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True [87]
2 years ago
9

A 2.0-kg object is lifted vertically through 3.00 m by a 150-N force. How much work is done on the object by gravity during this

process?
Physics
2 answers:
noname [10]2 years ago
6 0

Answer:

-58.8 J

Explanation:

The work done by a force is given by:

W=Fdcos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and the displacement.

In this problem, we are asked to find the work done by gravity, so we must calculate the magnitude of the force of gravity first, which is equal to the weight of the object:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

The displacement of the object is d = 3.00 m, while \theta=180^{\circ}, because the displacement is upward, while the force of gravity is downward; therefore, the work done by gravity is

W=Fdcos \theta=(19.6 N)(3.00 m)(cos 180^{\circ})=-58.8 J

And the work done is negative, because it is done against the motion of the object.


Nana76 [90]2 years ago
3 0

Answer:

-58.8 J

Explanation:

Work done on the object by gravity = Force due to gravity x Displacement

Force due to gravity = weight of the object

                                  = m g

                                  = (2.0 kg) (9.8 m/s²)

                                  = 19.6 N

The force due to gravity acts in downwards direction opposite the direction of the displacement. Thus, the angle between the force and the displacement is 180°.

Work done on the object by gravity

= 19.6 N x 3.00 m x cos 180° = -58.8 J

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Mice21 [21]

Answer:

The magnitude of the acceleration of the car is 35.53 m/s²

Explanation:

Given;

acceleration of the truck, a_t = 12.7 m/s²

mass of the truck, m_t = 2490 kg

mass of the car, m_c = 890 kg

let the acceleration of the car at the moment they collided = a_c

Apply Newton's third law of motion;

Magnitude of force exerted by the truck = Magnitude of force exerted by the car.

The force exerted by the car occurs in the opposite direction.

F_c = -F_t\\\\m_ca_c = -m_t a_t\\\\a_c =- \frac{m_ta_t}{m_c} \\\\a_c = -\frac{2490 \times 12.7}{890} \\\\a_c = - 35.53 \ m/s^2

Therefore, the magnitude of the acceleration of the car is 35.53 m/s²

3 0
2 years ago
A test car carrying a crash test dummy accelerates from 0 to 30 m/s and then crashes into a brick wall. Describe the direction o
earnstyle [38]

Answer:

the direction of acceleration of the vehicle is the same direction of its velocity of car

s acceleration has the opposite direction to the car speed.

Explanation:

The initial acceleration of the car can be calculated with

          v = v₀ + a t

          a = (v-v₀) t

       

indicate that the initial velocity is zero (v₀ = 0 m / s)

         a = v / t

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the direction of acceleration of the vehicle is the same direction of its acceleration movement.

When the car collides with the wall, it exerts a force in the opposite direction that stops the vehicle, therefore this acceleration has the opposite direction to the car speed. But your module must be much larger since the distance traveled to stop is small

5 0
2 years ago
Betty weighs 420 n and she is sitting on a playground swing seat that hangs 0.40 m above the ground. tom pulls the swing back an
Svet_ta [14]
Weight = mass * gravity
420 = mass * 9.8
mass of Betty = 42.857 kg

Difference in height = 1 - 0.45 = 0.55 meters

Total energy = Kinetic energy + potential energy

At the highest point, the kinetic energy is zero while the potential energy is maximum, therefore, we can get the total energy as follows:
Total energy = 0 + mgh
Total energy = 42.857*9.8*0.55 = 231 Joules

At the lowest point, the potential energy is zero while the kinetic energy is maximum. Therefore:
Total energy = 0.5 * m * (v)^2 + 0
231 = 0.5 * (42.857) * (velocity)^2
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2 years ago
A student uses an electronic force sensor to study how much force the student’s finger can apply to a specific location. The stu
melisa1 [442]

Answer:

B. Trial 2

Explanation:

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Because the trial 2 student finger applied to largest force.

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Explanation:

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Nuclear reactions can change an atom of one element into an atom of another element.

Atoms of a given element can have different numbers of neutrons.

Atoms contain smaller particles: protons, neutrons, and electrons.

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