a 63 kg object needs to be lifted 7 meters in a matter of 5 seconds. approximately how much horsepower is required to achieve this task

Answers

Answer 1

Answer: Power is defined as the rate of doing work or the work per unit of time. The first step to solve this problem is by calculating the work which can be determined by the equation:

W = Fd

where:

F = force exerted = ma
d = distance traveled
m = mass of object
a = acceleration

Acceleration is equivalent to the gravitational constant (9.81 m/s^2) if the force exerted has a vertical direction such as lifting.

W = Fd = mad = 63(9.81)(7) = 4326.21 Joules

Now that we have work, we can calculate power.

P = W/t = 4325.21 J / 5 seconds = 865.242 J/s or watts

Convert watts to horsepower (1 hp = 745.7 watts)

P = 865.242 watts (1hp/745.7 watts) = 1.16 hp

Answer 2

Answer:

work = mgh

power = mgh/t = 63•9.8•7/5 = 864 watts

1 HP = 746 watts

so that is 834/746 = 1.2 HP

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A spring (k = 20 N/m) sits on a frictionless surface. The spring is compressed 20 cm and a 2 kg mass is placed before the spring and released from rest. How much time passes between the time that the mass is at the point x = 0 m and x = 2.0 m? A.) 4.34s B.) 1.42s C.) 2.48s D.)0.23s E.)none of the above

The initial kinetic energy imparted to a 0.020 kg bullet is 1200 J. (a) Assuming it accelerated down a 1.00 m rifle barrel, estimate the average power delivered to it during
the firing. (b) Neglecting air resistance, find the range of this projectile when it is fired
at an angle such that the range equals the maximum height attained.

Answers

Answer:

(a) Power= 207.97 kW

(b) Range= 5768.6 meter

Explanation:

Given,

Mass of bullet, $m=0.02 kg$

Kinetic energy imparted, $K=1200 J$

Length of rifle barrel, $d=1 m$

(a)

Let the speed of bullet when it leaves the barrel is $v$ .

Kinetic energy, $K=(1)/(2) mv^(2)$

$v=\sqrt{(2K)/(m) }$

$=\sqrt{(2*1200)/(0.02) }$

$=346.4m/s$

Initial speed of bullet, $u=0$

The average speed in the barrel, $v_a_v_g=(u+v)/(2)$

$=(0+346.4)/(2) \n=173.2 m/s$

Time taken by bullet to cross the barrel, $t=(d)/(v)$

$=(1)/(173.2)\n =0.00577 second$

Power, $P_a_v_g=(W)/(t)$

$=(1200)/(0.00577) \n=207.97kW$

(b)

In projectile motion,

Maximum height, $H_m=(v^2\sin^2\theta)/(2g) \n$

Range, $R=(v^2\sin2\theta)/(g)$

given that, $H_m=R$

then, $(v^2\sin^2\theta)/(2g)=(v^2\sin2\theta)/(g)\n\sin^2\theta=2\sin\theta\cos\theta\n\n\tan\theta=4\n\theta=\tan^-^14\n\theta=75.96^0\nR=(v^2\sin2\theta)/(g)\n=(346.4^2*\sin(2*75.96))/(9.8)\n5768.6 meter$

Name the apparatus used to measure friction force? PLEASE HELP AS SOON AS POSSIBLE

Answers

There is no apparatus for it. It is either use something like ruler and table and rub together or rub our hands, and friction force will be showed.

Balance KOH+AlCl3 --> Al(OH)3 +KCL

Answers

3KOH+AlCl3--> Al(OH)3+3KCl

Please select ALL THAT APPLYWhich countries listed below EXCLUSIVELY use the metric system?

United States
France
England
Canada

Answers

Answer:

B, C, D

Explanation:

Considering the United States is the only country that does not used the metric system; France, England, and Canada all use the metric system.

3. A block of mass m1=1.5 kg on an inclined plane of an angle of 12° is connected by a cord over a mass-less, frictionless pulley to a second block of a mass m2= 1.2 kg hanging vertically. If the hanging block falls 0.92min 1.23 s, what is the coefficient of friction between m1 and the inclined plane?

Please show work

Answers

Answer:

$\mu=0.377$

Explanation:

we need to start by drawing the free body diagram for each of the masses in the system. Please see attached image for reference.

