Magnetic field 0.02 mfrom a wire is 0.1 T. What is the magnitude of the magnetic field of 0.01 m from the same wire?

A concrete block is hung from an ideal spring that has a force constant of 100 N/m . The spring stretches 0.129 m .A- What is th

madam [21]

Answer:

1.31498 kg

0.72050 s

Explanation:

m = Mass of block

g = Acceleration due to gravity = 9.81 m/s²

k = Spring constant = 100 N/M

x = Displacement = 0.129 m

The force balance is

$mg=kx\\\Rightarrow m=\dfrac{kx}{g}\\\Rightarrow m=\dfrac{100\times 0.129}{9.81}\\\Rightarrow m=1.31498\ kg$

The mass of the block is 1.31498 kg

Time period is given by

$T=2\pi\sqrt{\dfrac{m}{k}}\\\Rightarrow T=2\pi\sqrt{\dfrac{1.31498}{100}}\\\Rightarrow T=0.72050\ s$

The period of oscillations is 0.72050 s

The time period does not depend on the acceleration due to gravity. It varies with the mass and the spring constant.

Hence, the time period would be the same

8 0

3 years ago

Abishek is a runner. He runs the 100 m sprint

konstantin123 [22]

Answer:

1.67m/s

Explanation:

Total Distance to be travelled by a Runner=100m

Time Taken=10*6s

Speed=Distance/Time

=100/10*6=10/6=1.67m/s

3 0

3 years ago

Read 2 more answers

On a typical clear day, the atmospheric electric field points downward and has a magnitude of approximately 103 N/C. Compare the

Nina [5.8K]

Answer:

a) FE = 0.764FG

b) a = 2.30 m/s^2

Explanation:

a) To compare the gravitational and electric force over the particle you calculate the following ratio:

$\frac{F_E}{F_G}=\frac{qE}{mg}$ (1)

FE: electric force

FG: gravitational force

q: charge of the particle = 1.6*10^-19 C

g: gravitational acceleration = 9.8 m/s^2

E: electric field = 103N/C

m: mass of the particle = 2.2*10^-15 g = 2.2*10^-18 kg

You replace the values of all parameters in the equation (1):

$\frac{F_E}{F_G}=\frac{(1.6*10^{-19}C)(103N/C)}{(2.2*10^{-18}kg)(9.8m/s^2)}\\\\\frac{F_E}{F_G}=0.764$

Then, the gravitational force is 0.764 times the electric force on the particle

b)

The acceleration of the particle is obtained by using the second Newton law:

$F_E-F_G=ma\\\\a=\frac{qE-mg}{m}$

you replace the values of all variables:

$a=\frac{(1.6*10^{-19}C)(103N/C)-(2.2*10^{-18}kg)(9.8m/s^2)}{2.2*10^{-18}kg}\\\\a=-2.30\frac{m}{s^2}$

hence, the acceleration of the particle is 2.30m/s^2, the minus sign means that the particle moves downward.

7 0

3 years ago

30 km/h is _________m/s? A.)8.3 B.)5.6 C.)13.9 D.)11.1

Juli2301 [7.4K]

Answer:

Correct Option :- A

Explanation:

3 0

2 years ago

A cat’s crinkle ball toy of mass 15g is thrown straight up with an initial speed of 3m/s. Assume in this problem that air drag i

stiks02 [169]

Answer:

P.E = 0.068 J = 68 mJ

Explanation:

First we need to find the height attained by the ball toy. For this purpose, we will be using 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = -9.8 m/s² (negative sign due to upward motion)

h = height attained by the ball toy = ?

Vf = Final Velocity = 0 m/s (since it momentarily stops at the highest point)

Vi = Initial Velocity = 3 m/s

Therefore,

2(-9.8 m/s²)h = (0 m/s)² - (3 m/s)²

h = (9 m²/s²)/(19.6 m/s²)

h = 0.46 m

Now, the gravitational potential energy of ball at its peak is given by the following formula:

P.E = mgh

P.E = (0.015 kg)(9.8 m/s²)(0.46 m)

P.E = 0.068 J = 68 mJ

4 0

3 years ago