 

Pmin = W sin( α +Θ )

= W[ sin α.Cos Θ + cos α.sin Θ]

= 490.5 [ (sin 30.cos40) + (cos30.sin 40)]

= 460.9 N

Minimum push required to move the box up the ramp is 460.9 N

7 0

3 years ago

In a typical golf swing the club is in contact with the ball for about 0.0010 s. If the .045 kg ball experiences a force of 4000

Musya8 [376]

Answer:

v = 88.89 [m/s]

Explanation:

To solve this problem we must use the principle of conservation of momentum which tells us that the initial momentum of a body plus the momentum added to that body will be equal to the final momentum of the body.

We must make up the following equation:

$F*t = m*v$

where:

F = force applied = 4000 [N]

t = time = 0.001 [s]

m = mass = 0.045 [kg]

v = velocity [m/s]

$4000*0.001=0.045*v\\v=88.89[m/s]$

5 0

3 years ago