The mass of the earth is 6x10^24 kg and its radius is 6400 km.What is the mass of a man weighing 977 N in a spring balance.

What causes some materials to have a magnetic fields?

Viktor [21]

Well it determines on how much magnetism a material has.

6 0

3 years ago

A spring stretches 0.018 m when a 2.8-kg object is suspendedfrom its end. How much mass should be attached to this spring sothat

agasfer [191]

Answer:

m = 4.29 kg

Explanation:

Given that,

Mass of the object, m = 2.8 kg

Stretching in the spring, x = 0.018 m

Frequency of vibration, f = 3 Hz

Let m is the mass of the object that is attached to the spring. When it is attached the gravitational force is balanced by the force on spring. It is given by :

$mg=kx$

$k=\dfrac{mg}{x}$

$k=\dfrac{2.8\times 9.8}{0.018}$

k = 1524.44 N/m

Since, $\omega=\sqrt{\dfrac{k}{m}}$

$m=\dfrac{k}{4\pi^2 f^2}$

$m=\dfrac{1524.44}{4\pi^2 \times 3^2}$

m = 4.29 kg

So, the mass that is attached to this spring is 4.29 kg. Hence, this is the required solution.

4 0

2 years ago

A point charge of -0.70 μC is fixed to one corner of a square. An identical charge is fixed to the diagonally opposite corner. A

MakcuM [25]

Answer:

The magnitude and algebraic sign of q is $14\sqrt{2}\ \mu C$

Explanation:

Given that,

Point charge = -0.70 μC[/tex]

We need to calculate the force for all charges

The electric force at first corner

$F_{1}=\dfrac{-k0.70\times10^{-6}q}{r^2}$

The electric force at opposite corner

$F_{3}=\dfrac{-k0.70\times10^{-6}q}{r^2}$

The net force is

$F=\sqrt{F_{1}^2+F_{2}^2}$

Put the value into the formula

$F=\sqrt{(\dfrac{-k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}$

The electric force at second corner

$F_{2}=\dfrac{-kq^2}{2r^2}$

The net force acting on either of the charges is zero.

So, $F=F'$

$\sqrt{(\dfrac{k0.70\times10^{-6}q}{r^2})^2+(\dfrac{-k0.70\times10^{-6}q}{r^2})^2}=\dfrac{kq^2}{2r^2}$

$\sqrt{2}\times\dfrac{0.70\times10^{-6}kq}{r^2}=\dfrac{kq^2}{2r^2}$

$q=14\sqrt{2}\ \mu C$

Hence, The magnitude and algebraic sign of q is $14\sqrt{2}\ \mu C$

8 0

2 years ago

According to the Guinness Book of World Records (1990 edition, p. 169), the highest rotary speed ever attained was 2010 m/s (450

Valentin [98]

When a body performs a uniform circular motion, the direction of the velocity vector changes at every moment. This variation is experienced by the linear vector, due to a force called centripetal, directed towards the center of the circle that gives rise to centripetal acceleration, the mathematical expression is given as,

$a = \frac{v^2}{r}$

Where,

v = Tangential Velocity

r = Radius

The linear velocity was 2010m/s in a radius of 0.159m, then the centripetal acceleration is

$a = \frac{2010^2}{0.159}$

$a = 2.54*10^7m/s^2$

Therefore the centripetal acceleration of the end of the rod is $2.54*10^7m/s^2$

7 0

2 years ago

A man claims that he can hold onto a 15.0-kg child in a head-on collision as long as he has his seat belt on. Consider this man

suter [353]

Answer:

F= 5195.625 N

Explanation:

To obtain the force needed to hold the child, we need to know the aceleration in which the car is breaking.

Aceleration is equal to velocity divided by the time of breaking

In international system, velocity [m/s] is

v= (62 mi/h)*(1609 m/mi)*(1 h/3600 s)

v= 27.71 m/s

Now, we part the velocity by the time that is 0.08 seconds

a= v/t= (27.71 m/s)/(0.08 s)

a= 346.375 m/ $s^{2}$

The force in agreement with the Newton's second law is

F=m*a = 15 Kg*346.375 m/ $s^{2}$

F= 5195.625 N

(Note: 1 N = 1 Kg*m/ $s^{2}$ )

3 0

2 years ago

The mass of the earth is 6x10^24 kg and its radius is 6400 km.What is the mass of a man weighing 977 N in a spring balance.​

The mass of the earth is 6x10^24 kg and its radius is 6400 km.What is the mass of a man weighing 977 N in a spring balance.