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3 years ago

WILL GIVE BRAINLIEST TO THE CORRECT ANSWER

Physics

1 answer:

LenKa [72]3 years ago

4 0

Answer:

See the answers below.

Explanation:

We can solve both problems using Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

∑F =m*a

where:

F = force [N] (units of newtons)

m = mass = 1000 [kg]

a = acceleration = 3 [m/s²]

$F = 1000*3\\F=3000[N]$

And the weight of any body can be calculated by means of the mass product by gravitational acceleration.

$W=m*g\\W=1000*9.81\\W=9810 [N]$

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Two objects are dropped from rest from the same height. Object A falls through a distance <img src="https://tex.z-dn.net/?f=d_A"

Alenkinab [10]

Answer:

The answer to your question is given below

Explanation:

Since both object A and B were dropped from the same height and the air resistance is negligible, both object A and B will get to the ground at the same time.

From the question, we were told that object A falls through a distance to dA at time t and object B falls through a distance of dB at time 2t.

Remember, both objects must get to the ground at the same time..!

Let the time taken for both objects to get to the ground be t.

Time A = Time B = t

But B falls through time 2t

Therefore,

Time A = Time B = 2t

Height = 1/2gt^2

For A:

Time = 2t

dA = 1/2 x g x (2t)^2

dA = 1/2g x 4t^2

For B

Time = t

dB = 1/2 x g x t^2

Equating dA and dB

dA = dB

1/2g x 4t^2 = 1/2 x g x t^2

Cancel out 1/2, g and t^2

4 = 1

4dA = dB

Divide both side by 4

dA = 1/4 dB

8 0

3 years ago

A neutron star is an extremely dense, rapidly spinning object that results from the collapse of a massive star at the end ofits

Westkost [7]

Answer:

(a). The rotational inertia is $5.72\times10^{39}\ kg m^2$

(b). The magnitude of the magnetic torque is $3.20\times10^{35}\ N-m$

Explanation:

Given that,

Mass of neutron $M_{n}= 13M_{s}$

Density of neutron $\rho=4.8\times10^{17}\ kg/m^3$

(a). We need to calculate the rotational inertia

Using formula of rotational inertia for sphere

$I=\dfrac{2}{5}MR^2$ ...(I)

We know that,

$\rho=\dfrac{M}{V}$

Put the value of volume

$\rho=\dfrac{3M_{n}}{4\pi R^3}$

$R^2=(\dfrac{3M_{n}}{4\pi\rho})^{\frac{2}{3}}$

Put the value of R in equation (I)

$I=\dfrac{2}{5}\times M_{n}\times(\dfrac{3M_{n}}{4\pi\rho})^{\frac{2}{3}}$

Put the value into the formula

$I=\dfrac{2}{5}\times(13\times2\times10^{30})^{\frac{5}{3}}\times(\dfrac{3}{4\pi\times(4.8\times10^{17})})^{\frac{2}{3}}$

$I=5.72\times10^{39}\ kg m^2$

The rotational inertia is $5.72\times10^{39}\ kg m^2$ .

(b). We need to calculate the magnitude of the magnetic torque

Using formula of torque

$\tau=I\times \alpha$

Put the value into the formula

$\tau=5.72\times10^{39}\times5.6\times10^{-5}$

$\tau=3.20\times10^{35}\ N-m$

The magnitude of the magnetic torque is $3.20\times10^{35}\ N-m$

Hence, (a). The rotational inertia is $5.72\times10^{39}\ kg m^2$

(b). The magnitude of the magnetic torque is $3.20\times10^{35}\ N-m$

4 0

3 years ago

A satellite of mass ????=238 kg is in circular orbit around the Earth at an altitude above the earth's surface equal to the eart

lesya [120]

Answer:

Explanation:

a) the speed=

V=√2gr²/2r

=√gr/2

√10*637/2

V=31850m/s²

Magnitude of the acceleration

a=Gm/r²

10*238/(6370)²

2380/12740

=0.187

8 0

3 years ago

A child with a weight of 230 N swings on a playground swing attached to 1.90 m long chains. What is the gravitational potential

nlexa [21]

Answer:

437 J

Explanation:

Parameters given:

Weight of child, W = 230 N

Height of swing, h = 1.9 m

Gravitational Potential Energy is given as:

P. E. = m*g*h = W*h

m = mass

h = height above the ground

W = weight

P. E. = 230 * 1.9

P. E. = 437 J

7 0

3 years ago

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DESPERATE WILL GIVE BRAINLIST AND THANKS

Bas_tet [7]

Answer:

true

Explanation:

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3 years ago