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kolbaska11 [484]
2 years ago
12

A baseball is thrown horizontally at a rate of 40m/s toward a home plate 18.4 m away. How far below the launch height is the bal

l when it reaches home plate
Physics
1 answer:
sammy [17]2 years ago
8 0

Answer:

<h3>1.03684m</h3>

Explanation:

Using the formula for calculating range expressed as;

R = U√2H/g where

R is the distance moves in horizontal direction = 18.4m

H is the height

U is the velocity of the baseball = 40m/s

g is the acceleration due to gravity = 9.8m/s²

Substitute the given parameters into the formula and calculate H as shown;

18.4 = 40√2H/9.8

18.4/40 = √2H/9.8

0.46 = √2H/9.8

square both sides;

(0.46)² = (√2H/9.8)²

0.2116 = 2H/9.8

2H = 9.8*0.2116

2H = 2.07368

H = 2.07368/2

H = 1.03684m

Hence the ball is 1.03684m below the launch height when it reached home plate.

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Assume you have an ice cube and also a small rock that is the same size and shape as the ice cube. Predict what would happen if
Step2247 [10]

Answer: The ice cube would float on top of the water and the rock would sink to the bottom.

Explanation: The ice cube has a smaller density than the rock which allows the ice cube to float but makes the rock sink to the bottom of the glass of water.

7 0
3 years ago
A total of 750 j of work was done when a force of 125 n was exerted on a box to move it. how far was the box moved?
Slav-nsk [51]
Work = Force times Distance
W = Fd

Given W = 750J, F = 125N;

750 = 125d

Solving for d:
d = 750/125
d = 6

The box moved a distance of 6 meters.
3 0
3 years ago
In an experiment, a variable, position-dependent force FC) is exerted on a block of mass 1.0 kg that is moving on a horizontal s
marshall27 [118]

Answer:

C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of Fr.

Explanation:

Yo want to prove the following equation:

W_N=\Delta K\\\\

That is, the net force exerted on an object is equal to the change in the kinetic energy of the object.

The previous equation is also equal to:

F(x)x-F_f=\frac{1}{2}m(v_f^2-v_o^2)    (1)

m: mass of the block

vf: final velocity

v_o: initial velocity

Ff: friction force

F(x): Force

x: distance

You know the values of vf, m and x.

In order to prove the equation (1) it is necessary that you have C The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F.  Thus you can calculate experimentally both sides of the equation.

8 0
3 years ago
A bag weighing 20 N is pushed horizontally a distance of 35 m across a
topjm [15]

Answer:

350J

Explanation:

Given parameters:

Weight of bag  = 20N

Distance moved horizontally  = 35m

Force applied  = 10N

Unknown:

Work done on the bag  = ?

Solution:

Work done is the force applied to move a body through given distance.

  Work done  = Force applied x distance

So;

 Work done  = 10 x 35  = 350J

6 0
2 years ago
A blue line with 5 orange tick marks then one red tick mark then 4 orange tick marks. The number zero is above the red tick mark
Digiron [165]

Answer:

16 cm

Explanation:

Given that,

The object begins from 0 and moves 3cm towards left side followed by 7 cm towards the right and then, 6 cm towards the left side.

Let the x-axis to be the +ve and on the right side and -ve on the left

Thus, displacement would be:

= 0 -3 + 7 -6

= -2 cm

This implies that the object displaces 2cm towards the left.

While the total distance covered by the object equal to,

= 0cm + 3cm + 7cm + 6cm

= 16 cm

Thus, <u>16 cm</u> is the total distance.

3 0
3 years ago
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