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3 years ago

After traveling for 6.0 seconds, a runner reaches 10m/s. What is the runner's acceleration?

Physics

1 answer:

luda_lava [24]3 years ago

4 0

After traveling for 6.0 seconds, a runner reaches 10m/s. What is the runner's acceleration? Answer: 1.67 m/s2

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2 years ago

10. Which of the following is an acceleration? a. 12 m/s2 down b. 5 m/s up c. 8N West

nadya68 [22]

Answer:

a. 12 m/s² down

Explanation:

Acceleration has units of length per time squared. Acceleration is a vector, so it also has a direction.

3 0

3 years ago

Q1: A runner is jogging in a straight line at a steady vr= 6.8 km/hr. When the runner is L= 2.4 km from the finish line, a bird

serious [3.7K]

Answer:

Q1: 3.2km

Q2: 4.8K

Explanation:

Q1:

So db is the distance of bird, and dr is the distance of runner

db = 2vr and the distance of bird is going to be 2 times greater than the runner.

formulas: db = 2vr & db = 2dr

db = 2dr
L + (L - x) = 2x
2L - x = 2x
2L = 3x
x = $\frac{2}{3}$ L

Insert it in x = $\frac{2}{3}$ L

$\frac{2}{3}$ (2.4km) = 1.6km

Now we use formula db = 2dr

db = 2L - x
db = 2(2.4km) - 1.6km
db = 3.2km

Q2:

Formulas: Vr = L /Δt & Vb = db/Δt

Vr = L/ Δt ⇒ Δt = $\frac{L}{Vr}$
$\frac{2.4km}{6.8km/hr}$
$\frac{6}{17}hr$

(Km cancel each other)

Vb = db/Δt ⇒ db = VbΔt
13.6km/hr $(\frac{6}{17}hr )$
4.8km

(hr cancel each other)

Hope it helps you :)

6 0

3 years ago

Amy swims 500 m (0.5 Km) in 6 minutes (0.1 hour). What was her speed in Kilometers per hour? A. 8.3 Km/hr B. 5 Km/hr C. 50 Km/hr

kondor19780726 [428]

Speed = Distance ÷ Time so divide .5 km by .1h. .5 km÷.1h=5 km/h, so the answer is B. 5km/h.

8 0

3 years ago

Read 2 more answers

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton

pishuonlain [190]

Hello!

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.

Data:

Hooke represented mathematically his theory with the equation:

F = K * Δx

On what:

F (elastic force) = 2 N

K (elastic constant) = 4 N/cm

Δx (deformation or elongation of the elastic medium or distance from a spring) = ?

Solving:

$F = K * \Delta{x}$