What is the acceleration of an object with a mass of 367 kg when a net force of 183 N is applied to it

Answers

Answer 1

Answer:

[tex]0.5m/s^{2}[/tex]

Explanation:

F= m x a

a= [tex]\frac{F}{m}[/tex] =  [tex]\frac{183 N}{367 kg}[/tex]  = [tex]\frac{0.4986 N}{kg}[/tex] = [tex]\frac{0.4986 kg*m/s^{2} }{kg}[/tex] = [tex]0.5 m/s^{2}[/tex]

Hopefully, this is right!


Related Questions

calculate the velocity of an object moved around a circle with a radius of 1.65m and an acceleration of 3.5 m/s^2​

Answers

Answer:

2.4m/s

Explanation:

which force is a contact force

Answers

contact forces include push, pull and friction

A projectile is launched horizontally at a speed of 40 meters per second from a platform located a vertical distance h above the ground. The projectile strikes the ground after time at horizontal distance from the base of the platform. [Neglect friction.]

Calculate the vertical distance, h, if the projectile's total time of flight is 4.5 seconds. You must show work to earn full credit with the minimum 3 steps:
1. Equation used
2. Substitution with correct units
3. Final answer with correct units

Answers

Answer:1

Explanation:t=rad2h/g

The launch of projectiles allows to find the height of the body launched horizontally is:

The height is 99.2 m

Projectile launching is an application of kinematics where there is no acceleration on the x-axis and the y-axis is the gravity acceleration.

In the attachment we see a diagram of the movement, as the body is thrown horizontally, its initial vertical speed is zero.

           y = y₀ + [tex]v_o_y[/tex] t - ½ g t²

Where y, y₀i are the current and initial heights, respectively, [tex]v_{oy}[/tex] the vertical initial velocity, g the acceleration of gravity and t the time.

           0 =y₀ + 0 - ½ g t²

           y₀ = ½ g t²

let's calculate

           y₀ = ½ 9.8 4.5²

           y₀ = 99.2 m

          y₀ = h = 99.2 m

This is the initial height of the object when it is thrown.

In conclusion using the projectile launch we can find the height of the horizontally launched body is:

The height is 99.2 m

Learn more here:  brainly.com/question/10903823

The graph shows the speeds of two bicyclists on their way to school. what is the speed of bicycle B

Answers

Answer: 0.3

Explanation: Jusy took it and got a 100

Answer: 0.3

Explanation: This is right trust me I answered this and it was correct!!

ASAP it’s timed!!
An object has a mass of 5 kg. What force is needed to accelerate it at 6 m/s? (Formula: Fema)
0.89 N
1.2 N
11 N
30 N

Answers

Answer:

30N is the correct answer

30 N is the answer
Step by step: 30N

Man has never directly reached the inner layers of the Earth. But what phenomena allow to determine its chemical composition?
A)Surface wave propagation.
B)Plate displacement
C)Emanation of lava in volcanoes
D)Changes in the Earth's landscape
E)Waves formed in tidal waves

Answers

I think letter A.Surface wave propagation

What is also known as watered carbons

Answers

The name carbohydrate means "watered carbon" or carbon with attached water molecules. Many carbohydrates have empirical formuli which would imply about equal numbers of carbon and water molecules. For example, the glucose formula C6H12O6 suggests six carbon atoms and six water molecules.

Answer:

The name carbohydrate means "watered carbon" or carbon with attached water molecules. Many carbohydrates have empirical formuli which would imply about equal numbers of carbon and water molecules. For example, the glucose formula C6H12O6 suggests six carbon atoms and six water molecules.

The gradual temperature change from one season to the next is due to what? view attachment

Answers

Global warming jk part of the earth is either facing the sun or part of the sun or away from the sun

How does a jack spread out the work over a large distance?

Answers

Each complete rotation of a jack handle applies a small force over a large distance. A small force exerted over a large distance becomes a large force exerted over a short distance. Each rotation lifts the car only a very short distance.

if 56 j of work energy is required to lift a dumbbell to a height of 7m, what is the weight of the dumbbell?​

Answers

Answer:

8 N

Explanation:

work=force*distance

force=work/distance

=56/7

=8

An 85kg mass accelerates at 5.5m/s2 what force was applied

Answers

Answer:

467.5 N

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 85 × 5.5

We have the final answer as

467.5 N

Hope this helps you

In consideration of the above diagram, a projectile is launched at 45 with an initial horizontal velocity (V0x) of 3 m/s. Calculate the Vox which is the actual launch velocity at 45 in m/s.

Answers

CFISD too huh? :D

also tryna get the answer

A car turns from a road into a parking lot and into an available parking space. The car's initial velocity is 4 m/s [E 45° N]. The car's velocity just before the driver decreases speed is 4 m/s [E 10° N]. The turn takes 3s. What's the average acceleration of the car during the turn? The answer should have directions with an angle.

Answers

Answer:

Explanation:

From the given information:

The car's initial velocity = 4 m/s in the direction of east 45° due north

We can therefore express the vector of this component form as:

v₁ = (4 m/s) (cos(45º)i + sin(45º)j)

v₁ = (2.83 m/s)i + (2.83 m/s)j

Similarly, the car's final velocity = 4 m/s in the direction of the east side 10º north

v₂ = (4 m/s) (cos(10º) i + sin(10º) j)

v₂ = (3.94 m/s) i + (0.695 m/s) j

From the first equation of motion

v = u + at

Making acceleration "a" the subject of the formula, we have:

a = (v - u )/t

a = (v₂ - v₁)/t

a =  (0.370 m/s²) + (-0.711 m/s²)

The magnitude of the avg. acceleration is:

[tex]|| a||= \sqrt{(0.370 m/s^2)^2 + (-0.711 m/s^2)^2)[/tex]

[tex]|| a||= 0.8015 \ m/s^2[/tex]

And;

The direction can be determined by taking the tangent of the acceleration:

i.e.

[tex]tan(\theta) = \dfrac{-0.711 m/s^2}{ 0.370 m/s^2}[/tex]

[tex]tan(\theta) = -1.9216[/tex]

[tex]\theta = tan^{-1} ( -1.9216 )[/tex]

[tex]\mathbf{\theta = -62.51 ^0}[/tex]

Thus, the direction of the angle is approximately  S 62.51º E

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