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krok68 [10]
3 years ago
15

A 4.0 kg object will have a weight of approximately 14.8 N on Mars. What is the gravitational field strength on Mars?

Physics
2 answers:
zhannawk [14.2K]3 years ago
8 0
Field strength =force/unit mass
= 14.8N/ 4kg = 3.7N/kg
Shalnov [3]3 years ago
8 0

Answer:

The gravitational field strength on Mars is 3.7 N/kg.

Explanation:

Given that,

Mass of the object, m = 4 kg

Weight of the object on Mars, F = W = 14.8 N

We need to find the gravitational field strength on Mars. It is given by gravitational force per unit mass of an object. Mathematically, it is given by

G=\dfrac{F}{m}

G=\dfrac{14.8\ N}{4\ kg}

G = 3.7 N/kg

So, the gravitational field strength on Mars is 3.7 N/kg. Hence, this is the required solution.

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The dial of a scale looks like this: 00.0kg. A physicist placed a spring on it. The dial read 00.6kg. He then placed a metal cha
saveliy_v [14]

Answer:

d. The scale's resolution is too low to read the change in mass

Explanation:

If we want to find the change in energy of the spring, we will have to use the Hooke's Law. Hooke's Law states that:

F = kx

since,

w = Fd

dw = Fdx

integrating and using value of F, we get:

ΔE = (0.5)kx²

where,

ΔE = Energy added to spring

k = spring constant

x = displacement

The spring constant is typically in range of 4900 to 29400 N/m.

So if we take the extreme case of 29400 N/m and lets say we assume an unusually, extreme case of 1 m compression, we get the value of energy added to be:

ΔE = (0.5)(29400 N/m)(1 m)²

ΔE = 1.47 x 10⁴ J

Now, if we convert this energy to mass from Einstein's equation, we get:

ΔE = Δmc²

Δm = ΔE/c²

Δm = (1.47 x 10⁴ J)/(3 x 10⁸ m/s)²

<u>Δm =  4.9 x 10⁻¹³ kg</u>

As, you can see from the answer that even for the most extreme cases the value of mass associated with the additional energy is of very low magnitude.

Since, the scale only gives the mass value upto 1 decimal place.

Thus, it can not determine such a small change. So, the correct option is:

<u>d. The scale's resolution is too low to read the change in mass</u>

8 0
3 years ago
ich of the following is not accurate when describing solids? A. The amount of pressure exerted by a solid is solely dependent on
Dafna11 [192]
The amount of solid does not affect how you are describing the solid so a is the answer
5 0
3 years ago
Our milky way galaxy is 100000 lyly in diameter. a spaceship crossing the galaxy measures the galaxy's diameter to be a mere 1.
Sidana [21]

The speed of the spaceship relative to the galaxy is 0.99999995c.

A light-year measures distance rather than time (as the name might imply). A light-year is a distance a light beam travels in one year on Earth, which is roughly 6 trillion miles (9.7 trillion kilometers). One light-year equals 5,878,625,370,000 miles. Light moves at a speed of 670,616,629 mph (1,079,252,849 km/h) in a vacuum.We multiply this speed by the number of hours in a year to calculate the distance of a light-year (8,766).

The Milky way galaxy is 100,000 light years in diameter.

The galaxy's diameter is a mere 1. 0 ly.

We know that ;

L = L_0 \sqrt{1-\frac{v^2}{c^2} }

L = 1 light year

L₀ = 100,000 light year

1 = 100,000 \sqrt{1-\frac{v^2}{c^2} }

1 = 100,000 \sqrt{1-\frac{v^2}{(3*10^8)^2} }

\frac{1}{100,000}  = \sqrt{1-\frac{v^2}{c^2} }

v = 0.999999995 c

Therefore, the speed of the spaceship relative to the galaxy is 0.99999995c.

Learn more about a light year here:

brainly.com/question/17423632

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5 0
1 year ago
List the 5 Greek variables used to measure rotational motion and their counterparts for translational motion
Leni [432]

Explanation:

pizs cake and they dont

4 0
3 years ago
A 2.55 kg block starts from rest on a rough inclined plane that makes an angle of 60 degree with the horizontal. the coefficient
lorasvet [3.4K]

Answer:

Change in mechanical energy = work done by friction

so it is equal to

W = -8.16 J

Explanation:

As we know that change in mechanical energy must be equal to the work done by non conservative forces only

So here when block moves down the inclined plane then the work done by friction force is given as

W = F.d

here we have

F = \mu F_n

here we know that

F_n = mg cos\theta

so we have

F_n = 2.55(9.81)(cos60)

F_n = 12.5 N

Now the friction force on the block is given as

F_f = \mu F_n

F_f = 0.25 \times 12.5

F_f = 3.13 N

now work done by the friction is given as

W = -(3.13)(2.61)

W = -8.16 J

7 0
3 years ago
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