HELP PLEASE!!!!!!!!!!!!!!!

An electronic line judge camera captures the impact of a 57.0-g tennis ball traveling at 32.2 m/s with the side line of a tennis

mote1985 [20]

Answer:

average acceleration is 1.365 × $10^{4}$ m/s²

Explanation:

given data

initila speed u = -32.2 m/s

final speed v = 21.6 m/s

time taken t = 0.00394 s

solution

we get here average acceleration that will be express as

v = u + at ..........................1

put here value and we get

21.6 = -32.2 + a × 0.00394

solve it we get

a = 1.365 × $10^{4}$ m/s²

so average acceleration is 1.365 × $10^{4}$ m/s²

3 0

3 years ago

State the law of universal gravitation, and use examples to explain how changes in mass and changes in distance affect gravitati

juin [17]

F(g)= Gm1m2/ r^2 If mass is increased, so will the force of gravity because it is in direct relationship with the gravitational force, but if distance is increased, the force of gravity will decrease because it is indirectly related ( since it is on the bottom of the equation)

6 0

2 years ago

Tom kicks a soccer ball on a flat, level field giving it an initial speed of 20 m/s at an angle of 35 degrees above the horizont

SVEN [57.7K]

Answer:

(a) 2.34 s

(b) 6.71 m

(c) 38.35 m

(d) 20 m/s

Explanation:

u = 20 m/s, theta = 35 degree

(a) The formula for the time of flight is given by

$T = \frac{2 u Sin\theta }{g}$

$T = \frac{2 \times 20 \times Sin35 }{9.8}$

T = 2.34 second

(b) The formula for the maximum height is given by

$H = \frac{u^{2} \times Sin^{2}\theta }{2g}$

$H = \frac{20^{2} \times Sin^{2}35 }{2 \times 9.8}$

H = 6.71 m

(c) The formula for the range is given by

$R = \frac{u^{2} \times Sin 2\theta }{g}$

$R = \frac{20^{2} \times Sin 2 \times 35}{9.8}$

R = 38.35 m

(d) It hits with the same speed at the initial speed.

8 0

3 years ago

NEED HELP!!!! 11 POINTS!!!!

Anna007 [38]

Answer:

11

Explanation:

5 0

2 years ago

Read 2 more answers

A jet transport has a weight of 2.25 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel

Rudik [331]

Answer:

Explanation:

Given that,

Weight of jet

W = 2.25 × 10^6 N

It is at rest on the run way.

Two rear wheels are 16m behind the front wheel

Center of gravity of plane 10.6m behind the front wheel

A. Normal force entered on the ground by front wheel.

Taking moment about the the about the real wheel.

Check attachment for better understanding

So,

Clock wise moment = anti-clockwise moment

W × 5.4 = N × 16

2.25 × 10^6 × 5.4 = 16•N

N = 2.25 × 10^6 × 5.4 / 16

N = 7.594 × 10^5 N

B. Normal force on each of the rear two wheels.

Using the second principle of equilibrium body.

Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces

ΣFy = 0

Nr + Nr + N — W = 0

2•Nr = W—N

2•Nr = 2.25 × 10^6 — 7.594 × 10^5

2•Nr = 1.491 × 10^6

Nr = 1.491 × 10^6 / 2

Nr = 7.453 × 10^5 N

6 0

3 years ago