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3 years ago

If you know that the rock fell 4.9 m in 1 s , how far did it drop in the first 0.5 s after you dropped it?

Physics

2 answers:

iren [92.7K]3 years ago

8 0

Answer

1.225 m

Explanation

The formula for getting the distance cover in this case is;

s = ut + 1/2 gt²

Where u⇒intial velocity

t⇒time taken

g ⇒ acceleration due to gravity

But in this case u = 0, t = 1 and s =4.9.

So, the first thin is to calculate the value of g.

S = 1/2 gt²

4.9 = 1/2 g × 1²

g = 4.9 × 2

= 9.8 m/s².

Now calculate the distance after 0.5 s.

S = 1/2 × 9.8 × 0.5²

= 1.225 m

klio [65]3 years ago

7 0

When an object is free-falling, no other force is acting upon it but the gravitational force. Because of this, the equations of motion are simplified. We can determine first the initial velocity:

v = √2gy = √2(9.81)(4.9) = 9.805 m/s

Then, we use this to the equation below:
y = vt + 1/2*at²
y = (9.805)(0.5) + 1/2(9.81)(0.5)²
y = 6.13 m

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If a car is traveling at a speed of 45 miles per hour, how far can it travel in 40 minutes?

dusya [7]

Answer:

30 miles

Explanation:

Step 1:

Divide -> 45/60= .75 miles/minute

Step 2:

Multiply -> .75 x 40= 30

3 0

3 years ago

Read 2 more answers

When a voltage difference is applied to a piece of metal wire, a 5.0 mA current flows through it. If this metal wire is now repl

zheka24 [161]

Answer:

I = 21.13 mA ≈ 21 mA

Explanation:

I₁ = 5 mA

L₁ = L₂ = L

V₁ = V₂ = V

ρ₁ = 1.68*10⁻⁸ Ohm-m

ρ₂ = 1.59*10⁻⁸ Ohm-m

D₁ = D

D₂ = 2D

S₁ = 0.25*π*D²

S₂ = 0.25*π*(2*D)² = π*D²

If we apply the equation

R = ρ*L / S

where (using Ohm's Law):

R = V / I

we have

V / I = ρ*L / S

If V and L are the same

V / L = ρ*I / S

then

(V / L)₁ = (V / L)₂ ⇒ ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂

S₁ = 0.25*π*D² and

S₂ = 0.25*π*(2*D)² = π*D²

we have

ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)

⇒ I₂ = 4*ρ₁*I₁ / ρ₂

⇒ I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m

⇒ I₂ = 21.13 mA

5 0

3 years ago

The volume flow rate in an artery supplying the brain is 4.00 10-6 m3/s. (a) If the radius of the artery is 4.50 10-3 m, determi

mr_godi [17]

Answer:

(a) 0.063 m/s

(b) 1.01 m/s

Explanation:

rate of volume flow, V = 4 x 10^-6 m^3/s

(a) radius, r = 4.5 x 10^-3 m

Let the speed of blood is v.

So, V = A x v

where A be the area of crossection of artery

4 x 10^-6 = 3.14 x 4.5 x 10^-3 x 4.5 x 10^-3 x v

v = 0.063 m/s

Thus, the speed of flow of blood is 0.063 m/s .

(b) Now r' = r / 4 = 4.5 /4 x 10^-3 m = 1.125 x 10^-3 m

Let the speed is v'.

So, V = A' x v'

4 x 10^-6 = 3.14 x 1.125 x 10^-3 x 1.125 x 10^-3 x v'

v' = 1.01 m/s

Thus, the speed of flow of blood is 1.01 m/s .

8 0

3 years ago

An airplane propeller is 2.68 m in length (from tip to tip) and has a mass of 107 kg. When the airplane's engine is first starte

telo118 [61]

Answer:

a)30.14 rad/s2

b)43.5 rad/s

c)60633 J

d)42 kW

e)84 kW

Explanation:

If we treat the propeller is a slender rod, then its moments of inertia is

$I =\frac{mL^2}{12} = \frac{107*2.68^2}{12} = 64.04 kgm^2$

a. The angular acceleration is Torque divided by moments of inertia:

$\alpha = \frac{T}{I} = \frac{1930}{64.04} = 30.14 rad/s^2$

b. 5 revolution would be equals to $10\pi$ rad, or 31.4 rad. Since the engine just got started

$\omega^2 = 2\alpha\theta = 2*30.14*31.4 = 1893.5$

$\omega = \sqrt{1893.5} = 43.5 rad/s$

c. Work done during the first 5 revolution would be torque times angular displacement:

$W = T*\theta = 1930 * 31.4 = 60633 J$

d. The time it takes to spin the first 5 revolutions is

$t = \frac{\omega}{\alpha} = \frac{43.5}{30.14} = 1.44 s$

The average power output is work per unit time

$P = \frac{W}{t} = \frac{60633}{1.44} = 41991 W$ or 42 kW

e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:

$P_i = T*\omega = 1930*43.5=83983 W$ or 84 kW

7 0

3 years ago

A -1.12 μC charge is placed at the center of a conducting spherical shell, and a total charge of +8.65 μC is placed on the shell

Lisa [10]

Answer: 7.53 μC

Explanation: In order to explain this problem we have to use the gaussian law so we have:

∫E.dS=Qinside/εo we consider a gaussian surface inside the conducting spherical shell so E=0

Q inside= 0 = q+ Qinner surface=0

Q inner surface= 1.12μC so in the outer surface the charge is (8.65-1.12)μC=7.53μC

7 0

3 years ago