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3 years ago

A plane is landing. It started at 400m/s and ended up at 50m/s after 30 seconds. What is it's acceleration?

Physics

1 answer:

quester [9]3 years ago

4 0

Give u = start velocity

v = end velocity

v = u + at

50 = 400 + a*30

30a = -350

a = -116.67 m/ $s^{2}$

**Why the accecleration is negative number**

Because displacement, velocity, and acceleration are VECTOR QUANTITIES.

Vector Quantity must have direction.

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What are precautions in a simple pendulum experiment

monitta

Explanation:

1. draught

2. Parallax error

3. angle if displacement

4. air resistance or any form of obstruction

6 0

3 years ago

What would Hubble's constant be if we found one galaxy moving away at 30,000 km/s at a distance of 600 Mpc?

lora16 [44]

Answer:

H₀ = 1.6 x 10⁻¹⁸ s⁻¹

Explanation:

The Hubble's Constant can be found by the following formula:

$v = H_o D\\\\H_o = \frac{v}{D}$

where,

H₀ = Hubble's Constant = ?

v = speed of galaxy = 30000 km/s = 3 x 10⁷ m/s

D = Distacance = 600 Mpc = (6 x 10⁸ pc)(3.086 x 10¹⁶ m/1 pc)

D = 18.52 x 10²⁴ m

Therefore,

$H_o = \frac{3\ x\ 10^7\ m/s}{18.52\ x\ 10^{24}\ m}$

H₀ = 1.6 x 10⁻¹⁸ s⁻¹

3 0

3 years ago

A 4-kg object is moving with a speed of 5 m/s at a height of 2 m. The kinetic

tatyana61 [14]

Hello!

$\large\boxed{KE = 50J}$

Use the formula for kinetic energy:

$KE = \frac{1}{2}mv^{2}$

Plug in the given mass and velocity:

$KE = \frac{1}{2} (4)5^{2}$

Simplify:

$KE = \frac{1}{2} (100)\\\\KE = 50 J$

7 0

3 years ago

A 0.89 m aqueous solution of an ionic compound with the formula mx has a freezing point of -3.0 ∘c . van't hoff factor?

BigorU [14]

Answer is: Van't Hoff factor (i) for this solution is 1,81.
Change in freezing point from pure solvent to solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
b - molality, moles of solute per kilogram of solvent.
b = 0,89 m.
ΔT = 3°C = 3 K.
i = 3°C ÷ (1,86 °C/m · 0,89 m).

i = 1,81.

5 0

3 years ago

A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis

bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

$B=\frac{\mu_o I_r}{2\pi r}$ (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

$I_r=JA_r$

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by $A_r=\pi(r^2-R_1^2)$

The current density is given by the total area and the total current:

$J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}$

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current = 4 A

Then, the current in the wire for a distance r is:

$I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}$ (2)

You replace the last result of equation (2) into the equation (1):

$B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})$

Finally. you replace the values of all parameters:

$B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T$

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0

3 years ago