We have identified in green the forces on the blocks due to acceleration of gravity ( $w_1$ and $w_2$ ) which equal the product of the block's mass times "g".

On the second block ( $m_2$ ), there are just two forces acting: the block's weight ( $m_2\,*\,g$ ) and the tension (T) of the string. We know that this block is being accelerated since it has fallen 0.92 m in 1.23 seconds. We can find its acceleration with this information, and then use it to find the value of the string's tension (T). We would need both these values to set the systems of equations for block 1 in order to find the requested coefficient of friction.

To find the acceleration of block 2 (which by the way is the same acceleration that block 1 has since the string doesn't stretch) we use kinematics of an accelerated object, making use of the info on distance it fell (0.92 m) in the given time (1.23 s):

$x_f-x_i=v_i\,t-(1)/(2) a\,t^2$ and assume there was no initial velocity imparted to the block:

$x_f-x_i=v_i\,t-(1)/(2) a\,t^2\n-0.92\,m=0\,-(1)/(2) a\,(1.23)^2\na=(0.92\,*\,2)/(1.23^2) \na=1.216 \,(m)/(s^2)$

Now we use Newton's second law in block 2, stating that the net force in the block equals the block's mass times its acceleration:

$F_(net)=m_2\,a\nw_2-T=m_2\,a\nm_2\,g-T=m_2\,a\nm_2\,g-m_2\,a=T\nm_2\,(g-a)=T\n1.2\,(9.8-1.216)\,N=T\nT=10.3008\,N$

We can round this tension (T) value to 10.3 N to make our calculations easier.

Now, with the info obtained with block 2 (a - 1.216 $(m)/(s^2)$ , and T = 10.3 N), we can set Newton's second law equations for block 1.

To make our study easier, we study forces in a coordinate system with the x-axis parallel to the inclined plane, and the y-axis perpendicular to it. This way, the motion in the y axis is driven by the y-component of mass' 1 weight (weight1 times cos(12) -represented with a thin grey trace in the image) and the normal force (n picture in blue in the image) exerted by the plane on the block. We know there is no acceleration or movement of the block in this direction (the normal and the x-component of the weight cancel each other out), so we can determine the value of the normal force (n):

$n-m_1\,g\,cos(12^o)=0\nn=m_1\,g\,cos(12^o)\nn=1.5\,*\,9.8\,cos(12^o)\nn=14.38\,N$

Now we can set the more complex Newton's second law for the net force acting on the x-axis for this block. Pointing towards the pulley (direction of the resultant acceleration $a$ ), we have the string's tension (T). Pointing in the opposite direction we have two forces: the force of friction (f ) with the plane, and the x-axis component of the block's weight (weight1 times sin(12)):

$F_(net)=m_1\,a\nT-f-w_1\,sin(12)=m_1\,a\nT-w_1\,sin(12)-m_1\,a=f\nf=[10.3-1.5\,*\,9.8\,sin(12)-1.5\,*1.216]\,N\nf=5.42\,N$

And now, we recall that the force of friction equals the product of the coefficient of friction (our unknown $\mu$ ) times the magnitude of the normal force (14.38 N):

$f=\mu\,n\n5.42\,N=\mu\,*\,14.38\,N\n\mu=(5.42)/(14.38)\n\mu=0.377$

with no units.

While watching a recent science fiction movie, one Klingon spaceship blows up a Droid spaceship with a laser gun. The Klingon crew watches out the port window and covers their ears to muffle the noise from the explosion. Scientifically, this scene is inaccurate because A)
light can not reflect off objects in the vacuum of space.

B)
they would not be able to see anything outside the window.

C)
lasers can not be transmitted through the vacuum of space.

D)
the sound of the explosion would not be transmitted back to their ship.

Answers

Answer:

D) the sound of the explosion would not be transmitted back to their ship.

Explanation:

Sound is a mechanical wave. A mechanical wave requires a medium to travel and transfer energy. It travels be compression and rarefaction of the medium. Since in space it is vacuum, the sound energy by the explosion of the spaceship would not be able to travel through the space, the Klingon crew are not required to cover their ears to muffle the noise. This scene would be inaccurate scientifically.

D. is the right answer because his pressure is very bad out there in the air.good luck

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