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GenaCL600 [577]

3 years ago

14

How much time would it take for the sound of thunder to travel 2000 meters of sound travels at the speed of 330 meters per secon

d

1 answer:

SpyIntel [72]3 years ago

5 0

Time = (distance covered) / (speed)

Time = (2,000 m) / (330 m/s)

<em>Time = 6.06 seconds</em>

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When a rope is wrapped around a wheel (or pulley) and pulled, the lever arm is the ________ Blank 1 of the wheel and the angle b

m_a_m_a [10]

Answer:

the DISTANCE between the lever arm and the force is always 90º

Explanation:

In this exercise, you are asked to complete the missing words so that the phrase makes sense.

note that the torque is

τ = F x r

where bold indicates vectors

When the rope is pulled, the DISTANCE between the lever arm and the force is always 90º

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2 years ago

What causes the electric charges to flow from one end of the battery to the other? a balance in electric potential a balance in

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8 0

3 years ago

Read 2 more answers

Two cars A and B are 100m apart moving towards each other with

maxonik [38]

Let car A's starting position be the origin, so that its position at time <em>t</em> is

A: <em>x</em> = (40 m/s) <em>t</em>

and car B has position at time <em>t</em> of

B: <em>x</em> = 100 m - (60 m/s) <em>t</em>

<em />

They meet when their positions are equal:

(40 m/s) <em>t</em> = 100 m - (60 m/s) <em>t</em>

(100 m/s) <em>t</em> = 100 m

<em>t</em> = (100 m) / (100 m/s) = 1 s

so the cars meet 1 second after they start moving.

They are 100 m apart when the difference in their positions is equal to 100 m:

(40 m/s) <em>t</em> - (100 m - (60 m/s) <em>t</em>) = 100 m

(subtract car B's position from car A's position because we take car A's direction to be positive)

(100 m/s) <em>t</em> = 200 m

<em>t</em> = (200 m) / (100 m/s) = 2 s

so the cars are 100 m apart after 2 seconds.

3 0

2 years ago

A mass attached to a 50.0 cm long string starts from rest and is rotated 40 times in one minute before reaching a final angular

ch4aika [34]

To solve this problem it is only necessary to apply the kinematic equations of angular motion description, for this purpose we know by definition that,

$\theta = \frac{1}{2}\alpha t^2 +\omega_0 t + \theta_0$

Where,

$\theta =$ Angular Displacement

$\alpha =$ Angular Acceleration

$\omega_0 =$ Angular velocity

$\theta_0 =$ Initial angular displacement

For this case we have neither angular velocity nor initial angular displacement, then

$\theta = \frac{1}{2}\alpha t^2$

Re-arrange for $\alpha,$

$\alpha = \frac{2\theta}{t^2}$

Replacing our values,

$\alpha = \frac{2(40rev*\frac{2\pi rad}{1rev})}{60^2}$

$\alpha = 0.139rad/s$

Therefore the ANgular acceleration of the mass is $0.139rad/s^2$

4 0

2 years ago

A 200 kg weather rocket is loaded with 100 kg of fuel and fired straight up. It accelerates upward at 35 m/s2 for 35 s , then ru

r-ruslan [8.4K]

Answer:

T = 295.57 s

Explanation:

given,

mass of the rocket = 200 Kg

mass of the fuel = 100 Kg

acceleration = 35 m/s²

time, t = 35 s

time taken by the rocket to hit the ground, = ?

Final speed of the rocket when fuel is empty

using equation of motion

v = u + a t

v = 0 + 35 x 35

v = 1225 m/s

height of the rocket where fuel is empty

v² = u² + 2 a s

1225² = 0 + 2 x 35 x h₁

h₁ = 21437.5 m

After 35 s the rocket will be moving upward till the final velocity becomes zero.

Now, using equation of motion to find the height after 35 s

v² = u² + 2 g h₂

0² = 1225² + 2 x (-9.8) h₂

h₂ = 76562.5 m

total height = h₁ + h₂

= 76562.5 m + 21437.5 m = 98000 m

now, time taken by before the rocket hit the ground

using equation of motion

$s = u t +\dfrac{1}{2}at^2$

$-13500 = 1225 t -\dfrac{1}{2}\times 9.8 \times t^2$

negative sign is used because the distance travel by the rocket is downward.

4.9 t² - 1225 t - 13500 = 0

$t = \dfrac{-(-1225)\pm \sqrt{1225^2 - 4\times 4.9 \times (-13500)}}{2\times 4.9}$

t = 260.57 s

neglecting the negative sign

total time the rocket was in air

T = t₁ + t₂

T = 35 + 260.57

T = 295.57 s

Time for which rocket was in air is equal to 295.57 s.

6 0

3 years ago