When light hits a RED WAGON some of the energy is absorbed and ______

Neutron stars, such as the one at the center of the Crab Nebula, have about the same mass as our sun but have a much smaller dia

Crank

Answer:

Wn = 9.14 x 10¹⁷ N

Explanation:

First we need to find our mass. For this purpose we use the following formula:

W = mg

m = W/g

where,

W = Weight = 675 N

g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²

m = Mass = ?

Therefore,

m = (675 N)/(9.8 m/s²)

m = 68.88 kg

Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:

gn = (G)(Mn)/(Rn)²

where,

gn = acceleration due to gravity on surface of neutron star = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg

Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m

Therefore,

gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)

gn = 13.27 x 10¹⁵ m/s²

Now, my weight on neutron star will be:

Wn = m(gn)

Wn = (68.88)(13.27 x 10¹⁵ m/s²)

Wn = 9.14 x 10¹⁷ N

3 0

2 years ago

A painter stands a horizontal platform which has a mass of 20kg and is 5m long, the platform is suspended by two vertical ropes

Lerok [7]

Answer:

R = 715.4 N

L = 166.6 N

Explanation:

ASSUME the painter is standing right of center

Let L be the left rope tension

Let R be the right rope tension

Sum moments about the left end to zero. Assume CCW moment is positive

R[5] - 20(9.8)[5/2] - 70(9.8)[5/2 + 2] = 0

R = 715.4 N

Sum moments about the right end to zero

20(9.8)[5/2] + 70(9.8)[5/2 - 2] - L[5] = 0

L = 166.6 N

We can verify by summing vertical forces

116.6 + 715.4 - (70 + 20)(9.8) ?=? 0

0 = 0 checks

If the assumption about which side of center the paint stood is incorrect, the only difference would be the values of L and R would be swapped.

5 0

2 years ago

A 1 800-kg pile driver is used to drive a steel I-beam into the ground. The pile driver falls 4.60 m before coming into contact

N76 [4]

Answer:

F = 614913.88 N

Explanation:

We are given;

Mass of pile driver; m = 1800 kg

Height of fall of pole driver; h = 4.6 m

Depth driven into beam; d = 13.6 cm = 0.136 m

Now, from energy equations and applying to this question, we can write that;

Workdone = Change in potential energy

Formula for workdone is; W = F × d

While the average potential energy here is; W = mg(h + d)

Thus;

Fd = mg(h + d)

Where F is the average force exerted by the beam on the pile driver while in bringing it to rest.

Making F the subject, we have;

F = mg(h + d)/d

F = 1800 × 9.81 × (4.6 + 0.136)/0.136

F = 614913.88 N

7 0

2 years ago

............ llloooo llopp

Elenna [48]

Answer:

lllooooooooooo llopp

3 0

3 years ago

A sound from a source has an intensity of 270 dB when it is 1 m from the source.

Rufina [12.5K]

Remember that sound intensity decreases in inverse proportion to the distance squared. So, to solve this we are going to use the inverse square formula: $\frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2$
where
$I_{2}$ is the intensity at distance 2
$I_{1}$ is the intensity at distance 1
$d_{2}$ is distance 2
$d_{1}$ is distance 1

We can infer for our problem that $I_{1}=270$ , $d_{1}=1$ , and $d_{2}=3$ . Lets replace those values in our formula to find $I_{2}$ :
$\frac{I_{2} }{I_{1}}= (\frac{d{2} }{d_{1}})^2$
$\frac{I_{2} }{270} =( \frac{1}{3} )^2$
$\frac{I_{2} }{270} = \frac{1}{9}$
$I_{2}= \frac{270}{9}$
$I_{2}=30$ dB

We can conclude that the intensity of the sound when is 3 m from the source is 30 dB.

8 0

2 years ago

When light hits a RED WAGON some of the energy is absorbed and _______.