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olya-2409 [2.1K]
2 years ago
10

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

Physics
1 answer:
Pie2 years ago
3 0

Answer:

Gravitational field strength =weight/mass

Explanation:

14.8N/4.0kg

3.7N/kg

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Which of the following changes would make a heat engine waste more energy as heat
Komok [63]
A decrease in it's operating temperature would make a heat engine less efficient. This is because in order to operate, a heat engine needs to be hot and maintain that temperature.
4 0
3 years ago
A system gains 1500 J of heat, while the internal energy of the system increases by 4500 J and the volume decreases by . Assume
Assoli18 [71]

Answer:

Hence the pressure is 3\times 10^5 Pa

Explanation:

Given data

Q=1500 J   system gains heat

ΔV=- 0.010 m^3     there is a decrease in volume

ΔU= 4500 J        internal energy decrease

We know work done is

W= Q- ΔU

=1500-4500= -3000 J

The change in the volume at constant pressure is

ΔV= W/P

there fore P = W/ΔV= -3000/-0.01= 3×10^5

Hence the pressure is 3\times 10^5 Pa

3 0
3 years ago
A solid yellow stripe is to be painted in the middle of a certain highway. If 1 gallon of paint covers an area of p square feet
marin [14]

Answer:

option A is correct

Explanation:

Given:

The length to be painted = m miles

The width to be painted = t inches

Area painted in 1 gallon = p square feet

Converting the every given dimension in feet, we have

length to be painted = m × 5280 feet

width to be painted = t/12 feet

area to be painted = (m × 5280 feet) × t/12 feet

now, applying the unitary method, we have

p square feet is painted ⇒ 1 gallon

1 square feet is painted ⇒ 1/p gallon

(m × 5280 feet) × t/12 feet  square feet is painted ⇒ [(m × 5280 feet) × t/12 feet ] × 1/p gallon

thus, we get the gallons of paint required as 5820 mt/12p

hence option A is correct

3 0
3 years ago
A space vehicle of mass m has a speed v. At some instant, it separates into two pieces, each of mass 0.5m. One of the pieces is
Damm [24]

Answer:

W = ½ m v²

Explanation:

In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation

We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved

initial instant. before separation

        p₀ = m v

final attempt. after separation

       p_{f} = m /2  0 + m /2 v_{f}

       p₀ = p_{f}

       m v = m /2 v_{f}

       v_{f}= 2 v

this is the speed of the second part of the ship

now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body

     

initial energy

         K₀ = ½ m v²

final energy

        K_{f} = ½ m/2  0 + ½ m/2 v_{f}²

        K_{f} = ¼ m (2v)²

        K_{f} = m v²

         

the expression for work is

         W = ΔK = K_{f} - K₀

         W = m v² - ½ m v²

         W = ½ m v²

5 0
3 years ago
The mass of an electron is _______. What is it?
SOVA2 [1]
If a mass of a neutron is 1 the electron mass is 0.00054386734 and it's charge is negative. Hope this helps! ;)
5 0
3 years ago
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