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dybincka [34]
3 years ago
6

During a baseball game, a baseball is struck at ground level by a batter. The ball leaves the baseball bat with an initial veloc

ity v0 = 38 m/s at an angle θ = 41° above horizontal. Let the origin of the Cartesian coordinate system be the ball's position the instant it leaves the bat. Air resistance may be ignored throughout this problem.Calculate the horizontal distance xmax in meters the ball has traveled when it returns to ground level.
Physics
1 answer:
ale4655 [162]3 years ago
7 0

Answer:

145.8m

Explanation:

The toss distance is given by:

x=v_0^2*\frac{sin(2\phi)}{g} ,(g=9.8m/s^2)

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The greatest speed recorded by a baseball thrown by a pitcher was 162.3 km / h, obtained by Nolan Ryan in 1974. If the ball leav
Pepsi [2]

Answer:

0.96 m

Explanation:

First, convert km/h to m/s.

162.3 km/h × (1000 m/km) × (1 hr / 3600 s) = 45.08 m/s

Now find the time it takes to move 20 m horizontally.

Δx = v₀ t + ½ at²

20 m = (45.08 m/s) t + ½ (0 m/s²) t²

t = 0.4436 s

Finally, find how far the ball falls in that time.

Δy = v₀ t + ½ at²

Δy = (0 m/s) (0.4436 s) + ½ (-9.8 m/s²) (0.4436 s)²

Δy = -0.96 m

The ball will have fallen 0.96 meters.

3 0
3 years ago
If we start with 1.000 g of cobalt-60, 0.675 g will remain after 3.00 yr. this means that the of is _____
Tju [1.3M]
<span>Cobalt-60 is undergoing a radioactivity decay.

The formula of the decay is n=N(1/2)</span>∧(T/t).
<span>Where N </span>⇒ original mass of cobalt
<span>           n </span>⇒ remaining mass of cobalt after 3 years
          T ⇒ decaying period
           t ⇒ half-life of cobalt.

So,
0.675 = 1 × 0.5∧(3/t)
log 0.675 = log 0.5∧(3/t)
3/t = log 0.675 ÷log 0.5
 3/t= 0.567

t = 3÷0.567
  = 5.290626524

the half-life of Cobalt-60 is 5.29 years. 

<span>           
</span><span>
</span>
8 0
3 years ago
A block with mass m 2.00 kg is placed against a spring on a frictionless incline with angle 30 degrees (Fig. B-43). (The block i
guajiro [1.7K]

Answer:

Explanation:

a )

The stored elastic energy of compressed spring

= 1 / 2 k X²

= .5 x 19.6 x (.20)²

= .392 J

b ) The stored potential energy will be converted into gravitational potential energy of the block earth system when the block will ascend along the incline . So change in the gravitational potential energy will be same as stored elastic potential energy of the spring that is .392 J .

c ) Let h be the distance along the incline which the block ascends.

vertical height attained ( H ) =h sin30

= .5 h

elastic potential energy = gravitational energy

.392 = mg H

.392 = 2 x 9.8 x .5 h

h = .04 m

4 cm .

=

7 0
3 years ago
Why Study Science? Help me out
Mashutka [201]
All but the last one!
6 0
1 year ago
Read 2 more answers
He power output of a cyclist moving at a constant speed of 5.3 m/s on a level road is 120 w. (a) what is the force exerted on th
EleoNora [17]
P=F*V
F=P/V
F=120/5.3
F=22.6N

8 0
3 years ago
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