How do you know when a penalty has been called?

In a $100$ meter track event, Alice runs at a constant speed and crosses the finish line $5$ seconds before Beatrice does. If it

Liono4ka [1.6K]

Answer:

10s

Explanation:

If it took Beatrice 25 seconds to complete the race

Distance = 100 meter

Beatrice speed = 100/25

= 4m/s

If Alice runs at a constant speed and crosses the finish line $5$ seconds, she must have completed the race in 20s (25 -5).

Her speed where constant

= 100/20

= 5 m/s

It would take Alice

= 50/5

= 10s

It would take Alice 10s to run $50$ meters.

5 0

3 years ago

Can anyone help me? (physics)

Masja [62]

Answer:

The initial velocity of the golf is 15.7 m/s.

The direction of the golf is 57°.

Explanation:

The following data were obtained from the question:

Time of flight (T) = 2.7 secs

Range (R) = 23 m

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =.?

Direction (θ) =.?

T = 2U Sine θ /g

2.7 = 2 × U × Sine θ /9.8

Cross multiply

2.7 × 9.8 = 2 × U × Sine θ

26.46 = 2 × U × Sine θ

Divide both side by 2 × Sine θ

U = 26.46 /2 Sine θ

U = 13.23 / Sine θ ... (1)

R = U² Sine 2θ /g

23 = U² Sine 2θ / 9.8

U = 13.23 / Sine θ

23 = (13.23/ Sine θ)² Sine 2θ / 9.8

23 = (175.0329 / Sine² θ) × Sine 2θ / 9.8

23 = 17.8605/Sine² θ × Sine 2θ

Recall:

Sine 2θ = 2SineθCosθ

23 = 17.8605/ Sine² θ × 2SineθCosθ

23 = 17.8605/ Sine θ × 2Cosθ

23 = 35.721 Cos θ /Sine θ

Cross multiply

23 × Sine θ = 35.721 Cos θ

Divide both side by 23

Sine θ = 35.721 Cos θ /23

Sine θ = 1.5531 × Cos θ

Divide both side by Cos θ

Sine θ /Cos θ = 1.5531

Recall:

Sine θ /Cos θ = Tan θ

Sine θ /Cos θ = 1.5531

Tan θ = 1.5531

Take the inverse of Tan

θ = Tan¯¹ (1.5531)

θ = 57°

Therefore, the direction of the golf is 57°

Thus, the initial velocity can be obtained as follow:

U = 13.23 / Sine θ

θ = 57°

U = 13.23 / Sine 57

U = 13.23/0.8387

U = 15.7 m/s

Therefore, the initial velocity of the golf is 15.7 m/s

8 0

3 years ago

A rectangular tank is filled to a depth of 10m with freshwater and open to air at atmospheric pressure.

charle [14.2K]

Answer:

1. 39068.07 N

2. 19534.036 N

Explanation:

depth of water h = 10 m

atmospheric pressure Patm = 101325 Pa

density of water p = 1000 kg/m^3

acceleration due to gravity g = 9.81 m/s^2

pressure due to depth of water = pgh

P = 1000 x 9.81 x 10 = 98100 Pa

total pressure on the bottom of the tank is Patm + p = 101325 + 98100 = 199425 Pa

Left plug has diameter = 50 cm = 0.5 m

radius = 0.5/2 = 0.25 m

height = 1 cm = 0.01 m

height below tank surface = 10 - 0.01 = 9.99

pressure at this depth = 1000 x 9.81 x 9.99 = 98001.9 Pa

total pressure = Patm + P = 101325 + 98001.9 = 199326.9 Pa

surface area of plug = π $r^{2}$ = 3.142 x $0.25^{2}$ = 0.196 $m^{2}$

force required to lift left plug = pressure x area

F = 199326.9 x 0.196 = 39068.07 N

The right side is a hemisphere with the same diameter, therefore surface area is half of the left plug

A = 0.196/2 = 0.098 $m^{2}$

force F required to lift right plug = 199326.9 x 0.098 = 19534.036 N

6 0

3 years ago

Which of the following statements about floating objects is always true? a. The object’s density is greater than the density of

Yuki888 [10]

It should be C. If the object is denser than the fluid, it will sink. If it isn't, it will float

6 0

3 years ago

Read 2 more answers

What is the anomalous expansivity of water

lora16 [44]

Answer:

The anomalous expansion of water is an abnormal property of water whereby it expands instead of contracting when the temperature goes from 4o C to 0o C, and it becomes less dense. The density is maximum at 4 degree centigrade and decreases below that temperature as shown in graph.

Explanation:do you want me to explain it more??

6 0

2 years ago