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3 years ago

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A car slows from 22 m/s to 3.0 m/s at a constant rate of 2.1 m/s2. How many seconds are required before the car is traveling at

a forward velocity of 3.0 m/s?

1 answer:

Mumz [18]3 years ago

4 0

Answer: 9.05 s

<h2>Explanation: v = u +at</h2>

v = 3 m/s

u = 22 m/s

Deceleration = 2.1 m/s^2

∴ 3 = 22 - 2.1 ( t )

∴ t = 9 .05 s

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A block of height 120mm is placed on top of another block with a height of 1.50 m. What is the height of the two blocks together

pav-90 [236]

We will first convert all units to meters and then solve the problem.
We are given that:
1000 mm = 1 m
120 mm = ?? meters
using cross multiplication:
120 mm = (120*1) / 1000 = 0.12 m

Now, when the two objects are placed over each other, their total height is the result of summation of both heights, therefore:
total height = 0.12 + 1.5 = 1.62 m

Based on the above calculations, the correct choice is:
<span>b) 1.62 m </span>

6 0

4 years ago

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00

Nutka1998 [239]

Answer:

$6.99535\times 10^{-6}\ V/m$

Explanation:

P = Power Output = 1000 W

r = Radius = 35000000 m

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

c = Speed of light = $3\times 10^8\ m/s$

Intensity of Electric radiation is given by

$I=\dfrac{P}{A}\\\Rightarrow I=\dfrac{P}{4\pi r^2}\\\Rightarrow I=\dfrac{1000}{4\pi\times 35000000^2}\ W/m^2$

Intensity of Electric radiation is given by

$I=\dfrac{1}{2}c\epsilon_0E_0\\\Rightarrow E_0=\sqrt{\dfrac{2I}{c\epsilon_0}}\\\Rightarrow E_0=\sqrt{\dfrac{2\times \dfrac{1000}{4\pi\times 35000000^2}}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E_0=6.99535\times 10^{-6}\ V/m$

The amplitude of the electric field vector is $6.99535\times 10^{-6}\ V/m$

6 0

4 years ago

A bar-magnet with magnetic moment 2.5 Am^2 is placed in a homogeneous magnetic field (of 0.1 T that is directed along the z-axis

Angelina_Jolie [31]

Answer:

1)

Force on bar magnet = 0

Torque on bar magnet = 0

2)

Force on bar magnet = 0

Torque on bar magnet = 0.177 Nm

3)

Force on bar magnet = 0

Torque on bar magnet = 0.25 Nm

Explanation:

Part 1)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

$\tau = MBsin\theta$

when bar magnet is inclined along z axis along magnetic field

then we will have

$\tau = MBsin0 = 0$

Part 2)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

$\tau = MBsin\theta$

when bar magnet is pointing 45 degree with z axis then we will have

$\tau = MBsin45$

$\tau = (2.5)(0.1)sin45$

$\tau = 0.177 Nm$

Part 3)

net force on bar magnet in uniform magnetic field is always zero

Torque on bar magnet is given as

$\tau = MBsin\theta$

when bar magnet is pointing 90 degree with z axis then we will have

$\tau = MBsin90$

$\tau = (2.5)(0.1)sin90$

$\tau = 0.25 Nm$

8 0

4 years ago

Stars give off a tremendous amount of thermal and ________ energy.(1 point) chemical chemical electrical electrical nuclear nucl

Olegator [25]

I believe the answer is Nuclear energy

8 0

2 years ago

Two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the

Y_Kistochka [10]

Answer:

two sinusoidal waves, which are identical except for a phase shift, travel along in the same direction. The wave equation of the resultant wave is yR (x, t) = 0.70 m sin⎛ ⎝3.00 m−1 x − 6.28 s−1 t + π/16 rad⎞ ⎠ . What are the angular frequency, wave number, amplitude, and phase shift of the individual waves?

ω = 6.28 s − 1 ,

k = 3.00 m− 1 ,

φ = π rad,

A R = 2 A cos (φ 2 ) ,

A = 0.37 m

Explanation:

y1 ( x , t ) = A sin( k x − ω t +φ ) ,

y 2 ( x , t ) = A sin ( k x − ω t ) .

from the principle of superposition which states that when two or more waves combine, there resultant wave is the algebriac sum of the individual waves

y1 ( x , t ) = A sin( k x − ω t +φ ) , is generaL form of thw wave eqaution

A=amplitude

k=angular wave number

ω=angular frequency

φ =phase constant

k=2π/lambda

ω=2π/T

yR (x, t) = 0.70 m sin{3.00 m−1 x − 6.28 s−1 t + π/16 rad}....................*

two waves superposed to give the above, assuming they are moving in the +x direction

y1 ( x , t ) = A sin( k x − ω t +φ ) , .....................1

y 2 ( x , t ) = A sin ( k x − ω t ) ...........................2

adding the two equation will give

A sin( k x − ω t +φ )+A sin ( k x − ω t ) .................3

A( sin( k x − ω t +φ )+ sin ( k x − ω t ) ),......................4

similar to the following trigonometry identity

sina+sinb=2cos(a-b)/2sin(a+b)/2

let a= ( k x − ω t

b=k x − ω t +φ )

y(x,t)=2Acos(φ/2)sin(k x − ω t +φ/2)

k=3m^-1

lambda=2π/k=2.09m

ω=6.28= T=2π/6.28

T=1s

φ/2=π/16

φ=π/8rad

amplitude

2Acos(φ/2)=0.70 m

A=0.7/2cos(π/8)

A=0.37 m

6 0

4 years ago