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wariber [46]
3 years ago
5

A 0.750 kg ball is thrown vertically upward with an initial speed of 5.00 m/s. If the altitude at which the ball is defined to h

ave zero gravitational potential energy is 1.00 m below its initial position, a simple calculation shows that its initial gravitational potential energy is 7.35 J. What is the ball's gravitational potential energy at 0.50 m above its initial position
Physics
1 answer:
Lelechka [254]3 years ago
4 0

Answer:

12.267367282991963672828w9

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A block of weight mg sits on an inclined plane as shown in (Figure 1) . A force of magnitude F1 is applied to pull the block up
svlad2 [7]

Answer:

W = (F1 - mg sin θ) L,   W = -μ  mg cos θ L

Explanation:

Let's use Newton's second law to find the friction force. In these problems the x axis is taken parallel to the plane and the y axis perpendicular to the plane

Y Axis  

       N - W_{y} =

       N = W_{y}

X axis

       F1 - fr - Wₓ = 0

       fr = F1 - Wₓ

Let's use trigonometry to find the components of the weight

     sin θ = Wₓ / W

     cos θ = W_{y} / W

      Wₓ = W sin θ

      W_{y} = W cos θ

We substitute

      fr = F1 - W sin θ

Work is defined by

        W = F .dx

        W = F dx cos θ

The friction force is parallel to the plane in the negative direction and the displacement is positive along the plane, so the Angle is 180º and the cos θ= -1

         

        W = -fr x

        W = (F1 - mg sin θ) L

Another way to calculate is

         fr = μ N

         fr = μ W cos θ

the work is

         W = -μ  mg cos θ L

4 0
3 years ago
Altitude is the angle measured above ____.<br><br> North Pole<br> horizon<br> equator<br> zenith
juin [17]
Altitude is the angle measured above the horizon
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What is the insertion for the brachialis
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Coronoid process of the ulna

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Ben hit a 0.25kg nail into a wood with a 5.2kg hammer. If the hammer moves witht the speed of 52 m/s and two fifth of its kineti
Art [367]

Answer:

Explanation:

Mass of nails is 0.25kg

Mass of hammer 5.2kg

Speed of hammer is =52m/s

Then, Ben kinetic energy is given as

K.E= ½mv²

K.E= ½×5.2×52²

K.E= 7030.4J

Given that, two-fifth of kinetic energy is converted to internal energy

Internal energy (I.E) = 2/5 × K.E

Internal energy (I.E) = 2/5 × 7030.4

I.E=2812.16J.

Energy increase is total Kinetic energy - the internal energy

∆Et= K.E-I.E

∆Et= 7030.4 - 2812.16

∆Et= 4218.24J

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