Ask question

Ask question

All categories

4 years ago

13

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket

. The bolt hits the ground 6.10s later. What was the rocket's acceleration?

1 answer:

Olenka [21]4 years ago

7 0

Answer:

1.3m/s²

Explanation:

Given values in the first part: acceleration a, time t₁:

1) velocity v₀ $= at_1$

2) height h₀ $=\frac{1}{2}at_1^2$

Given values in the second part: acceleration -g, time t₂:

3) height h $= -\frac{1}{2}gt_2^2+v_0t_2+h_0$

Combining equations 1,2,3 and setting h to zero:

$0=-\frac{1}{2}gt_2^2+(at_1)t_2+\frac{1}{2}at_1^2\\ 0=a(t_1t_2+\frac{1}{2}t_1^2)-\frac{1}{2}gt_2^2$

Solve for a with t₁ = 4s and t₂=2.1s:

$a=\frac{1}{2}gt_2^2(\frac{1}{t_1t_2+\frac{1}{2}t_1^2})$

You might be interested in

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

konstantin123 [22]

Answer:

$I=2.71\times 10^{-5}\ A$

Explanation:

A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

Let given is,

The diameter of a parallel plate capacitor is 6 cm or 0.06 m

Separation between plates, d = 0.046 mm

The potential difference across the capacitor is increasing at 500,000 V/s

We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :

$C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}$ , r is radius

Let I is the displacement current. It is given by :

$I=C\dfrac{dV}{dt}$

Here, $\dfrac{dV}{dt}$ is rate of increasing potential difference

So

$I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A$

So, the value of displacement current is $2.71\times 10^{-5}\ A$ .

4 0

3 years ago

At the equator, the radius of the Earth is approximately 6370 km. A plane flies at a very low altitude at a constant speed of v

Cerrena [4.2K]

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From there we will define the distance as the circumference of the earth (approximate as a sphere). With the speed given in the statement we will simply clear the equations below and find the time.

$R= 6370*10^3 m$

$v = 239m/s$

$a = 16.5m/s^2$

The circumference of the earth would be

$\phi = 2\pi R$

Velocity is defined as,

$v = \frac{x}{t}$

$t = \frac{x}{v}$

Here $x = \phi$ , then

$t = \frac{\phi}{v} = \frac{2\pi (6370*10^3)}{239}$

$t = 167463.97s$

Therefore will take 167463.97 s or 1 day 22 hours 31 minutes and 3.97seconds

4 0

3 years ago

2. A car with mass 2,500 kg is travelling at 15 m/s. What is the car's kinetic energy?

BabaBlast [244]

Answer:

c, 280,000J

Explanation:

You should use the equation KE (Kinetic Energy) = $\frac{1}{2} mv^2$

m = 2500

v = 15

KE = $\frac{1}{2} * 2500 * (15)^2\\$

KE = 281250 Joules.

This is consistent with the answer B, as all of the parameters given to you are rounded to 2 significant figures so naturally, the answer should also be rounded to 2 sig figs.

KE = 280,000 J

7 0

3 years ago

During normal beating, the heart creates a maximum 4.00-mV potential across 0.300 m of a person’s chest, creating a 1.00-Hz elec

erik [133]

Answer:

(a). The maximum electric field strength is 0.0133 V/m.

(b). The maximum magnetic field strength in the electromagnetic wave is $4.433\times10^{-11}\ T$

(c). The wavelength of the electromagnetic wave is $3\times10^{8}\ m$

Explanation:

Given that,

Maximum potential = 4.00 mV

Distance = 0.300\ m

Frequency = 1.00 Hz

(a). We need to calculate the maximum electric field strength

Using formula of the potential difference

$\Delta V=Ed$

$E=\dfrac{\Delta V}{d}$

$E=\dfrac{4.00\times10^{-3}}{0.300}$

$E=0.0133\ V/m$

(b). We need to calculate the maximum magnetic field strength in the electromagnetic wave

Using formula of the maximum magnetic field strength in the electromagnetic wave

$B=\dfrac{E}{c}$

Put the value into the formula

$B=\dfrac{0.0133}{3\times10^{8}}$

$B=4.433\times10^{-11}\ T$

(c). We need to calculate the wavelength of the electromagnetic wave

Using formula of wavelength

$c=f\lambda$

$\lambda=\dfrac{c}{f}$

Put the value into the formula

$\lambda=\dfrac{3\times10^{8}}{1.00}$

$\lambda=3\times10^{8}\ m$

Hence, (a). The maximum electric field strength is 0.0133 V/m.

(b). The maximum magnetic field strength in the electromagnetic wave is $4.433\times10^{-11}\ T$

(c). The wavelength of the electromagnetic wave is $3\times10^{8}\ m$

4 0

3 years ago

Poorly treated municipal wastewater is discharged to a stream. The river flow rate upstream of the discharge point is Qu/s = 8.7

Musya8 [376]

Answer

given,

Q u = 8.7 m³/s

Q d= 0.9 m³/s

BOD concentration = 50 mg/L

a) BOD concentration at the down stream

$C_{down}=\dfrac{0.9\times 50}{8.7+0.9}$

$C_{down}=\dfrac{0.9\times 50}{9.6}$

= 4.69 mg/L

b) discharge = 9.6 m³/s

cross sectional area = 10 m²

velocity steam = $\dfrac{8.7+0.9}{10}$

= 0.96 m/s

time taken to move 50 km down stream = $\dfrac{50 \times 1000}{0.96}$

= 52083.3 s

= $\dfrac{52083.3}{3600\times 24}$

= 0.6 days

now,

$C_t=C_0e^{-kt}$

$C_t=4.6875\ e^{-0.2\times 0.6}$

$C_t = 4.16 mg/L$

5 0

4 years ago