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faust18 [17]
3 years ago
12

A 55 newton force applied on an object moves the object 10 meters in the same direction as the force. What is the value of work

done on the object?
A.
0 joules
B.
5.5 joules
C.
-5.5 joules
D.
5.5 × 102 joules
E.
-5.5 × 102 joules
Physics
2 answers:
ryzh [129]3 years ago
5 0

Answer:

D

Explanation:

Work done refers to the product of the Force and the distance moved by the object in the same direction as the force. The formula is given by:

W = F × D

W is the Work Done in J

F is the Force in N

D is the parallel distance in m.

Hence by applying the Formula, we get:

W = 55(10)

= 550

Hence <u>answer is D.</u>

bearhunter [10]3 years ago
5 0

Answer:

D.

Explanation:

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The average kinetic energy of water molecules is greatest in
user100 [1]
<h2>Answer: in a gaseous state </h2>

The average kinetic energy of the water molecules is greater in its gaseous state (in the form of water steam).

This is because in the gaseous state the water molecules are well separated from each other and can move freely in all the available space they have; because there are no cohesion forces that bond them.

In contrast to the liquid and solid state, in which the molecules have less movement.

3 0
3 years ago
Two students have fitted their scooters with the same engine. Student A and his
sammy [17]

The force exerted by student A with his scooter is 306 N and that of student B is 204 N.

<h3>Force applied by each student</h3>

The force exerted by each student is calculated from Newton's second law of motion.

F = ma

where;

  • m is mass
  • a is acceleration

F(A) = 127.5 x 2.4

F(A) = 306 N

F(B) = 120 x 1.7

F(B) = 204 N

Thus, the force exerted by student A with his scooter is 306 N and that of student B is 204 N.

Learn more about force here: brainly.com/question/12970081

#SPJ1

3 0
2 years ago
The average intensity of sunlight at the top of the earth's atmosphere in 1390 w/m2. what is the maximum energy that a 34-m x 46
slavikrds [6]
Panel surface area =34m×46m=1,564m^2
total power =1564m^2×1390w/m^2
=2173960watts
now you must calculate total energy.
Energy = power×Time
However time must be in seconds so we multiply 2hrs×60min×60s=7200seconds
7200s×2173960w =15,652,512,000 joules of energy
7 0
3 years ago
Two asteroids collide and stick together. The first asteroid has a mass of 15\times 10^3\,\mathrm{kg}15×10 3 kg and is initially
statuscvo [17]

Answer:

Final speed is 900.06 m/s at 0.2215^{\circ}  

Solution:

As per the question:

Mass of the first asteroid, m = 15\times 10^{3}\kg

Mass of the second asteroid, m' = 20\times 10^{3}\kg

Initial velocity of the first asteroid, v = 770 m/s

Initial velocity of the second asteroid, v' = 1020 m/s

Angle between the two initial velocities, \theta = 20^{\circ}

Now,

Since, the velocities and hence momentum are vector quantities, then by the triangle law of vector addition of 2 vectors A and B, the resultant is given by:

\vec{R} = \sqrt{A^{2} + 2ABcos\theta + B^{2}}

Thus applying vector addition and momentum conservation, the final velocity is given by:

(m + m')v_{final} = \sqrt{(mv)^{2} + 2(mv)(m'v')cos20^{\circ} + (m'v')^{2}}                               (1)

Now,

(m +m')v_{final} = (35\times 10^{3})v_{final}

(mv)^{2} = (15\times 10^{3}\times 770)^{2} = 1.334\times 10^{14}

(m'v')^{2} = (20\times 10^{3}\times 1020)^{2} = 4.16\times 10^{14}

2(mv)(m'v')cos20^{\circ} = 2(15\times 10^{3}\times 770)(20\times 10^{3}\times 1020)cos20^{\circ} = 4.43\times 10^{14}

Now, substituting the suitable values in eqn (1), we get:

v_{final} = 900.06\ m/s

Now, the direction for the two vectors is given by:

\theta = sin^{- 1} \frac{m'v'sin20^{\circ}}{(m + m')v_{final}}

\theta = sin^{- 1} \frac{20\times 10^{3}\times 1020sin20^{\circ}}{(35\times 10^{3})\times 900.06} = 0.2215^{\circ}

5 0
3 years ago
Three astronauts outside a spaceship and that they decide to play catch. All the astronauts weigh the same on Earth and are equa
guajiro [1.7K]

Answer:

Well concluding there is no gravity their motions would be slow and lightweighted. Let's say they were playing on Earth it would approximately take around 5 to 6 minutes even less, so in space it will approximately take around 10 to 12 minutes may be more but this is just my opinion after using my calculator! Hope this helped!

5 0
3 years ago
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