I don't understand this question at all, can I please get some help?

Suppose that a sled is accelerating at a rate of 2 m/s^2. if the net force is tripled and the mass is doubled, then what is the

Gwar [14]

So new acceleration is 3 m/s^2

6 0

3 years ago

A 200-turn flat coil of wire 30.0 cm in diameter acts as an antenna for FM radio at a frequency of 100 MHz. The magnetic field o

dalvyx [7]

Answer:

a) 8.4×10^-12W

b) 5.65 ×10^ -3V

c) 1.01 ×10^ -12 F

Explanation: Given:

Radius r = d/2 = 30/2=15cm= 0.15m

Maximum feild strength = Bo= 10^-12 T

Permiability of vacuum, Uo = 4pi×10^-7

I = CBo/2Uo = (3 ×10^8×(10^-12)^2)/ (4 ×3.142×10^-7)

I = 1.19×10^-10W/m^2

The intensity I = > P/A = p = IA = I(4pi^2)

P = 1.19×10^-10 × 3.142 × (0.15^2)

P = 8.4 × 10^-12W

b) Given: f = 100Mhz= 100×10^6hz

Number of turns in the coil, N = 200

Time of cycle, T = 1/f = 1/ 100×10^6

T = 10^-8 seconds

Change in t= 1/4T = 1/4×10^-8 = 2.5 ×10^-9sec

E = (N×change in B×A)/ change in t

E = (NBopir^2)/change in t

E = (200 ×10^-12× 3.142 ×(0.15^2))/ (2.5×10^-9)

E = 5.65 ×10^-3V

c) Self inductance L = 2.5UH=2.5×10^-6H

Resonant frequency, fo= 1/(2pisqrt(LC))

C= 1/(4×3.142×( 2.5×10^-6)× (100×10^6)^2

C = 1.01×10^-12F

3 0

4 years ago

Read 2 more answers

What units is time measured in

Artist 52 [7]

Answer:

The base unit for time is the second (the other SI units are: metre for length, kilogram for mass, ampere for electric current, kelvin for temperature, candela for luminous intensity, and mole for the amount of substance). The second can be abbreviated as s or sec.

Explanation:

6 0

3 years ago

Name two examples of deposition.

Rashid [163]

The reverse of deposition is sublimation and hence sometimes deposition is called desublimation. One example of deposition is the process by which, in sub-freezing air, water vapor changes directly to ice without first becoming a liquid.

8 0

4 years ago

Read 2 more answers

A car of 1000 kg with good tires on a dry road can decelerate (slow down) at a steady rate of about 5.0 m/s2 when braking. If a

vredina [299]

(a) 4.0 s

The acceleration of the car is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time interval

For this car, we have

v = 0 (the final speed is zero since the car comes to a stop)

u = 20 m/s is the initial velocity

$a=-5.0 m/s^2$ is the deceleration of the car

Solving the equation for t, we find the time needed to stop the car:

$t=\frac{v-u}{a}=\frac{0-(20 m/s)}{-5.0 m/s^2}=4 s$

(b) 40 m

The stopping distance of the car can be calculated by using the equation

$v^2 - u^2 = 2ad$

where

v = 0 is the final velocity

u = 20 m/s is the initial velocity

a = -5.0 m/s^2 is the acceleration of the car

d is the stopping distance

Solving the equation for d, we find

$d=\frac{v^2-u^2}{2a}=\frac{0^2-(20 m/s)^2}{2(-5.0 m/s^2)}=40 m$

(c) $-5.0 m/s^2$

The deceleration is given by the problem, and its value is $-5.0 m/s^2$ .

(d) 5000 N

The net force applied on the car is given by

$F=ma$

where

m is the mass of the car

a is the magnitude of the acceleration

For this car, we have

m = 1000 kg is the mass

$a=5.0 m/s^2$ is the magnitude of the acceleration

Solving the formula, we find

$F=(1000 kg)(5.0 m/s^2)=5000 N$

(e) $2.0\cdot 10^5 J$

The work done by the force applied by the car is

$W=Fd$

where

F is the force applied

d is the total distance covered

Here we have

F = 5000 N

d = 40 m (stopping distance)

So, the work done is

$W=(5000 N)(40 m)=2.0\cdot 10^5 J$

(f) The kinetic energy is converted into thermal energy

Explanation:

when the breaks are applied, the wheels stop rotating. The car slows down, as a result of the frictional forces between the brakes and the tires and between the tires and the road. Due to the presence of these frictional forces, the kinetic energy is converted into thermal energy/heat, until the kinetic energy of the car becomes zero (this occurs when the car comes to a stop, when v = 0).

8 0

4 years ago