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tatiyna
3 years ago
8

How does an inclined plane affect the effort needed to move a load vertically?

Physics
1 answer:
VashaNatasha [74]3 years ago
5 0

If we pull an object vertically upwards then we need to apply a force which is equal in the magnitude of the weight of the object

F = mg

now when we pull the same object upwards along an inclined plane with angle then we require a force which will balance the component of weight along the inclined

so it is given as

F' = mgsin\theta

so as if we compare the two forces we can say that since the value of sine is always less than 1 for an angle less than 90 degree

so in the 2nd case when we pull the object along the inclined plane it will require less effort

so correct answer is

<em>A. reduce effort</em>

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he drove 3 kilometers every minute

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Which of the following is a part of the digestive system
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The Pancreas

<em>Is an inflammation organ that lies in the lower part of the stomach which plays a big part in the digestive system.</em>

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3 years ago
Suppose the ball is thrown from the same height as in the PRACTICE IT problem at an angle of 32.0°below the horizontal. If it st
scoray [572]

The figure of the problem is missing: find in attachment.

(a) 1.64 s

The ball follows a projectile motion path. The horizontal displacement is given by

x(t) = v_0 cos \theta t

where

v_0 is the initial speed

t is the time

\theta=32.0^{\circ} is the angle below the horizontal

We can rewrite this equation as

t=\frac{x(t)}{v_0 cos \theta} (1)

The vertical displacement instead is given by

y(t) = -v_0 sin \theta t - \frac{1}{2}gt^2 (2)

where

g=9.8 m/s^2 is the acceleration of gravity

Substituting (1) into (2),

y(t) = -x(t) tan \theta - \frac{1}{2}gt^2

We know that for t = time of flight, the horizontal displacement is

x(t) =50.8 m

We also know that the vertical displacement is

y(t) = -45 m

Substituting everything into the equation, we can find the time of flight:

\frac{1}{2}gt^2=-y -x tan \theta\\t=\sqrt{\frac{2(-y-xtan \theta)}{g}}=\sqrt{\frac{2(-(-45)-50.8 tan 32.0^{\circ})}{9.8}}=1.64 s

(b) 36.5 m/s

We can now find the initial speed directly by using the equation for the horizontal displacement:

x(t) = v_0 cos \theta t

where we have

x = 50.8 m

\theta=32.0^{\circ}

Substituting the time of flight,

t = 1.64 s

We find:

v_0 = \frac{x}{t cos \theta}=\frac{50.8}{(1.64)(cos 32.0^{\circ})}=36.5 m/s

(c) 47.1 m/s at 48.8 degrees below the horizontal

As the ball follows a projectile motion, its horizontal velocity does not change, so its value remains equal to

v_x = v_0 cos \theta = (36.5)(cos 32.0^{\circ})=31.0 m/s

The initial vertical velocity is instead

u_y = -v_0 sin \theta = -(36.5)(sin 32.0^{\circ})=-19.3 m/s

And it changes according to the equation

v_y = u_y -gt

So at t = 1.64 s (when the ball hits the ground),

v_y = -19.3 - (9.8)(1.64)=-35.4 m/s

So the impact speed is:

v=\sqrt{v_x^2+v_y^2}=\sqrt{(31.0)^2+(-35.4)^2}=47.1 m/s

While the direction is:

\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-35.4}{31.0})=-48.8^{\circ}

8 0
3 years ago
The change in electric potential energy per unit charge is
zepelin [54]

A) amperes

B) current.  

C) gravity.  

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6 0
3 years ago
Read 2 more answers
An average man is 80 k (800 N) and the area of the shoes he is wearing is 0.0092 m2. What is the pressure he is exerting on the
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Answer:

P = 86956.52 Pa

Explanation:

Data:

  • F = 800 N
  • A = 0.0092 m²
  • P = ?

Use the formula:

  • \boxed{\bold{P=\frac{F}{A}}}

Replace and solve:

  • \boxed{\bold{P=\frac{800\ N}{0.0092\ m^{2}}}}
  • \boxed{\bold{P=86956.52\ Pa}}}

The pressure it exerts on the ground is <u>86956.52 Pascal.</u>

Greetings.

7 0
2 years ago
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