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Alik [6]
3 years ago
7

se the model for projectile motion, assuming there is no air resistance and g = 32 feet per second per second. A baseball, hit 3

feet above the ground, leaves the bat at an angle of 45° and is caught by an outfielder 3 feet above the ground and 300 feet from home plate. What is the initial speed of the ball, and how high does it rise?
Physics
1 answer:
guajiro [1.7K]3 years ago
8 0

Answer:

b

Explanation:

just took the test

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A rock falls from a tower that is 320 feet high. As it is​ falling, its height is given by the formula h equals 320 minus 16 t s
RSB [31]

Answer:

a. 4.5secs

Explanation:

From the question, the equation describing the height is given by

h(t)=320-16t^{2}\\

at the point when the rock hit the ground, the height,h will be zero.

Hence we can have

h(t)=320-16t^{2}\\\\at  h(t)=0\\0=320-16t^{2}\\hence \\320=16t^{2}\\t^{2}=\frac{320}{16}\\ t^{2}=20\\t=\sqrt{20}\\ t=4.47secs\\t=4.5sec\\

hence the rock will hit the ground in 4.5secs

4 0
4 years ago
A student sees a newspaper ad for an apartment that has 1330 square feet of floor space. How many square meters of area are ther
hichkok12 [17]
One square meter equals 10.7 feet. 

1330 ft^2 * \frac{1m}{10.7ft^2} = 123.561
8 0
4 years ago
A small block has constant acceleration as it slides down a frictionless incline. The block is released from rest at the top of
Alchen [17]

Answer:

The speed of the block when it is 5.00 m from the top of the incline is 3.04 m/s

Explanation:

given information:

s = 7.80 m

v = 3.8 m/s

if s = 5 m, v?

first we have to find the acceleration of the block using the following equation:

v² = v₀² + 2as, v₀²  = 0 thus

3.8² = 2 a 7.8

a = 0.93

so, if s = 5m the final speed is

v² = 2 (0.93) (5)

   = 9.26

v = √9.26

  = 3.04 m/s

5 0
4 years ago
6. A baseball player hits a 0.155-kilogram fastball traveling at -44.0 m/s into center field at a speed of 50.0
Zolol [24]

Answer:

3,237.78N

Explanation:

According to Newton's second law of motion

F = mass * acceleration

Since a = v-u/t

F = m(v-u)/t

Given

Mass m = 0.155kg

v = 50.0m/s

u = -44.0m/s

Time t = 0.00450secs

Substitute

F = 0.155(50-(-44))/0.00450

F = 0.155(50+44)/0.00450

F = 0.155(94)/0.00450

F = 14.57/0.00450

F = 3,237.78N

hence he hit the baseball with a force of 3,237.78N

5 0
3 years ago
A cylinder contains 3.0 L of oxygen at 310 K and 2.5 atm. The gas is heated, causing a piston in the cylinder to move outward. T
Alex_Xolod [135]

Answer:

The gas pressure is: 1.55 atm.

Explanation:

We need to use the equation that relate the variables given at the exercise (pressure, temperature and volume) from the ideal gas law formula, when the mass is constant we can reduce the expretion PV=nRT to \frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2} } solving to P2 we get:\frac{P_{1}V_{1}T_{2}}{T_{1}V_{2}}=P_{2} replace the values P_{2}=\frac{2.5*3*610}{9.5*310} =1.55(atm).

5 0
3 years ago
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