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horrorfan [7]
3 years ago
9

What force is required to give an object with mass 2000 kg an acceleration of 3.5 m/s2

Physics
2 answers:
ch4aika [34]3 years ago
8 0
According to Newton's Second Law:
F = m*a = 2000 kg*3.5 m/s^2 = 7000 N
Alex787 [66]3 years ago
4 0

<u>Answer:</u> The force exerted on the object is 7000 N

<u>Explanation:</u>

Force is defined as the push or pull on an object with some mass that causes change in its velocity.

It is also defined as the mass multiplied by the acceleration of the object.

Mathematically,

F=m\times a

where,

F = force exerted on the object

m = mass of the object = 2000 kg

a = acceleration of the object = 3.5m/s^2

Putting values in above equation, we get:

F=2000kg\times 3.5m/s^2\\\\F=7000N

Hence, the force exerted on the object is 7000 N

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A researcher studying the nutritional value of a new candy places a 4.60 g sample of the candy inside a bomb calorimeter and com
blagie [28]

Answer:

4500.5 nutritional calories per gram

Explanation:

Heat lost by the new candy = heat gained by the bomb calorimeter.

Heat gained by the bomb calorimeter = c×ΔT

where c = heat capacity of the calorimeter = 32.20 KJ/K = 32200 J/K

ΔT = change in temperature = 2.69°C = 2.69 K.

Heat gained by the bomb calorimeter = 32200 × 2.69 = 86618 J

Heat lost by the new candy = heat gained by the bomb calorimeter = 86618 J = 20702.2 calories

4.60 g of the new candy lost this amount of calories by undergoing combustion,

The amount of calories per g = 20702.2 calories/4.6 g = 4500.5 calories per gram

8 0
3 years ago
Net force needed to accelerate a 1000-kg car at 0.5g
just olya [345]
It would be 2000N ( newtons )
6 0
3 years ago
1. A sprinter races in the 100 meter dash. It takes him 10 second to reach the finish line
poizon [28]

Answer:

v = 10 m/s

Explanation:

Given that,

Distance covered by a sprinter, d = 100 m

Time taken by him to reach the finish line, t = 10 s

We need to find his average velocity. We know that velocity is equal to the distance covered divided by time taken. So,

v = d/t

v=\dfrac{100\ m}{10\ s}\\\\v=10\ m/s

Hence, his average velocity is 10 m/s.

6 0
2 years ago
A 1.8-kg object is attached to a spring and placed on frictionless, horizontal surface. A force of 40 N stretches a spring 20 cm
Sergio [31]

Answer:

a) k = 200 N/m

b) E = 4 J

c) Δx = 6.3 cm

Explanation:

a)

  • In order to find force constant of the spring, k, we can use the the Hooke's Law, which reads as follows:

       F = - k * \Delta x (1)

  • where F = 40 N and Δx =- 0.2 m (since the force opposes to the displacement from the equilibrium position, we say that it's a restoring force).
  • Solving for k:

       k =- \frac{F}{\Delta x} =-\frac{40 N}{-0.2m} = 200 N/m (2)

b)

  • Assuming no friction present, total mechanical energy mus keep constant.
  • When the spring is stretched, all the energy is elastic potential, and can be expressed as follows:

        U = \frac{1}{2}* k* (\Delta x)^{2} (3)

  • Replacing k and Δx by their values, we get:

       U = \frac{1}{2}* k* (\Delta x)^{2} = \frac{1}{2}* 200 N/m* (0.2m)^{2} = 4 J (4)

c)

  • When the object is oscillating, at any time, its energy will be part elastic potential, and part kinetic energy.
  • We know that due to the conservation of energy, this sum will be equal to the total energy that we found in b).
  • So, we can write the following expression:

        \frac{1}{2}* k* \Delta x_{1} ^{2} + \frac{1}{2} * m* v^{2}  = \frac{1}{2}*k*\Delta x^{2}   (5)

  • Replacing the right side of (5) with (4), k, m, and v by the givens, and simplifying, we can solve for Δx₁, as follows:

        \frac{1}{2}* 200N/m* \Delta x_{1} ^{2} + \frac{1}{2} * 1.8kg* (-2.0m/s)^{2}  = 4J   (6)

⇒      \frac{1}{2}* 200N/m* \Delta x_{1} ^{2}   = 4J  - 3.6 J = 0.4 J (7)

⇒     \Delta x_{1}   = \sqrt{\frac{0.8J}{200N/m} } = 6.3 cm (8)

6 0
2 years ago
A mass of .1539 kg moves down a 5 meter ramp in 2 seconds. What
enot [183]

Answer:

Velocity=2.5m/s

KE=4809.375J

Explanation:

Velocity=5m/2s=2.5m/s

KE=½×1539kg×(2.5m/s)²

KE=4809.375J

4 0
3 years ago
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