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Savatey [412]
3 years ago
12

What is the effeciency of the machine that has an input of 120 and an output of 96?

Physics
1 answer:
Rama09 [41]3 years ago
6 0
Efficiency  = output divided by input
96/120 = 0.8
or can say 80% as well

You might be interested in
1.) A projectile is launched at 15 degrees. The landing height is the same as the launch position. At what other angle can the p
allsm [11]
1) The general equations of motion of the projectile on the x and y axis are:
x(t) = v_0 \cos \alpha t
y(t)=v_0 \sin \alpha t -  \frac{1}{2}gt^2
where v0 is the initial velocity, \alpha is the angle with respect to the ground, and g=9.81 m/s^2 is the gravitational acceleration. We can see that the motion of the projectile is an uniform motion on the x-axis and an uniformly accelerated motion on the y-axis.

First, we need to find what is the total horizontal displacement of the projectile when it is launched with an angle of 15^{\circ}. To do that, we need to find first the time t at which the projectile lands to the ground, and we can find it by requiring y(t)=0:
v_0 \sin \alpha t -  \frac{1}{2}gt^2 =0
t( v_0 \sin \alpha -  \frac{1}{2} gt)=0
that has two solutions: t=0 (beginning of the motion) and
t= \frac{2 v_0 \sin \alpha}{g}
and this is the time after which the projectile lands to the ground. If we substitute this value into the equation for x(t), we find the total horizontal displacement of the projectile:
x_1=v_0 \cos \alpha t = v_0 \cos \alpha ( \frac{2 v_0 \sin \alpha }{g} )= \frac{2 v_0^2}{g} \sin \alpha \cos \alpha
with \alpha=15^{\circ}.

If we call \beta the other angle at which the projectile reaches the same horizontal displacement, the total horizontal displacement in this case is
x_2 =  \frac{2 v_0^2}{g} \sin \beta \cos \beta
Since the horizontal displacement should be the same in the two cases, we can write x1=x2, which becomes:
\sin \alpha \cos \alpha = \sin \beta \cos \beta
Now let's remind that \cos \theta= \sin (90^{\circ} -\theta) so that we can rewrite the equation as
\sin \alpha \sin (90^{\circ}-\alpha) = \sin \beta \sin (90^{\circ}-\beta)
and using \alpha=15^{\circ}:
\sin 15^{\circ} \sin (75^{\circ}) = \sin \beta \sin (90^{\circ}-\beta)
and we can see that there are two values of \beta that satisfy the equation: \beta=\alpha=15^{\circ} and \beta=75^{\circ}, which is the solution of our problem.

2) The vertical velocity of the ball at the very top of its trajectory is zero. In fact, the very top of the trajectory is the point where the ball starts to go down, so it means it is the moment when the the direction of the vertical velocity of the ball is changing from upward to downward, so it must be the moment when the vertical velocity is zero.
7 0
3 years ago
PLEASE PROVIDE AN EXPLANATION<br><br> THANK YOU!
Rus_ich [418]

Answer:

(a) 0.993 s

(b) 14.0 N/m

(c) -3.02 m/s

(d) -6.01 m/s²

Explanation:

(a) The block's position can be modeled as a cosine wave:

x(t) = A cos(ωt)

where A is the amplitude (in this case, 50 cm) and ω is the angular frequency.

At t = 0.200 s, x(t) = 15.0 cm.

15.0 cm = 50.0 cm cos((0.200 s) ω)

0.3 = cos((0.2 s) ω)

1.266 rad = (0.2 s) ω

ω = 6.33 rad/s

The period is:

T = (2π rad) (1 s / 6.33 rad)

T = 0.993

(b) For a spring-mass system, ω = √(k/m).  The mass of the block is 0.350 kg, so:

ω = √(k/m)

6.33 rad/s = √(k / 0.350 kg)

6.33 rad/s = √(k / 0.350 kg)

40.1 rad/s² = k / 0.350 kg

k = 14.0 N/m

(c) Energy is conserved:

EE₀ = EE + KE

½ kx₀² = ½ kx² + ½ mv²

kx₀² = kx² + mv²

(14.0 N/m) (0.50 m)² = (14.0 N/m) (0.15 m)² + (0.35 kg) v²

v = -3.02 m/s

Alternatively, we can take the derivative of our position equation:

v(t) = -Aω sin(ωt)

v = -(0.50 m) (6.33 rad/s) sin((6.33 rad/s) (0.2 s))

v = -3.02 m/s

(d) Sum of forces on the block:

∑F = ma

-kx = ma

a = -kx / m

a = -(14.0 N/m) (0.15 m) / (0.350 kg)

a = -6.01 m/s²

Alternatively, we can take the derivative of our velocity equation:

a(t) = -Aω² cos(ωt)

a = -(0.50 m) (6.33 rad/s)² cos((6.33 rad/s) (0.2 s))

a = -6.01 m/s²

6 0
3 years ago
Read 2 more answers
FORCE AND DISPLACEMENT AT AN ANGLE A sailor pulls a boat a distance of 30.0 m along a dock using a rope that makes a 25.0° angle
Bingel [31]

Answer: 6117.58 J

Explanation:

We know that W=Fd*cos(theta) where theta is the angle between the displacement and the force.

In this case, we are given that F=225 N, d=30 m, and theta=25 degrees.

Plugging all this in we get

W=225*30*cos(25)=6117.58 J

7 0
3 years ago
A person jumping out of speeding bus goes forward why?​
kobusy [5.1K]

Answer:

gravity because its gravity and you fall and die

4 0
2 years ago
two bowling balls each have a mass of 8kg. if they are 2 m apart, what is the gravitational force between them?
Scilla [17]

Answer:

1 x 10^-9 N

Explanation:

F = Gm²/d² = 6.674e-11(8²)/2² = 1.06784e-9

4 0
2 years ago
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