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Naddik [55]
3 years ago
11

A golfer tees off and hits a golf ball at a speed of 31 m/s and at an angle of 35 degrees. What is the vertical velocity compone

nt of the ball? Round the answer to the nearest tenth of a m/s?
Physics
1 answer:
Mariulka [41]3 years ago
4 0
Depends on how strong you are. People can hit it anywhere between a few inches and 500 yards.
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Which objects can be electrically polarized?
SIZIF [17.4K]

neutral objects is correct

5 0
3 years ago
Read 2 more answers
A 6.99-g bullet is moving horizontally with a velocity of +341 m/s, where the sign + indicates that it is moving to the right (s
Ratling [72]

Answer:

a). 1.218 m/s

b). R=2.8^{-3}

Explanation:

m_{bullet}=6.99g*\frac{1kg}{1000g}=6.99x10^{-3}kg

v_{bullet}=341\frac{m}{s}

Momentum of the motion the first part of the motion have a momentum that is:

P_{1}=m_{bullet}*v_{bullet}

P_{1}=6.99x10^{-3}kg*341\frac{m}{s} \\P_{1}=2.3529

The final momentum is the motion before the action so:

a).

P_{2}=m_{b1}*v_{fbullet}+(m_{b2}+m_{bullet})*v_{f}}

P_{2}=1.202 kg*0.554\frac{m}{s}+(1.523kg+6.99x10^{-3}kg)*v_{f}

P_{1}=P_{2}

2.529=0.665+(1.5299)*v_{f}\\v_{f}=\frac{1.864}{1.5299}\\v_{f}=1.218 \frac{m}{s}

b).

kinetic energy

K=\frac{1}{2}*m*(v)^{2}

Kinetic energy after

Ka=\frac{1}{2}*1.202*(0.554)^{2}+\frac{1}{2}*1.523*(1.218)^{2}\\Ka=1.142 J

Kinetic energy before

Kb=\frac{1}{2}*mb*(vf)^{2}\\Kb=\frac{1}{2}*6.99x10^{-3}kg*(341)^{2}\\Kb=406.4J

Ratio =\frac{Ka}{Kb}

R=\frac{1.14}{406.4}\\R=2.8x10^{-3}

3 0
4 years ago
Need help asap if I fail this one I fail my grade so please tell me the right awser
bazaltina [42]
If you look up "distance vs time graph" you can pull up images on google and check them out. That's what I did and on the x-axis (horizontal line) it said "time." I'd go with time. 
6 0
3 years ago
The strength of an electromagnet CANNOT be increased by?
zmey [24]

Answer:

reversing the current

Explanation:

4 0
3 years ago
A boy shoves his stuffed toy zebra down a frictionless chute. It starts at a height of 1.69 m above the bottom of the chute with
Veseljchak [2.6K]

Answer:

x = 6.94 m

Explanation:

For this exercise we can find the speed at the bottom of the ramp using energy conservation

Starting point. Higher

            Em₀ = K + U = ½ m v₀² + m g h

Final point. Lower

            Em_{f} = K = ½ m v²

            Em₀ = Em_{f}

            ½ m v₀² + m g h = ½ m v²

            v² = v₀² + 2 g h

             

Let's calculate

             v = √(1.23² + 2 9.8 1.69)

             v = 5.89 m / s

In the horizontal part we can use the relationship between work and the variation of kinetic energy

            W = ΔK

            -fr x = 0- ½ m v²  

               

Newton's second law

              N- W = 0

     

The equation for the friction is

               fr = μ N

               fr = μ m g

We replace

             μ m g x = ½ m v²

             x = v² / 2μ g

Let's calculate

            x = 5.89² / (2 0.255 9.8)

            x = 6.94 m

6 0
4 years ago
Read 2 more answers
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