What is the value of work done on an object when a

10-newton force moves it 30 meters and the angle between
the force and the displacement is 25°?

10 * 30 cos 25

Work is component of force in direction of motion times distance moved.

By the way I have no idea what ICP is and I just happened to stumble on your question. Identify your subject next time please.

Icp is integrated chemistry physics.....

also answer be wrong

To calculate the work done on an object, we can use the formula:

Work = Force × Displacement × cos(θ)

Where:
- Force is the magnitude of the applied force (in newtons),
- Displacement is the distance the object is moved in the direction of the force (in meters), and
- θ (theta) is the angle between the force vector and the displacement vector (in degrees).

In this particular case, we are given:
- Force = 10 newtons,
- Displacement = 30 meters, and
- θ = 25°.

To find the value of work done, we can substitute these values into the formula:

Work = 10 newtons × 30 meters × cos(25°)

Now, let's calculate the value of cos(25°). To do this, we can use a calculator or a mathematical software tool. The cosine of 25° is approximately 0.9063.

Substituting this value back into the formula:

Work = 10 newtons × 30 meters × 0.9063

Now we can calculate the multiplication:

Work ≈ 271.89 joules

Therefore, the value of work done on the object is approximately 271.89 joules.