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3 years ago

What is the time required for an object starting at rest to fall freely 500 meters near earth surface

2 answers:

7 0

Without air resistance . . .

Distance = 1/2 g T²

500 m = (1/2) (9.8 m/s²) (T²)

500 m = (4.9 m/s²) (T²)

Divide each side by (4.9 m/s²) :

T² = (500 m) / (4.9 m/s²)

= 102.04 s²

T = √ (102.04 s²)

= 10.1 seconds

Gnom [1K]3 years ago

3 0

Using kinematics, approximately 10.1 seconds

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3 years ago

At time t=0 a positively charged particle of mass m=3.57 g and charge q=9.12 µC is injected into the region of the uniform magne

larisa [96]

Answer:

10.78 s

Explanation:

The force on the charge is computed by using the equation:

$F^{\to}= qE^{\to} +q (v^{\to} + B^{\to}) \\ \\ F^{\to} = (9.12 \times 10^{-6}) *278 (-\hat k) +9.12 *10^{-6} *2.1 *0.18 (\hat i * \hat k) \\ \\ F^{\to} = -2.535 *10^{-3} \hat k -3.447*10^{-6} \hat j$

F = ma

∴

$a ^{\to}= \dfrac{F^{\to}}{m}$

$a ^{\to}= \dfrac{-1}{3.57\times 10^{-3}}(2.535*10^{-3}\hat k + 3.447*10^{-6} \hat j)$

$a ^{\to}=-0.710 \hat k -9.656*10^{-4} \hat j$

At time t(sec; the partiCle velocity becomes $v(t) = 3.78 v_o$

The velocity of the charge after the time t(sec) is expressed by using the formula:

$v^{\to}= v_{o \ \hat i} + a^{\to }t \\ \\ \implies (2.1)\hat i -0.710 t \hat k -9.656 \times 10^{-4} t \hat j = 3.78 v_o \\ \\ \implies (2.1)^2 +(0.710\ t)^2+ (9.656 *10^{-4}t )^2 = (3.78 *2.1^2 \\ \\ \implies 4.41 +0.5041 t^2 +9.324*10^{-7} t^2 = 63.012 \\ \\ \implies 4.41 +0.5041 t^2 = 63.012\\ \\ 0.5041t^2 = 63.012-4.41 \\ \\ t^2 = \dfrac{58.602}{0.5041} \\ \\ t^2 = 116.25 \\ \\ t = \sqrt{116.25} \\ \\ \mathbf{t = 10.78 \ s}$

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3 years ago

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2 years ago

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4 years ago

Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car

lapo4ka [179]

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

$A_a(t) = 5m/s^2$

To get the velocity, we integrate over time:

$V_a(t) = (5m/s^2)*t + V_0$

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

$V_a(t) = (5m/s^2)*t$

To get the position equation we integrate again over time:

$P_a(t) = 0.5*(5m/s^2)*t^2 + P_0$

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

$P_a(t) = 0.5*(5m/s^2)*t^2$

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

$V_b(t) =20m/s$

To get the position equation, we can integrate:

$P_b(t) = (20m/s)*t + P_0$

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

$P_b(t) = (20m/s)*t + 100m$

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

$P_a(t) = P_b(t)$

We can solve this for t.

$0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0$

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

$t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)} }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}$

We only care for the positive solution, which is:

$t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s$

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

$P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m$

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4) What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

$V_a(t) = (5m/s^2)*11.48s = 57.4 m/s$

7 0

3 years ago