All categories

3 years ago

please go to my profile and answer the work and energy questions, i only have 20 minutes left to finish it

Physics

1 answer:

Alexeev081 [22]3 years ago

7 0

Answer:

Okay I'll do it right now

Explanation:

You might be interested in

A vehicle that goes from 5m/s to 45m/s in 8s. what is its acceleration?

GaryK [48]

Answer: 5m/s^2

Explanation:

V= 45m/s

U = 5m/s

t = 8s

a =?

V = u + at

45 = 5 + 8a

8a = 45 — 5

8a = 40

a = 40 / 8

a = 5m/s^2

3 0

3 years ago

Convert the speed of light 3.0x10^8 m/s to km/day

Aleksandr [31]

Answer: $2.592 \times 10^{8}km/day$

$1 m = 0.001 km\\ 1 s= 1.157\times10^{-5} days\\ 1 m/s = \frac {0.001}{1.157\times10^{-5}} km/day = 86.4 km/day \\ 3.0\times 10^{8} m/s = 3.0\times 10^{8} m/s \times \frac {86.4 km/day}{1m/s} =2.592 \times 10^{8}km/day$

7 0

3 years ago

Read 2 more answers

n electromagnetic wave in vacuum has an electric field amplitude of 611 V/m. Calculate the amplitude of the corresponding magnet

enot [183]

Answer:

The corresponding magnetic field is

Explanation:

From the question we are told that

The electric field amplitude is $E_o = 611\ V/m$

Generally the magnetic field amplitude is mathematically represented as

$B_o = \frac{E_o }{c }$

Where c is the speed of light with a constant value

$c = 3.0 *0^{8} \ m/s$

$B_o = \frac{611 }{3.0*10^{8}}$

$B_o = 2.0 4 *10^{-6} \ Vm^{-2} s$

Since 1 T is equivalent to $V m^{-2} \cdot s$

$B_o = 2.0 4 *10^{-6} \ T$

6 0

3 years ago

Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha

mixer [17]

The density of seawater at a depth where the pressure is 500 atm is $1124kg/m^3$

Explanation:

The relationship between bulk modulus and pressure is the following:

$B=\rho_0 \frac{\Delta p}{\Delta \rho}$

where

B is the bulk modulus

$\rho_0$ is the density at surface

$\Delta p$ is the variation of pressure

$\Delta \rho$ is the variation of density

In this problem, we have:

$B=2.3\cdot 10^9 N/m^2$ is the bulk modulus

$\rho_0 =1100 kg/m^3$

$\Delta p = p-p_0 = 500 atm - 1 atm = 499 atm = 5.05\cdot 10^7 Pa$ is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)

Therefore, we can find the density of the water where the pressure is 500 atm as follows:

$\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3$

Learn more about pressure in a fluid:

brainly.com/question/9805263

#LearnwithBrainly

7 0

3 years ago

Mrs. Howel, the principal, can expel a student for misconduct. In terms of social pressure, Mrs. Howel is capable of eliciting

dybincka [34]

The answer would be D)Apathy

8 0

3 years ago