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katen-ka-za [31]
3 years ago
10

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}

<u>H₀ = 1.6 x 10⁻¹⁸ s⁻¹</u>

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: V<span>an'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.
</span><span>b = 0,89 m.
ΔT = 3°C = 3 K.
i = </span>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
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